[PHP-DB] undefined function error
I have a web site that connects to a mysql database using this connect script - ? $db=mysql_connect(localhost,***,***); mysql_select_db(simradusa,$db); ? When I access the page it gives me this error - Fatal error: Call to undefined function: mysql_connect() in /usr/local/www/data-dist/connect on line 1 Using mysql-4.1.10a for FreeBSD-5.3 with php-4.3.9 Any idea why I'm getting this error? (the same connect string works fine on a differant web server running FreeBSD-5.2.1/php-4.3.6/mysql-3.23.49) Thanks, Chip -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
[PHP-DB] regex question
I am writing some regex on a php form to validate input and have come up with this, for example - if (empty($first) || !eregi(^[A-Za-z]+[- ]?[A-Za-z]+$, $first)) for checking a persons name. This allows a single name, or a hyphenated double name, or a non-hyphenated double name, and it works. My question is this - why is the third set followed by a '+' optional? I thought the + meant the preceding group is present 1 or more times. The ? means the preceding group is present 0 or more times. Why is it that when I put a ? in the place of the + after the last ] if (empty($first) || !eregi(^[A-Za-z]+[- ]?[A-Za-z]?$, $first)) the regex is broken? Regards, Chip -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
[PHP-DB] odd results when running this
Okay, so I have got the php bits working in this web page but it is
running 3 times every time the page loads, and I see no reason why (I have
opened this page in two differant browsers). The code for the entire page
is pasted below, it includes some calls to include files which are not
included in this message of course (and this file is itself included in
another file), the page can be seen at this address:
http://www.simradusa.com/index-test.php
If anyone can see any reason for the script to run 3 times please let me
know.
Thanks,
Chip
==
table width=100% border=0 cellpadding=5 cellspacing=0
align=center
summary=table that contains the body of the page
tr
td
div align=top
p class=welcomeWelcome to Simrad, Inc/p
p class=paratext
Simrad is one of the world's largest manufacturers of marine electronics
for the yachting,
fishery and commercial marine markets, offering sales and service
worldwide. A complete
range of products from Simrad is available for many different types of
marine activity,
including auto steering, navigation, hydro acoustic and communication
equipment. Ranging
from pleasure boating, through fishing and into commercial craft. Please
select the type
of activity that you are interested in using the links on the left.
/p
br /br /br /
p class=paratextRead a
href=SimradNews/Spring2004/SimradNews-p1.html target=_newspan
class=redSimrad/spanspan class=blueNews/span/a - Your
quarterly guide to today's marine electronics (opens in a new window)br
/
!-- Or download the .pdf version a
href=SimradNews/Spring2004/SimradNews.phphere/a (765Kbytes) --/p
br /
p class=c2img src=images/sim-net-icon.gif alt=ANNOUNCING THE NEW
SIMNET NETWORK border=0 title=SAWDUST helm - all Simrad products
//p
p
a href=#
onclick=window.open('simnet.html','simnet','toolbar=no,width=450,height=500,left=0,top=0,screenX=0,screenY=0,status=no,scrollbars=yes,resize=yes');return
false
onmouseover=window.status='ANNOUNCING THE NEW SIMNET NETWORK';return
true;
onmouseout=window.status='ANNOUNCING THE NEW SIMNET NETWORK'; return
true;
title=Read more about the SimNet class=special-linksemANNOUNCING
THE NEW SIMNET NETWORK/p
/div
/td
tdbr /
p class=titleWhat's New/p
?
$sql=select * from hotspots;
$result = mysql_query($sql);
while ( $row = mysql_fetch_array($result))
{
printf(pimg src=\hotspots/latest1.gif\ alt=\%s\
title=\%s\/pp class=\footnote\%s/p, $row[alt-text],
$row[title-text], $row[desc-text]);
printf(pimg src=\hotspots/latest2.gif\ alt=\%s\
title=\%s\/pp class=\footnote\%s/p, $row[alt-text],
$row[title-text], $row[desc-text]);
printf(pimg src=\hotspots/latest3.gif\ alt=\%s\
title=\%s\/pp class=\footnote\%s/p, $row[alt-text],
$row[title-text], $row[desc-text]);
}
?
/td
/tr
/table
table summary=company blurb align=center border=0 cellspacing=0
cellpadding=10 width=80%
tr
td colspan=2
div class=co-blurb
hr width=75% /
p class=paratexta href=contact/simradbackground.php
onmouseover=window.status='Some Background information about
Simrad';return true;
title=Some Background information about Simrad
class=special-linksSimrad/a has built its name on
products which do what they are designed to do, whatever the conditions.
As part of a
href=http://www.kongsberg.com/;
onmouseover=window.status='The Kongsberg Maritime Corporate Home
Page';return true;
title=Our corporate parent company home page target=_blank
class=special-linksKongsberg Maritime/a, the largest
manufacturer of marine electronics in the world, Simrad manufactures
products which are
developed and built on the same principles as the company's sophisticated
electronics
for the demanding commercial shipping and fisheries market./p
/div
/td
/tr
/table
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RE: [PHP-DB] odd results when running this
Thanks for the tips.
I find this confusing - I have 3 statements pulling 3 differant images, so
I don't understand why the script would run 3 times if there is not loop
telling it to do so. Shouldn't it just load each image and then stop?
Doesn't the while loop stop when it hits the last row?
--
Chip
Bastien Koert [EMAIL PROTECTED] wrote on 01/28/2005 08:37:58 AM:
it runs three times because there are three rows of data and you are
setting
the same thing for each set of data.
try this
?
$sql=select * from hotspots;
$result = mysql_query($sql);
$x = 1;
while ( $row = mysql_fetch_array($result))
{
printf(pimg src=\hotspots/latest$x.gif\ alt=\%s\
title=\%s\/pp class=\footnote\%s/p, $row[alt-text],
$row[title-text], $row[desc-text]);
$x++;
}
?
bastien
From: Chip Wiegand [EMAIL PROTECTED]
To: PHP DB php-db@lists.php.net
Subject: [PHP-DB] odd results when running this
Date: Fri, 28 Jan 2005 07:41:30 -0800
Okay, so I have got the php bits working in this web page but it is
running 3 times every time the page loads, and I see no reason why (I
have
opened this page in two differant browsers). The code for the entire
page
is pasted below, it includes some calls to include files which are not
included in this message of course (and this file is itself included in
another file), the page can be seen at this address:
http://www.simradusa.com/index-test.php
If anyone can see any reason for the script to run 3 times please let
me
know.
Thanks,
Chip
==
table width=100% border=0 cellpadding=5 cellspacing=0
align=center
summary=table that contains the body of the page
tr
td
div align=top
p class=welcomeWelcome to Simrad, Inc/p
p class=paratext
Simrad is one of the world's largest manufacturers of marine
electronics
for the yachting,
fishery and commercial marine markets, offering sales and service
worldwide. A complete
range of products from Simrad is available for many different types of
marine activity,
including auto steering, navigation, hydro acoustic and communication
equipment. Ranging
from pleasure boating, through fishing and into commercial craft.
Please
select the type
of activity that you are interested in using the links on the left.
/p
br /br /br /
p class=paratextRead a
href=SimradNews/Spring2004/SimradNews-p1.html target=_newspan
class=redSimrad/spanspan class=blueNews/span/a - Your
quarterly guide to today's marine electronics (opens in a new
window)br
/
!-- Or download the .pdf version a
href=SimradNews/Spring2004/SimradNews.phphere/a (765Kbytes)
--/p
br /
p class=c2img src=images/sim-net-icon.gif alt=ANNOUNCING THE
NEW
SIMNET NETWORK border=0 title=SAWDUST helm - all Simrad products
//p
p
a href=#
onclick=window.open('simnet.html','simnet','toolbar=no,width=450,
height=500,left=0,top=0,screenX=0,screenY=0,status=no,
scrollbars=yes,resize=yes');return
false
onmouseover=window.status='ANNOUNCING THE NEW SIMNET NETWORK';return
true;
onmouseout=window.status='ANNOUNCING THE NEW SIMNET NETWORK'; return
true;
title=Read more about the SimNet
class=special-linksemANNOUNCING
THE NEW SIMNET NETWORK/p
/div
/td
tdbr /
p class=titleWhat's New/p
?
$sql=select * from hotspots;
$result = mysql_query($sql);
while ( $row = mysql_fetch_array($result))
{
printf(pimg src=\hotspots/latest1.gif\ alt=\%s\
title=\%s\/pp class=\footnote\%s/p, $row[alt-text],
$row[title-text], $row[desc-text]);
printf(pimg src=\hotspots/latest2.gif\ alt=\%s\
title=\%s\/pp class=\footnote\%s/p, $row[alt-text],
$row[title-text], $row[desc-text]);
printf(pimg src=\hotspots/latest3.gif\ alt=\%s\
title=\%s\/pp class=\footnote\%s/p, $row[alt-text],
$row[title-text], $row[desc-text]);
}
?
/td
/tr
/table
table summary=company blurb align=center border=0
cellspacing=0
cellpadding=10 width=80%
tr
td colspan=2
div class=co-blurb
hr width=75% /
p class=paratexta href=contact/simradbackground.php
onmouseover=window.status='Some Background information about
Simrad';return true;
title=Some Background information about Simrad
class=special-linksSimrad/a has built its name on
products which do what they are designed to do, whatever the
conditions.
As part of a
href=http://www.kongsberg.com/;
onmouseover=window.status='The Kongsberg Maritime Corporate Home
Page';return true;
title=Our corporate parent company home page target=_blank
class=special-linksKongsberg Maritime/a, the largest
manufacturer of marine electronics in the world, Simrad manufactures
products which are
developed and built on the same principles as the company's
sophisticated
electronics
for the demanding commercial shipping and fisheries market./p
/div
/td
/tr
/table
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RE: [PHP-DB] odd results when running this
Thanks, I did use your code as is, and it works perfectly.
Regards,
Chip
Bastien Koert [EMAIL PROTECTED] wrote on 01/28/2005 09:56:35 AM:
What you should do then is to compare some value to only write the image
data if that value matchesin my code snippet, since you are writing
the
same data, i just loop thru it and increment the image counter to output
the
correct image. A better way to manage this might be to put the image
name in
the db with all the other data as a separate field...then you just fill
in
the image dataand reduce the code to pretty much what i had...
bastien
From: Chip Wiegand [EMAIL PROTECTED]
To: php-db@lists.php.net
Subject: RE: [PHP-DB] odd results when running this
Date: Fri, 28 Jan 2005 09:17:35 -0800
Thanks for the tips.
I find this confusing - I have 3 statements pulling 3 differant images,
so
I don't understand why the script would run 3 times if there is not
loop
telling it to do so. Shouldn't it just load each image and then stop?
Doesn't the while loop stop when it hits the last row?
--
Chip
Bastien Koert [EMAIL PROTECTED] wrote on 01/28/2005 08:37:58
AM:
it runs three times because there are three rows of data and you are
setting
the same thing for each set of data.
try this
?
$sql=select * from hotspots;
$result = mysql_query($sql);
$x = 1;
while ( $row = mysql_fetch_array($result))
{
printf(pimg src=\hotspots/latest$x.gif\ alt=\%s\
title=\%s\/pp class=\footnote\%s/p, $row[alt-text],
$row[title-text], $row[desc-text]);
$x++;
}
?
bastien
From: Chip Wiegand [EMAIL PROTECTED]
To: PHP DB php-db@lists.php.net
Subject: [PHP-DB] odd results when running this
Date: Fri, 28 Jan 2005 07:41:30 -0800
Okay, so I have got the php bits working in this web page but it is
running 3 times every time the page loads, and I see no reason why
(I
have
opened this page in two differant browsers). The code for the
entire
page
is pasted below, it includes some calls to include files which are
not
included in this message of course (and this file is itself
included in
another file), the page can be seen at this address:
http://www.simradusa.com/index-test.php
If anyone can see any reason for the script to run 3 times please
let
me
know.
Thanks,
Chip
==
table width=100% border=0 cellpadding=5 cellspacing=0
align=center
summary=table that contains the body of the page
tr
td
div align=top
p class=welcomeWelcome to Simrad, Inc/p
p class=paratext
Simrad is one of the world's largest manufacturers of marine
electronics
for the yachting,
fishery and commercial marine markets, offering sales and service
worldwide. A complete
range of products from Simrad is available for many different types
of
marine activity,
including auto steering, navigation, hydro acoustic and
communication
equipment. Ranging
from pleasure boating, through fishing and into commercial craft.
Please
select the type
of activity that you are interested in using the links on the left.
/p
br /br /br /
p class=paratextRead a
href=SimradNews/Spring2004/SimradNews-p1.html target=_newspan
class=redSimrad/spanspan class=blueNews/span/a - Your
quarterly guide to today's marine electronics (opens in a new
window)br
/
!-- Or download the .pdf version a
href=SimradNews/Spring2004/SimradNews.phphere/a (765Kbytes)
--/p
br /
p class=c2img src=images/sim-net-icon.gif alt=ANNOUNCING
THE
NEW
SIMNET NETWORK border=0 title=SAWDUST helm - all Simrad
products
//p
p
a href=#
onclick=window.open('simnet.html','simnet','toolbar=no,width=450,
height=500,left=0,top=0,screenX=0,screenY=0,status=no,
scrollbars=yes,resize=yes');return
false
onmouseover=window.status='ANNOUNCING THE NEW SIMNET
NETWORK';return
true;
onmouseout=window.status='ANNOUNCING THE NEW SIMNET NETWORK';
return
true;
title=Read more about the SimNet
class=special-linksemANNOUNCING
THE NEW SIMNET NETWORK/p
/div
/td
tdbr /
p class=titleWhat's New/p
?
$sql=select * from hotspots;
$result = mysql_query($sql);
while ( $row = mysql_fetch_array($result))
{
printf(pimg src=\hotspots/latest1.gif\ alt=\%s\
title=\%s\/pp class=\footnote\%s/p, $row[alt-text],
$row[title-text], $row[desc-text]);
printf(pimg src=\hotspots/latest2.gif\ alt=\%s\
title=\%s\/pp class=\footnote\%s/p, $row[alt-text],
$row[title-text], $row[desc-text]);
printf(pimg src=\hotspots/latest3.gif\ alt=\%s\
title=\%s\/pp class=\footnote\%s/p, $row[alt-text],
$row[title-text], $row[desc-text]);
}
?
/td
/tr
/table
table summary=company blurb align=center border=0
cellspacing=0
cellpadding=10 width=80%
tr
td colspan=2
div
RE: [PHP-DB] storing images in database
Thanks for all the tips guys. I'll keep the last couple for future
reference.
--
Chip
Gareth Heyes [EMAIL PROTECTED] wrote on 01/26/2005 05:30:45 AM:
if(isset($_GET['id'])) {
$id=$_GET['id'];
$query = select bin_data, filetype from binary_data where id=$id;
This is a really bad example, anybody can inject your query with
malicious sql commands.
Never trust user supplied data.
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[PHP-DB] storing images in database
I have stored a .jpg image in a database, then when I make a sql statement to display that image on a web page all I get is the cryptic code in place of the image. I am storing it in a row configured as a blob, mime type image/jpeg and binary (using phpMyAdmin). What am I doing wrong? Regards, Chip -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
RE: [PHP-DB] storing images in database
Thanks Bastien,
In testing this I have added the code samples to a page and have it
working except the path statement is not correct. For now, I've just added
all the code to one page, rather than including a second page. The
statement - echo 'img src=id=$id'; is resulting in this error - The
requested URL /id=$id was not found on this server. Any suggestions?
Thanks,
Chip
Bastien Koert [EMAIL PROTECTED] wrote on 01/25/2005 09:45:39 AM:
the best way to do this is to move the image processing code to a
separate
page and include it like this
echo 'img src=./path/to/image.php?id=$id';
then the image page looks like this:
?php
if($_GET['id']) {
$id = $_GET['id'];
// you may have to modify login information for your database server:
@MYSQL_CONNECT(localhost,root,password);
@mysql_select_db(binary_data);
$query = select bin_data,filetype from binary_data where id=$id;
$result = @MYSQL_QUERY($query);
$data = @MYSQL_RESULT($result,0,bin_data);
$type = @MYSQL_RESULT($result,0,filetype);
Header( Content-type: $type);
echo $data;
};
?
bastien
From: Chip Wiegand [EMAIL PROTECTED]
To: PHP DB php-db@lists.php.net
Subject: [PHP-DB] storing images in database
Date: Tue, 25 Jan 2005 09:11:07 -0800
I have stored a .jpg image in a database, then when I make a sql
statement
to display that image on a web page all I get is the cryptic code in
place
of the image. I am storing it in a row configured as a blob, mime type
image/jpeg and binary (using phpMyAdmin). What am I doing wrong?
Regards,
Chip
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RE: [PHP-DB] storing images in database
I have done it an easier way, and probably a better way all-around anyway.
I am storing the images in a directory and have the script call the
file/alt text/title text and a description text in a paragraph below the
image. It works quite well this way.
What I'm doing is on this home page there is a place for a 'hotspot', a
special mention area for the latest news about a particular item. So I
have a directory and in it will store an image file called 'latest.gif',
so any new image that gets put here will overwrite the existing image.
This will be fine for the purposes and the site.
Here is the code I have used -
?
include connect;
$sql=select * from hotspots;
$result = mysql_query($sql);
while ( $row = mysql_fetch_array($result))
{
printf(pimg src=\hotspots/latest.gif\ alt=\%s\
title=\%s\/pp class=\footnote\%s/p, $row[alt-text],
$row[title-text], $row[desc-text]);
}
?
I'm sure there are many ways to do this sort of thing, but this is quick
and easy, and works.
Thanks guys,
--
Chip
Bastien Koert [EMAIL PROTECTED] wrote on 01/25/2005 01:06:01 PM:
And how are you feeding the $id?are you setting a value for that
element?
In the sample code the default is the record_id that corresponds back to
the
id of the row with the image blob field.
Bastien
From: Chip Wiegand [EMAIL PROTECTED]
To: [EMAIL PROTECTED]
CC: PHP DB php-db@lists.php.net
Subject: RE: [PHP-DB] storing images in database
Date: Tue, 25 Jan 2005 12:57:40 -0800
Bastien Koert [EMAIL PROTECTED] wrote on 01/25/2005 12:46:12
PM:
yes goes back to the whole header problem which is why you are here.
If you could post the code, it would be simpler to help you...
Bastien
This is in the main page -
?
printf(pimg src=\image-src.php?id=$id\ alt=\hotspot
images\%s/p, $row[text]);
?
and this is in a new included page -
?
if($_GET['id']) {
$id = $_GET['id'];
$query = select * from hotspots where id=$id;
$result = @MYSQL_QUERY($query);
$data = @MYSQL_RESULT($result,0,image);
$type = @MYSQL_RESULT($result,0,type);
Header( Content-type: $type);
echo $data;
};
?
The database connection statements are in an include file called at the
top of the main page. In the first statement shown above the alt text
for
the image appears on the web page just fine, the image itself shows a
broken image icon. FWIW, I have the image stored in the database in a
blob
field, is that correct?
--
Chip
From: Chip Wiegand [EMAIL PROTECTED]
To: [EMAIL PROTECTED]
Subject: RE: [PHP-DB] storing images in database
Date: Tue, 25 Jan 2005 12:44:44 -0800
Bastien Koert [EMAIL PROTECTED] wrote on 01/25/2005
12:39:15
PM:
Its not src='id=$id' that will defnintely blow up
echo 'img src=./path/to/image.php?id=$id';
where $id is the id of the record you are trying to get the
image
to...
Bastien
So the code has to be a seperate included page I guess?
--
Chip
From: Chip Wiegand [EMAIL PROTECTED]
To: [EMAIL PROTECTED]
CC: php-db@lists.php.net
Subject: RE: [PHP-DB] storing images in database
Date: Tue, 25 Jan 2005 12:37:15 -0800
Thanks Bastien,
In testing this I have added the code samples to a page and
have it
working except the path statement is not correct. For now, I've
just
added
all the code to one page, rather than including a second page.
The
statement - echo 'img src=id=$id'; is resulting in this
error -
The
requested URL /id=$id was not found on this server. Any
suggestions?
Thanks,
Chip
Bastien Koert [EMAIL PROTECTED] wrote on 01/25/2005
09:45:39
AM:
the best way to do this is to move the image processing code
to
a
separate
page and include it like this
echo 'img src=./path/to/image.php?id=$id';
then the image page looks like this:
?php
if($_GET['id']) {
$id = $_GET['id'];
// you may have to modify login information for your
database
server:
@MYSQL_CONNECT(localhost,root,password);
@mysql_select_db(binary_data);
$query = select bin_data,filetype from binary_data where
id=$id;
$result = @MYSQL_QUERY($query);
$data = @MYSQL_RESULT($result,0,bin_data);
$type = @MYSQL_RESULT($result,0,filetype);
Header( Content-type: $type);
echo $data;
};
?
bastien
From: Chip Wiegand [EMAIL PROTECTED]
To: PHP DB php-db@lists.php.net
Subject: [PHP-DB] storing images in database
Date: Tue, 25 Jan 2005 09:11:07 -0800
I have stored a .jpg image in a database, then when I make
a
sql
statement
to display that image on a web page all I get is the
cryptic
code
in
place
of the image. I am storing it in a row configured as a
blob,
mime
type
[PHP-DB] query of two tables returns too many rows, many more than the two tables contain
I have two tables I want to get out the rows that are different between them. The results I am getting is almost 50,000 rows, but the two tables, combined, contain only about 600 rows total. Here is the select statement - SELECT dealers.account_no, dealers.DealerName, blackgate_users.User_Name, blackgate_users.DealerName FROM dealers, blackgate_users WHERE dealers.account_no NOT LIKE blackgate_users.User_Name in these tables the dealers.account_no is the same data as the blackgate_users.User_Name dealers.DealerName is the same data as the blackgate_users.DealerName I just want the rows that are in the dealers table but not in the blackgate_users table. Thanks for any help, Chip Wiegand Computer Services Simrad, Inc 425-778-8821 425-771-7211 (FAX) -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
RE: [PHP-DB] query of two tables returns too many rows, many more than the two tables contain
Gryffyn, Trevor [EMAIL PROTECTED] wrote on 11/11/2004 01:39:37 PM: This is a common join issue. If you don't specify ANYTHING to connect between the two tables, it'll do one row from the first table, then ALL the rows from the second. Row #2 from the first, then ALL the rows from the second. snip Just read the last thing you wrote.. Let me revise what I said then. You say you want everything where there's an entry in Dealers, but no corresponding entry in blackgate_users eh? That's an outer join. Try this (syntax happy with SQL Server, I don't use MySQL a lot so it might be slightly different): SELECT dealers.account_no, dealers.DealerName, blackgate_users.User_Name, blackgate_users.DealerName FROM dealers left join blackgate_users on dealers.DealerName = blackgate_users.DealerName WHERE blackgate_users.DealerName is null Thanks for the help. That gets me much closer. I did a count in both tables and figured there should be 121 rows returned by the query. The above select statement gets me 141 rows returned. With a little sleuthing around in there I will probably figure out what the extra 10 rows are. Thanks you very much. Regards, Chip What this says is take everything in Dealers, left join it against blackgate_users (left join says to take everything from the left side.. And match against the right side but leave NULL entries where there's no match). Then we tell it that the field we want to compare is DealnerName in both cases. The WHERE clause says only show us where DealerName is null (meaning no corresponding record in blackgate_users). I think that'll do it for ya. -TG -Original Message- From: Chip Wiegand [mailto:[EMAIL PROTECTED] Sent: Thursday, November 11, 2004 4:28 PM To: PHP DB Subject: [PHP-DB] query of two tables returns too many rows, many more than the two tables contain I have two tables I want to get out the rows that are different between them. The results I am getting is almost 50,000 rows, but the two tables, combined, contain only about 600 rows total. Here is the select statement - SELECT dealers.account_no, dealers.DealerName, blackgate_users.User_Name, blackgate_users.DealerName FROM dealers, blackgate_users WHERE dealers.account_no NOT LIKE blackgate_users.User_Name in these tables the dealers.account_no is the same data as the blackgate_users.User_Name dealers.DealerName is the same data as the blackgate_users.DealerName I just want the rows that are in the dealers table but not in the blackgate_users table. Thanks for any help, Chip Wiegand Computer Services Simrad, Inc 425-778-8821 425-771-7211 (FAX) -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
[PHP-DB] I'm not getting any mailing list messages
I just subscribed to the list yesterday and have yet to receive any messages from the list. Any ideas? -- Chip Wiegand Computer Services Simrad, Inc 425-778-8821 425-771-7211 (FAX)
[PHP-DB] How to do an update with arithmetic from a user input
I have an existing database and web page showing the query results. I have
added another field to allow a user to input a two digit number. I want
that number to change the number in one of the existing columns for the
item it was input next to.
I have these columns -
Part Number Description QuantityList Price Put on
hold
I want the Quantity column changed according to the number the user input
in the Put on hold column.
I'm stuck in the query to write, trying to do
$sql1 = update refurbs set qty(qty - $qty) order by 'description';
but that doesn't work. I have pasted in the main part of the code below...
Thanks for any help you can provide,
Chip Wiegand
Computer Services
Simrad, Inc
425-778-8821
425-771-7211 (FAX)
BTW, thanks for the replys to the first message, I was expecting the list
to be much more busy.
===
form action=? PHP_SELF ? method=post name=refurbs
table summary= border=1 cellpadding=5 cellspacing=0
width=90% align=center
tr
td class=small bgcolor=#ffPart Number/td
td class=small bgcolor=#ffDescription/td
td class=small bgcolor=#ffQuantity/td
td class=small bgcolor=#ffList Price/td
td class=small bgcolor=#ffSpecial Price/td
td class=small bgcolor=#ffPut on Hold
/td
/tr
?
if (isset($submit)):
$sql = update refurbs set qty(qty - '$buy') order by 'description';
$result = mysql_query($sql);
$Colors = array('#ffcc99','#ffcc33');
$I = 0;
while ( $row = mysql_fetch_array($result))
{
printf(tr bgcolor= . $Colors[ $I ++ % count( $Colors ) ] .
td%s\n/td\n, $row[part_number]);
printf(td%s\n/td\n, $row[description]);
printf(td%s\n/td\n, $row[qty]);
printf(td%s\n/td\n, $row[list]);
printf(td%s\n/td\n, $row[special]);
printf(tdinput type=\text\ name=\buy\ size=\2\ maxlength=\2\
/td/tr\n);
}
?
/table
table summary= border=0 cellpadding=5 cellspacing=0
width=90% align=center
tr
tdbr /div align=centerinput type=submit name=submit
value=Submit //div/td
/tr
/table
/form
Re: [PHP-DB] How to do an update with arithmetic from a user input
Justin Patrin [EMAIL PROTECTED] wrote on 08/12/2004 12:55:24 PM:
On Thu, 12 Aug 2004 12:46:19 -0700, Chip Wiegand
[EMAIL PROTECTED] wrote:
I have an existing database and web page showing the query results. I
have
added another field to allow a user to input a two digit number. I
want
that number to change the number in one of the existing columns for
the
item it was input next to.
I have these columns -
Part Number Description QuantityList Price Put on
hold
I want the Quantity column changed according to the number the user
input
in the Put on hold column.
I'm stuck in the query to write, trying to do
$sql1 = update refurbs set qty(qty - $qty) order by 'description';
Perhaps:
$sql1 = update refurbs set qty = qty - $qty;
Also,
1) you have no WHERE there, so this will update *all* records.
2) why are you trying to order an update??
Thanks for the quick reply. The order by stuff is gone now, it was an
'artifact'. hehe. Anyway, I added the where clause so my query looks like
this:
$sql = update refurbs set qty = qty - $buy where part_number =
$part_number;
But when I try to run this I get an error -
Parse error: parse error, unexpected $ in
/usr/local/www/data-dist/auth_dealers/refurbs-test.php on line 102
I don't see an rogue $'s in the code.
(Eventually the form will submit to another page which will email the info
to a person here in our office).
--
Chip
Now the code looks like this -
===
form action=? PHP_SELF ? method=post name=refurbs
table summary= border=1 cellpadding=5 cellspacing=0
width=90% align=center
tr
td class=small bgcolor=#ffPart Number/td
td class=small bgcolor=#ffDescription/td
td class=small bgcolor=#ffQuantity/td
td class=small bgcolor=#ffList Price/td
td class=small bgcolor=#ffSpecial Price/td
td class=small bgcolor=#ffPut on Hold
/td
/tr
?
if (isset($submit)):
$sql = update refurbs set qty = qty - $buy where part_number =
$part_number;
$result = mysql_query($sql);
$Colors = array('#ffcc99','#ffcc33');
$I = 0;
while ( $row = mysql_fetch_array($result))
{
printf(tr bgcolor= . $Colors[ $I ++ % count( $Colors ) ] .
td%s\n/td\n, $row[part_number]);
printf(td%s\n/td\n, $row[description]);
printf(td%s\n/td\n, $row[qty]);
printf(td%s\n/td\n, $row[list]);
printf(td%s\n/td\n, $row[special]);
printf(tdinput type=\text\ name=\buy\ size=\2\ maxlength=\2\
/td/tr\n);
}
?
/table
table summary= border=0 cellpadding=5 cellspacing=0
width=90% align=center
tr
tdbr /div align=centerinput type=submit name=submit
value=Submit //div/td
/tr
/table
/form
Re: [PHP-DB] How to do an update with arithmetic from a user input
Justin Patrin [EMAIL PROTECTED] wrote on 08/12/2004 01:37:35 PM:
Unless $part_number is a number, you're going to want to put quotes
around it. Well, actually, you want to run mysql_real_escape_string()
on it.
You also don't seem to have an end to that if statement. Usually,
people use curly braces around those:
if(true) {
//do something
}
Thanks, I'm making headway. This time when I entered a value and
submitted, the database value for the qty field was not changed, but
instead the description field was cleared. The database table has columns
titled qty as well as description and others.
Regards,
Chip
The newest version of the code is this -
form action=refurbs-send.php method=post name=refurbs
table summary= border=1 cellpadding=5 cellspacing=0
width=90% align=center
tr
td class=small bgcolor=#ffPart Number/td
td class=small bgcolor=#ffDescription/td
td class=small bgcolor=#ffQuantity/td
td class=small bgcolor=#ffList Price/td
td class=small bgcolor=#ffSpecial Price/td
td class=small bgcolor=#ffPut on Hold
/td
/tr
?
if (isset($submit)):
{
$sql = update refurbs set qty = qty - '$buy' where part_number =
'$part_number';
$result = mysql_query($sql);
}
else:
$sql = select * from refurbs order by description;
$result = mysql_query($sql);
$Colors = array('#ffcc99','#ffcc33');
$I = 0;
while ( $row = mysql_fetch_array($result))
{
printf(tr bgcolor= . $Colors[ $I ++ % count( $Colors ) ] .
td%s\n/td\n, $row[part_number]);
printf(td%s\n/td\n, $row[description]);
printf(td%s\n/td\n, $row[qty]);
printf(td%s\n/td\n, $row[list]);
printf(td%s\n/td\n, $row[special]);
printf(tdinput type=\text\ name=\buy\ size=\2\ maxlength=\2\
/td/tr\n);
}
?
/table
table summary= border=0 cellpadding=5 cellspacing=0
width=90% align=center
tr
tdbr /div align=centerinput type=submit name=submit
value=Submit //div/td
/tr
/table
/form
RE: [PHP-DB] address info, forms, maintanance
John W. Holmes [EMAIL PROTECTED] wrote on 06/04/2003 05:24:22 PM: Thanks to everyone for the suggestions. Got it fixed. Just added a couple lines of code - $sql1 = select * from endusers where name like '$name'; $result1 = mysql_query($sql1); $count1 = mysql_num_rows($result1); if ($count1 == 0 ) $sql = insert into endusers..blah blah blah Now it works fine, no more duplicate entries. But now you're doing two queries for every insert. If you simply made your columns unique and let the database handle it, you'd only have to do one INSERT. Then check affected_rows() or mysql_error() to see if either no rows were affected (no rows inserted) or the error mentions duplicate. If either is the case, the row wasn't inserted because of a unique constraint. If there is an error but it's not duplicate or whatever, then it's another error and you should show it. ---John W. Holmes... Thanks for the tips, I'll work on that. In the meantime, this database is pretty low useage, very specialized information for a small market segment, so for now it will be okay. It'll never have hundreds or thousands of hits per day. But I will look into making the suggested changes. Thanks, Chip -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
[PHP-DB] address info, forms, maintanance
I need to modify some company web pages that include a form for asking the company departments questions, such as service dept, etc. As the forms are currently built there is some field validation but no cookies. This info is saved in a mysql database, problem is, there are duplicate entries, every time an end-user posts a question. What kind of code do I need to add to these existing pages to prevent duplicate entries in the database? (Manually scanning through and deleting dupes is a real drag.) And what about cookies? Could I make use of these on these pages? I haven't worked with cookies yet, so am not sure where to start with that part. Thanks, -- Chip -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP-DB] address info, forms, maintanance
Dan Brunner [EMAIL PROTECTED] wrote on 06/04/2003 01:24:13 PM: Hello!! Are talking about the same Post Question??? On another words, what is being duplicated?? If not, you can use auto_increment. Dan This pertains to every time a person visits the page and fills in the form - today, tommorrow, next week, next month, whenever. The address info is being written to the database each time they send the form. The message body itself isn't saved to the database, only the name and address info. -- Chip On Wednesday, June 4, 2003, at 03:13 PM, [EMAIL PROTECTED] wrote: I need to modify some company web pages that include a form for asking the company departments questions, such as service dept, etc. As the forms are currently built there is some field validation but no cookies. This info is saved in a mysql database, problem is, there are duplicate entries, every time an end-user posts a question. What kind of code do I need to add to these existing pages to prevent duplicate entries in the database? (Manually scanning through and deleting dupes is a real drag.) And what about cookies? Could I make use of these on these pages? I haven't worked with cookies yet, so am not sure where to start with that part. Thanks, -- Chip -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP-DB] address info, forms, maintanance
Thanks to everyone for the suggestions. Got it fixed. Just added a couple lines of code - $sql1 = select * from endusers where name like '$name'; $result1 = mysql_query($sql1); $count1 = mysql_num_rows($result1); if ($count1 == 0 ) $sql = insert into endusers..blah blah blah Now it works fine, no more duplicate entries. Just for those who asked - this is a database of end-user names and addresses for a mailing list. The questions posted are kept by the department that responded to them. No one here has ever asked or implied a need to have the messages saved to the database, what they do with the message is their game, not mine. I'm just cleaning up their existing code a bit. -- Chip Miles Thompson [EMAIL PROTECTED] wrote on 06/04/2003 08:31:14 AM: Chip, Given that you have duplicate entries, then one field, or a combination of several fields, will probably be unique. Before adding a record execute a SELECT, using the values from the form against that combination. If the number of records returned is greater than 0, then execute an UPDATE query, otherwise an INSERT. Now, that information is so vague to be almost useless, for it begs a number of other questions, such as How many questions are end users allowed to ask? How are questions flagged for expiration or completion/solution? What kind of mix of questions do you allow? Cookies only store a bit of information on the user's browser. There's nothing magic about them - google for netscape cookies. I don't think they'd be much help to you here. HTH - Miles Thompson At 01:13 PM 6/4/2003 -0700, [EMAIL PROTECTED] wrote: I need to modify some company web pages that include a form for asking the company departments questions, such as service dept, etc. As the forms are currently built there is some field validation but no cookies. This info is saved in a mysql database, problem is, there are duplicate entries, every time an end-user posts a question. What kind of code do I need to add to these existing pages to prevent duplicate entries in the database? (Manually scanning through and deleting dupes is a real drag.) And what about cookies? Could I make use of these on these pages? I haven't worked with cookies yet, so am not sure where to start with that part. Thanks, -- Chip -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP-DB] Odd browser problem with sessions
On Wed, 29 Jan 2003 08:30:31 -0600 [EMAIL PROTECTED] wrote: IE 5.01 has an issue with its basic auth. There is no patch, only an upgrade path Gary Every I wish it were that simple, it happens in all versions of IE. I've tested it myself in the latest versions and it happens. I get at least one phone call a day from people asking why they can't log in, the screen just reloads. I suggest to them to upgrade to Mozilla, Netscape or Opera. -- Chip Sr. UNIX Administrator Ingram Entertainment (615) 287-4876 Pay It Forward mailto:[EMAIL PROTECTED] http://accessingram.com -Original Message- From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED]] Sent: Tuesday, January 28, 2003 2:02 PM To: PHP-DB List Subject: [PHP-DB] Odd browser problem with sessions I have a web site that is protected using sessions, and it works fine. Only occasionally when a end-user tries to log in using IE the login screen will just reload, every time they try to log in. I found that by closing IE (all windows) then trying again, it will go on to the proper page. This behavior appears to only effect IE, I saw it once on either Netscape or Mozilla (I don't recall which), and no one I've talked to on the phone has had this problem in any browser other than IE, and there have been many calls about this. Any idea what this might be caused by? Thanks, -- Chip Wiegand Computer Services Simrad, Inc www.simradusa.com [EMAIL PROTECTED] There is no reason anyone would want a computer in their home. --Ken Olson, president, chairman and founder of Digital Equipment Corporation, 1977 (Then why do I have 8? Somebody help me!) -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
[PHP-DB] Odd browser problem with sessions
I have a web site that is protected using sessions, and it works fine. Only occasionally when a end-user tries to log in using IE the login screen will just reload, every time they try to log in. I found that by closing IE (all windows) then trying again, it will go on to the proper page. This behavior appears to only effect IE, I saw it once on either Netscape or Mozilla (I don't recall which), and no one I've talked to on the phone has had this problem in any browser other than IE, and there have been many calls about this. Any idea what this might be caused by? Thanks, -- Chip Wiegand Computer Services Simrad, Inc www.simradusa.com [EMAIL PROTECTED] There is no reason anyone would want a computer in their home. --Ken Olson, president, chairman and founder of Digital Equipment Corporation, 1977 (Then why do I have 8? Somebody help me!) -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
[PHP-DB] query results page ignores style sheet reference
I have a site I have to convert to frames (*ugh*) (so I'm using php frames, better than html frames anyway) and came up with a little problem. On one page is a list of states, click on the state name and get a list of dealers in that state. This works. What doesn't work is the list of dealers page ignores the style sheet somewhat, so the original page looks great, the results page doesn't. What's happening is the original page has certain font sizes dictated by the style sheet, but the results page uses the browsers defaults for the fonts. Other style sheet attributes are obeyed by the results page. This is quite strange. Anyone seen this behavior before? -- Chip Wiegand Computer Services Simrad, Inc www.simradusa.com [EMAIL PROTECTED] There is no reason anyone would want a computer in their home. --Ken Olson, president, chairman and founder of Digital Equipment Corporation, 1977 (They why do I have 8? Somebody help me!) -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
[PHP-DB] database update question
I have a database with several hundred entries of file names that end with .pdf. I have converted all those docs to .zip, now I need to change all the entries in the database to .zip. I tried to use update table_name set col_name='%.zip' where col_name like '%.pdf' id = '11' but of course that changed the file name for id 11 to %.zip. Is there a way to change all the entries from .pdf to .zip without writing an update statement for each individual row? -- Chip Wiegand Computer Services Simrad, Inc www.simradusa.com [EMAIL PROTECTED] There is no reason anyone would want a computer in their home. --Ken Olson, president, chairman and founder of Digital Equipment Corporation, 1977 (They why do I have 8? Somebody help me!) -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
[PHP-DB] database update question
I have a database with several hundred entries of file names that end with .pdf. I have converted all those docs to .zip, now I need to change all the entries in the database to .zip. I tried to use update table_name set col_name='%.zip' where col_name like '%.pdf' id = '11' but of course that changed the file name for id 11 to %.zip. Is there a way to change all the entries from .pdf to .zip without writing an update statement for each individual row? -- Chip Wiegand Computer Services Simrad, Inc www.simradusa.com [EMAIL PROTECTED] There is no reason anyone would want a computer in their home. --Ken Olson, president, chairman and founder of Digital Equipment Corporation, 1977 (They why do I have 8? Somebody help me!) -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP-DB] a menu item from a database, linked to a list from a anothertable
Jason Wong [EMAIL PROTECTED] wrote on 11/13/2002 10:46:32 PM: On Thursday 14 November 2002 07:32, [EMAIL PROTECTED] wrote: Is this even possible? I have a menu that is generated from a table. What is your menu -- a set of a tags? or a selector box? I suspect you mean the latter ... Not a list box, but a menu of about 15 category titles, all shown on the page in a list, click on category get the category items listed. Both the category titles and the individual items will come from a database. I am trying this to make a faq's page - list the faq categories, click on the category and get the list of faq's. The first part is easy, the second part is easy, both seperately. I am wondering if I can make the first part (title) be a link to select the items for the category from a database. I've seen this done before, but the app I was looking at had so many include files referencing other include files that getting it all straight was near impossible. -- Chip Typically, the menu items are links to html pages. I would like to have the link bring up a list of items from another table in the database, not an html page. So I guess this would be a select statement linked from a select statement? Is this even possible? Yes, google - dynamic linked listbox php. -- Jason Wong - Gremlins Associates - www.gremlins.biz Open Source Software Systems Integrators * Web Design Hosting * Internet Intranet Applications Development * /* The early bird who catches the worm works for someone who comes in late and owns the worm farm. -- Travis McGee */ -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
[PHP-DB] a menu item from a database, linked to a list from a another table
Is this even possible? I have a menu that is generated from a table. Typically, the menu items are links to html pages. I would like to have the link bring up a list of items from another table in the database, not an html page. So I guess this would be a select statement linked from a select statement? Is this even possible? -- Chip Wiegand Computer Services Simrad, Inc www.simradusa.com [EMAIL PROTECTED] There is no reason anyone would want a computer in their home. --Ken Olson, president, chairman and founder of Digital Equipment Corporation, 1977 (They why do I have 8? Somebody help me!) -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP-DB] alter table statement from within php web page
Forget it, It works, I was entering a field name with a space in it, and that was causing the problem. Duh! -- Chip phpdbOn Tue, 2002-10-01 at 22:01, Chip Wiegand wrote: I would like to know how to add columns to an existing table. I've tried this - $qry = mysql_query(alter table exercises add column $fieldname char($length) not null) or die(sql_error()); but it doesn't work, I get this error - 1064: You have an error in your SQL syntax near 'lifts, char(25), not null' at line 1 I don't see any reference to this in the php manual or in the two php/mysql books that I have. Anybody know how to do this? -- Chip -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
[PHP-DB] alter table statement from within php web page
I would like to know how to add columns to an existing table. I've tried this - $qry = mysql_query(alter table exercises add column $fieldname char($length) not null) or die(sql_error()); but it doesn't work, I get this error - 1064: You have an error in your SQL syntax near 'lifts, char(25), not null' at line 1 I don't see any reference to this in the php manual or in the two php/mysql books that I have. Anybody know how to do this? -- Chip -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
[PHP-DB] big sessions problem - page refusing to load, but will if I skip the logonpage
I have a logon screen that starts a session for several other pages and it
has been working great,
over the weekend something happened. I checked the site from home and
after login the main
page would not load, just a white screen and view source showed absolutely
nothing. Now I'm in the
office and looking at the source code and I don't see anything wrong,
nothing changed. I can load the
pages by skipping the logon and typing in the complete url manually and
they all load just fine, soon
as I try to use the logon screen the first main page after it will not
load, if I enter the full url to another
protected page it will load, then I can use the links for the main page
and it too will now load. Just after
the logon, the main page will not load. I have tried this with both my own
login and a users login and
it is happening to all.
Any idea what could be causing this? I've gotta get this fixed ASAP,
perminently, this is a business web
site. The code for the login screen and the main page are pasted below.
Regards,
--
Chip Wiegand
Computer Services
Simrad, Inc
www.simradusa.com
[EMAIL PROTECTED]
There is no reason anyone would want a computer in their home.
--Ken Olson, president, chairman and founder of Digital Equipment
Corporation, 1977
(They why do I have 9? Somebody help me!)
=
Code for authorization login:
---
?php
//auth_user.php
include ../common_db.inc;
$register_script = register.php;
function auth_user($userid, $userpassword) {
global $default_dbname, $user_tablename;
$link_id = db_connect($default_dbname);
$query = SELECT username FROM $user_tablename
WHERE userid = '$userid'
AND userpassword = '$userpassword';
$result = mysql_query($query);
if(!mysql_num_rows($result)) return 0;
else {
$query_data = mysql_fetch_row($result);
return $query_data[0];
}
}
function login_form() {
global $PHP_SELF;
?
HTML
HEAD
TITLELogin/TITLE
/HEAD
BODY
FORM METHOD=POST ACTION=? echo $PHP_SELF ?
DIV ALIGN=CENTERCENTER
H3Please log in to access the page you requested./H3
TABLE BORDER=1 WIDTH=200 CELLPADDING=2
TR
TH WIDTH=18% ALIGN=RIGHT NOWRAPID/TH
TD WIDTH=82% NOWRAP
INPUT TYPE=TEXT NAME=userid SIZE=8
/TD
/TR
TR
TH WIDTH=18% ALIGN=RIGHT NOWRAPPassword/TH
TD WIDTH=82% NOWRAP
INPUT TYPE=PASSWORD NAME=userpassword SIZE=8
/TD
/TR
TR
TD WIDTH=100% COLSPAN=2 ALIGN=CENTER NOWRAP
INPUT TYPE=SUBMIT VALUE=LOGIN NAME=Submit
/TD
/TR
/TABLE
/CENTER/DIV
/FORM
/BODY
/HTML
?
}
session_start();
if(!isset($userid)) {
login_form();
exit;
}
else {
session_register(userid, userpassword);
$username = auth_user($userid, $userpassword);
if(!$username) {
session_unregister(userid);
session_unregister(userpassword);
echo centerpAuthorization failed.br .
Sorry, you must enter a valid userid and password combo./p
.
pThis site is restricted to Simrad Authorized Dealers
only.br\n;
echo Click a href=\$PHP_SELF\here/A to try again./p;
echo pIf you are a Simrad Authorized Dealer and \n;
echo you are not a member yet,brplease .
call Simrad, Inc and request your login information./p\n;
echo pIf you are not a Simrad Authorized Dealer please click on
thebr \n;
echo following link to return to our home page
\n;
echo a href=\../index.php\Simrad,
Inc/a/p/center\n;
exit;
}
// else
// {
//echo pa href=\dealers_page.php\Click here/a to continue;
//header( 'Location: dealers_page.php' );
// }
}
?
==
Code for the main page after the successful login:
-
?
include auth_user.php;
?
!DOCTYPE html PUBLIC -//W3C//DTD XHTML 1.0 Transitional//EN
http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd;
html xmlns=http://www.w3.org/1999/xhtml;
head
? include ../connect;
$sql = select * from main where cat = 'dealers' and page =
'dealers';
$result = mysql_query($sql);
while ($row=mysql_fetch_array($result))
{ printf(title%s/title, $row[name]); }
?
meta name=generator content=HTML Tidy, see www.w3.org /
meta name=target
content=td, tda, tda157, 157, tda 157, tdc, tdc338, tdc 338, tdl,
td-l, 1100, 1550, 1620, tb, tb548, tb 548, buoy, radio buoy, direction
finder, df, addf /
? include ../metatags_pages.inc; ?
style type=text/css
/*![CDATA[*/
td.small {font-size: 75%}
.red {color: red;
font-size: 75%; }
.black {color: black;
font-size: 75%; }
.big {font-size: 150% }
/*]]*/
/style
/head
body
!-- Code for Extreme Tracker --
a target
[PHP-DB] advise needed for 'authorized only' site
I have set up a section of my company site for use by authorized dealers only. I am currently using mysql authorization, which works for the first page, but if someone were to type in the url of an underlying page they would be able to get in without authorization. I know I could use .htaccess for handling this but with a minimum of 350 -400 users to keep track of that would be unwieldly to say the least, especially for my boss who doesn't have a clue about *nix and has never even heard of .htaccess. What other options do I have to keep the underlying pages from being accessed without the user being forced to go through the logon screen? Thanks, -- Chip Wiegand Computer Services Simrad, Inc www.simradusa.com [EMAIL PROTECTED] There is no reason anyone would want a computer in their home. --Ken Olson, president, chairman and founder of Digital Equipment Corporation, 1977 (They why do I have 9? Somebody help me!) -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP-DB] how do I echo this statement to the browser? -solved
Thanks for the tip, I solved it a little differently - sql1=select f_name, l_name, count(*) as 'Attendance' from kids group by l_name; sql2=select count(distinct kids_id) as 'total' from kids; I added a new column to the database called kids_id, so when a new kid is added he/she gets assigned a number, if a record is added for an existing kid the new record keeps the existing kids_id, making it simple to count only distinct kids_id's. The input form lists the kids_id, f_name, l_name as a link to edit their record, and delete for a delete link, in the left column, and in the right is the form for input/editing. The kids list is ordered by kids_id, making it easy to know what id's already exist, or don't yet. -- Chip On Sat, 2002-09-21 at 01:39, Ignatius Reilly wrote: This is a query requiring two different levels of aggregation. MySQL does not support it. You can reduce it to two different queries SELECT f_name, l_name, COUNT(*) AS Total_Kids FROM kids GROUP BY l_name and SELECT COUNT(*) AS Total_Rows FROM kids your result will have columns $row1['Total_Kids'] and $row2['Total_Rows'] When MySQL supports subqueries, you can write: SELECT f_name, l_name, COUNT(*) AS Total_Kids FROM kids GROUP BY l_name JOIN SELECT COUNT(*) AS Total_Rows FROM kids HTH Ignatius - Original Message - From: Chip Wiegand [EMAIL PROTECTED] To: phpdb [EMAIL PROTECTED] Sent: Saturday, September 21, 2002 7:18 AM Subject: [PHP-DB] how do I echo this statement to the browser? I want to get the first name, last name, total row count, and total time a name appears in the database. Something like this - select f_name, l_name, count(*) as 'Total Kids', count distinct l_name from kids group by l_name it doesn't work of course, the count distinct l_name part is completely wrong. The database has kids names in it appearing multiple times, whence the need for distinct, I want the total number of times it appears though. Plus, how do I echo the number for 'Total Kids' on the browser? I don't have a column heading to reference like with f_name and l_name. I s'pose this is clear as mud. Thanks for any help you all can provide, Chip -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
[PHP-DB] how do I echo this statement to the browser?
I want to get the first name, last name, total row count, and total time a name appears in the database. Something like this - select f_name, l_name, count(*) as 'Total Kids', count distinct l_name from kids group by l_name it doesn't work of course, the count distinct l_name part is completely wrong. The database has kids names in it appearing multiple times, whence the need for distinct, I want the total number of times it appears though. Plus, how do I echo the number for 'Total Kids' on the browser? I don't have a column heading to reference like with f_name and l_name. I s'pose this is clear as mud. Thanks for any help you all can provide, Chip -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
[PHP-DB] update db problem
I am making a web form for entering/editing/deleting items from a database,
the entering and deleting
parts work fine. The editing part is giving me some problems -
Here's the code -
?
include connect;
if ($submit) {
// if no ID then we're adding a new rma, else we're editing an existing
rma
if ($id) {
$sql = UPDATE rma SET rma_no='$rma_no',rma_name='$rma_name',rec_date
='$rec_date',rma_status='$rma_status' WHERE id='$id';
} else {
$sql = INSERT INTO rma (rma_no,rma_name,rec_date,rma_status) VALUES
('$rma_no','$rma_name','$rec_date','$rma_status');
}
// run SQL against the DB
$result = mysql_query($sql);
echo Record updated/edited!p a href=\rma_input.php\List
RMA's/a;
} elseif ($delete) {
// delete a record
$sql = DELETE FROM rma WHERE id=$id;
$result = mysql_query($sql);
echo Deleted!pa href=\rma_input.php\List RMA's/a/p;
} else {
// this part happens if we don't press submit
if (!$id) {
// print the list if there is not editing
$result = mysql_query(SELECT * FROM rma,$db);
while ($myrow = mysql_fetch_array($result)) {
printf(a href=\%s?id=%s\%s %s/a \n, $PHP_SELF, $myrow[id],
$myrow[rma_no], $myrow[rma_name], $myrow[rec_date], $myrow
[rma_status]);
printf(a href=\%s?id=%sdelete=yes\(DELETE)/abr,
$PHP_SELF, $myrow[id]);
}
}
if ($id) {
// editing so select a record
$sql = SELECT * FROM rma WHERE id=$id;
$result = mysql_query($sql);
$myrow = mysql_fetch_array($result);
$id = $myrow[id];
$rma_no = $myrow[rma_no];
$rma_name = $myrow[rma_name];
$rec_date = $myrow[rec_date];
$rma_status = $myrow[rma_status];
// print the id for editi
?
input type=hidden name=id value=?php echo $id ?
?
}
}
?
What is happening is when I click on an item on the list it will display
all the fields except rma_status (a textarea form field).
When I make any changes and press submit, the changes to fields other than
rma_status, those fields are updated or left
as is, but the rma_status textarea is wiped out.
Also, instead of updating the current item in the db, it is submitting
another row into the database, giving multiple rows of
the same rma, but with each one showing it's edits over time. This would be
fine for historical purposes, but not what I want.
What am I doing wrong?
--
Chip Wiegand
Computer Services
Simrad, Inc
www.simradusa.com
[EMAIL PROTECTED]
There is no reason anyone would want a computer in their home.
--Ken Olson, president, chairman and founder of Digital Equipment
Corporation, 1977
(They why do I have 9? Somebody help me!)
--
PHP Database Mailing List (http://www.php.net/)
To unsubscribe, visit: http://www.php.net/unsub.php
[PHP-DB] verify text entry in a textarea doesn't work
I have some error checking routines for my web forms, and they are working
great except for one
field, a textarea field. I am using the following code, but it does not
give an error when the field is
empty:
if (empty($feedback))
{
$errmsg .= font color='red'liYou might want to enter some
comments/li/font\n;
}
I use the same type of checking on input type=text boxes and it works
fine.
Any ideas why it doesn't work on a textarea field?
--
Chip Wiegand
Computer Services
Simrad, Inc
www.simradusa.com
[EMAIL PROTECTED]
There is no reason anyone would want a computer in their home.
--Ken Olson, president, chairman and founder of Digital Equipment
Corporation, 1977
(They why do I have 9? Somebody help me!)
--
PHP Database Mailing List (http://www.php.net/)
To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP-DB] verify text entry in a textarea doesn't work
Thankyou Jason, the problem was indeed in the html code. The code was
broken
into multiple lines and indented, by moving it all onto one line with no
spaces between the it now works. Also, when I tab or place the cursor
inside
the textarea, the cursor now is placed in the upper left corner, whereas
before
it was one tab furthor into the box, now I know why that happens also.
thanks,
--
Chip Wiegand
Computer Services
Simrad, Inc
www.simradusa.com
[EMAIL PROTECTED]
There is no reason anyone would want a computer in their home.
--Ken Olson, president, chairman and founder of Digital Equipment
Corporation, 1977
(They why do I have 9? Somebody help me!)
Jason Wong [EMAIL PROTECTED] wrote on 08/27/2002 09:24:35 AM:
On Tuesday 27 August 2002 23:26, [EMAIL PROTECTED] wrote:
I have some error checking routines for my web forms, and they are
working
great except for one
field, a textarea field. I am using the following code, but it does not
give an error when the field is
empty:
if (empty($feedback))
{
$errmsg .= font color='red'liYou might want to enter some
comments/li/font\n;
}
Presumably $errmsg is not appended to because empty($feedback) is FALSE.
IOW
$feedback is not empty, IOW $feedback contains something. What I'm
getting at
is when trying to debug some code, use a liberal dose of common-sense and
logic.
Print it out to see what it contains. Most likely it's whitespace. To
convince
yourself that it does contain something use strlen().
I use the same type of checking on input type=text boxes and it works
fine.
Any ideas why it doesn't work on a textarea field?
What HTML code are you using to display the textarea field? Make sure
there
are no spaces and/or newlines in there.
--
Jason Wong - Gremlins Associates - www.gremlins.com.hk
--
PHP Database Mailing List (http://www.php.net/)
To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP-DB] keeping form field data when reloading a form
Thanks, that works great. -- Chip Stuart McDonald [EMAIL PROTECTED] wrote on 08/20/2002 05:55:40 PM: Are you setting the variables in the form? Something like this - note the value field in the username field: form method=post action=index.php name=loginuser table summary=layout table to hold login data width=300 align=center cellpadding=1 cellspacing=0 border=0 bgcolor= #ff tr td align=leftp class=smallbUsername:/b/p/td td align=leftinput type=text name=username value=?php echo $username;? maxlength=50 size=15 style=width: 120px; font-size: 12px/td /tr tr td align=left p class=smallbPassword:/b/p/td td align=leftinput type=password name=password maxlength=50 size=15 style=width: 120px; font-size: 12px /tr tr td align=leftpnbsp; /p/td td align=leftpinput type=submit value=login name=login style=width: 50px; heigth: 18px; font-size: 12px/p/td /tr /table /form This way, depending on how youare refreshing the login form, if the $username variable passed your validation tests it should be set automatically into the form. Cheers Stuart - Original Message - From: [EMAIL PROTECTED] To: PHP_DB [EMAIL PROTECTED] Sent: Wednesday, August 21, 2002 12:58 AM Subject: [PHP-DB] keeping form field data when reloading a form I have my email checker working now, and all the other fields are checked also, but when I do enter some wrong data the form is reloaded with all blank fields. I'm sure there is a way to keep the existing data in the fields so the end-user will only have to change the incorrect fields, and not have to re-enter all the data. But how? -- Chip Wiegand Computer Services Simrad, Inc www.simradusa.com [EMAIL PROTECTED] There is no reason anyone would want a computer in their home. --Ken Olson, president, chairman and founder of Digital Equipment Corporation, 1977 (They why do I have 9? Somebody help me!) -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
[PHP-DB] form validation question - regex not working
I hope this isn't too far off topic -
I have a regex for validating email addresses -
if (empty($useremail) || !eregi(^([A-Za-z0-9\.\_-])+@([A-Za-z0-9\_-])+\.
([A-Za-z]{2,3})+$, $useremail))
Notice the {2,3} which is supposed to limit the last part to 2 or 3
letters, but I have been testing this and
it allows as many letters as I put in there, but not 1 only.
What's wrong?
Also, when using eregi do I need to specify A-Za-z or just a-z, since it is
case-insensitive?
--
Chip Wiegand
Computer Services
Simrad, Inc
www.simradusa.com
[EMAIL PROTECTED]
There is no reason anyone would want a computer in their home.
--Ken Olson, president, chairman and founder of Digital Equipment
Corporation, 1977
(They why do I have 9? Somebody help me!)
--
PHP Database Mailing List (http://www.php.net/)
To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP-DB] form validation question - regex not working
Jason Wong [EMAIL PROTECTED] wrote on 08/21/2002 10:00:23 AM:
On Thursday 22 August 2002 00:29, [EMAIL PROTECTED] wrote:
I hope this isn't too far off topic -
I have a regex for validating email addresses -
if (empty($useremail) || !eregi(^([A-Za-z0-9\.\_-])+@([A-Za-z0-9\_-])
+\.
([A-Za-z]{2,3})+$, $useremail))
Notice the {2,3} which is supposed to limit the last part to 2 or 3
letters, but I have been testing this and
it allows as many letters as I put in there, but not 1 only.
What's wrong?
Also, when using eregi do I need to specify A-Za-z or just a-z, since
it is
case-insensitive?
You're strongly advised not to write your own regex for validating email
addresses. Your regex (once you get it working) will invalidate a lot of
valid email addresses. Search archives, or google for some tried and
tested
regex which will do the job properly.
--
Jason Wong - Gremlins Associates - www.gremlins.com.hk
This is from a php/mysql book. I added the parenthesis simply to group the
sections, and added the {2,3} because a web site tutorial shows that will
limit the preceding section to that many characters (2 or 3 only in this
case).
The script works fine without the {2,3}, and I may have to use it that way,
since another response mentioned foreign addresses, I hadn't taken into
account.
I'd just like to know why it doesn't work, becuase the it's supposed to.
--
Chip
/*
The most important things, each person must do for himself.
*/
Interesting tag considering your respoonse that I shouldn't do this mysqlf,
eh? :-)
--
PHP Database Mailing List (http://www.php.net/)
To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP-DB] form validation question - regex not working
Jason Wong [EMAIL PROTECTED] wrote on 08/21/2002 12:10:02 PM: It's only checking for addresses of the form [EMAIL PROTECTED] As you can see, the list address ([EMAIL PROTECTED]), and my address ([EMAIL PROTECTED]) would be treated as invalid. Not very clever is it? Got it figured out, and foreign address do work now - if (empty($useremail) || !eregi(^([A-Za-z0-9\.\_-])+@([A-Za-z0-9\_-])+\.([A-Za-z])+\.([a-z])*$, $useremail)) -- Chip -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
[PHP-DB] what's wrong with this ereg?
I am trying an email verification function, like this:
if (empty($useremail)) || !eregi(^[A-Za-z0-9\_-]+@[A-Za-z0-9\_-]
+.[A-Za-z0-9\_-]+.*, $email))
{
$errmsg .= The email address appears to be invalid\n;
}
But it will not work, on one server I get this error:
Parse error: parse error in
/home/virtual/site109/fst/var/www/html/auth_dealers/user_input2.php on line
70
and on another server I get this error:
Parse error: parse error, unexpected T_BOOLEAN_OR in
/usr/local/apache/htdocs/auth_dealers/user_input2.php on line 70
What do I need to do to fix this?
--
Chip Wiegand
Computer Services
Simrad, Inc
www.simradusa.com
[EMAIL PROTECTED]
There is no reason anyone would want a computer in their home.
--Ken Olson, president, chairman and founder of Digital Equipment
Corporation, 1977
(They why do I have 9? Somebody help me!)
--
PHP Database Mailing List (http://www.php.net/)
To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP-DB] what's wrong with this ereg?
Steve Cayford [EMAIL PROTECTED] wrote on 08/20/2002 10:28:46 AM:
Well, one thing is you've got mismatched parentheses. You need another
another opening paren right after the if.
Also, don't you need to escape the last hyphen in your character sets?
As well as the . before the third character set?
-Steve
Got it fixed like this:
if (empty($useremail) || !eregi(^[A-Za-z0-9\.\_-]+@[A-Za-z0-9\_-]
+\.[A-Za-z0-9\_-]+$, $useremail))
--
Chip
On Tuesday, August 20, 2002, at 11:11 AM, [EMAIL PROTECTED] wrote:
I am trying an email verification function, like this:
if (empty($useremail)) || !eregi(^[A-Za-z0-9\_-]+@[A-Za-z0-9\_-]
+.[A-Za-z0-9\_-]+.*, $email))
{
$errmsg .= The email address appears to be invalid\n;
}
But it will not work, on one server I get this error:
Parse error: parse error in
/home/virtual/site109/fst/var/www/html/auth_dealers/user_input2.php on
line
70
and on another server I get this error:
Parse error: parse error, unexpected T_BOOLEAN_OR in
/usr/local/apache/htdocs/auth_dealers/user_input2.php on line 70
What do I need to do to fix this?
--
Chip Wiegand
Computer Services
Simrad, Inc
www.simradusa.com
[EMAIL PROTECTED]
There is no reason anyone would want a computer in their home.
--Ken Olson, president, chairman and founder of Digital Equipment
Corporation, 1977
(They why do I have 9? Somebody help me!)
--
PHP Database Mailing List (http://www.php.net/)
To unsubscribe, visit: http://www.php.net/unsub.php
--
PHP Database Mailing List (http://www.php.net/)
To unsubscribe, visit: http://www.php.net/unsub.php
--
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To unsubscribe, visit: http://www.php.net/unsub.php
[PHP-DB] keeping form field data when reloading a form
I have my email checker working now, and all the other fields are checked also, but when I do enter some wrong data the form is reloaded with all blank fields. I'm sure there is a way to keep the existing data in the fields so the end-user will only have to change the incorrect fields, and not have to re-enter all the data. But how? -- Chip Wiegand Computer Services Simrad, Inc www.simradusa.com [EMAIL PROTECTED] There is no reason anyone would want a computer in their home. --Ken Olson, president, chairman and founder of Digital Equipment Corporation, 1977 (They why do I have 9? Somebody help me!) -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP-DB] oddity with insert multiple input fields to multipletables
On Sat, 2002-08-17 at 19:49, John Coder wrote: This may or may not work but what the h***. If you have multiple insets commands test for ! isnull() on each input field ,i.e. variable. if (! isnull($input_field)) then insert_query as I said this will only work with an insert query for each field. I see how that would work with individual insert querys for individual fields, but in my case there are four input fields per query, so it won't work, even then it is possible for one field to be left empty in each query (the comments field). And that is okay. If I try this - if (!isnull($date_am || $exercise_am || $reps_am)) then mysql_query($sql_am) or die (Error in this query $sql : .mysql_error()); I get this error - Parse error: parse error, unexpected T_STRING in /usr/local/apache/htdocs/workout-abs.php on line 111 I have 4 fields, 3 will always be filled, 1 is optional. I thought I would try to test for any of the 3 being empty but it didn't work. -- Chip On Sat, 2002-08-17 at 17:08, Chip Wiegand wrote: I have a web page interface to a mysql database. In this web page I have about a dozen form input fields. On submit these are submitted to multiple tables, a differant table for each input field. If I leave any fields blank, and insert only some of the fields, the database will insert an empty row to all the effected tables that didn't have any data from the input fields. I have another page that generates graphs from the tables, the empty rows show up as breaks in the graphs lines, if I manually delete all the empty rows the graphs work fine. What do I need to do to prevent these empty rows from being written to the tables? -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP-DB] oddity with insert multiple input fields tomultipletables
On Sat, 2002-08-17 at 21:25, Jason Wong wrote:
You have to check them before inserting.
if ($value !== ) {
insert_value();
}
--
Jason Wong - Gremlins Associates - www.gremlins.com.hk
Okay, so I tried this -
if ($date_am || $exercise_am || $reps_am !== )
{
$sql_am = insert into absmachine (today,exercise,reps,comments)
values ('$date_am','$exercise_am','$reps_am','$comments_am');
mysql_query($sql_am) or die (Error in this query $sql :
.mysql_error());
}
and this -
$sql_am = insert into absmachine (today,exercise,reps,comments)
values ('$date_am','$exercise_am','$reps_am','$comments_am');
if ($date_am || $exercise_am || $reps_am !== )
{
mysql_query($sql_am) or die (Error in this query $sql :
.mysql_error());
}
But I still get empty rows inserted into the tables.
--
Chip
-Original Message-
From: Chip Wiegand [mailto:[EMAIL PROTECTED]]
Sent: Sunday, 18 August 2002 7:08 AM
To: phpdb
Subject: [PHP-DB] oddity with insert multiple input fields to multiple
tables
I have a web page interface to a mysql database. In this web page I have
about a dozen form input fields. On submit these are submitted to
multiple tables, a differant table for each input field. If I leave any
fields blank, and insert only some of the fields, the database will
insert an empty row to all the effected tables that didn't have any data
from the input fields.
I have another page that generates graphs from the tables, the empty
rows show up as breaks in the graphs lines, if I manually delete all the
empty rows the graphs work fine.
What do I need to do to prevent these empty rows from being written to
the tables?
Open Source Software Systems Integrators
* Web Design Hosting * Internet Intranet Applications Development *
/*
Between grand theft and a legal fee, there only stands a law degree.
*/
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Re: [PHP-DB] oddity with insert multiple input fields tomultipletables
On Sun, 2002-08-18 at 01:20, Joni Järvinen wrote: if (!isnull($date_am || $exercise_am || $reps_am)) then mysql_query($sql_am) or die (Error in this query $sql : .mysql_error()); Try: if(!isnull($date_am) !isnull($exercise_am) !isnull($reps_am)) This should IMO work :) -Joni- Here's the error that results from using that: Fatal error: Call to undefined function: isnull() in /usr/local/apache/htdocs/workout-abs.php on line 111 -- Chip -- // Joni Järvinen // [EMAIL PROTECTED] // http://www.reactorbox.org/~wandu -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP-DB] oddity with insert multiple input fields tomultipletables
On Sun, 2002-08-18 at 01:20, Joni Järvinen wrote:
if (!isnull($date_am || $exercise_am || $reps_am)) then
mysql_query($sql_am) or die
(Error in this query $sql : .mysql_error());
Try:
if(!isnull($date_am) !isnull($exercise_am) !isnull($reps_am))
This should IMO work :)
-Joni-
I was looking at the php manual and noticed the function is actually
is_null, not isnull. Yet it still does not work, I tried this:
$sql_am = insert into absmachine (today,exercise,reps,comments)
values ('$date_am','$exercise_am','$reps_am','$comments_am');
if(!is_null($date_am))
mysql_query($sql_am) or die (Error in this query $sql :
.mysql_error());
And it still inserts the row no matter what, as long as submit is
pressed a row is written, with or without any data.
--
Chip
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Re: [PHP-DB] oddity with insert multiple input fields - olved
On Sun, 2002-08-18 at 01:20, Joni Järvinen wrote:
if (!isnull($date_am || $exercise_am || $reps_am)) then
mysql_query($sql_am) or die
(Error in this query $sql : .mysql_error());
Try:
if(!isnull($date_am) !isnull($exercise_am) !isnull($reps_am))
This should IMO work :)
-Joni-
Got it working like this -
$sql_am = insert into absmachine (today,exercise,reps,comments)
values ('$date_am','$exercise_am','$reps_am','$comments_am');
if(!empty($date_am) !empty($exercise_am) !empty($reps_am)
mysql_query($sql_am) or die (Error in this query $sql :
.mysql_error());
Thanks to everyone for pointing me in the right direction.
--
Chip
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[PHP-DB] oddity with insert multiple input fields to multiple tables
I have a web page interface to a mysql database. In this web page I have about a dozen form input fields. On submit these are submitted to multiple tables, a differant table for each input field. If I leave any fields blank, and insert only some of the fields, the database will insert an empty row to all the effected tables that didn't have any data from the input fields. I have another page that generates graphs from the tables, the empty rows show up as breaks in the graphs lines, if I manually delete all the empty rows the graphs work fine. What do I need to do to prevent these empty rows from being written to the tables? -- chip w www.wiegand.org [EMAIL PROTECTED] (my web server died and the hard drive crashed hard, and is running from an incomplete backup on another machine, what a drag) :-( -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
[PHP-DB] user authentication problem
I have this script which is giving me a couple of problems -
1 - I cannot use include ../connect.inc, I get headers already sent error
2 - I can use this script once, but if I try to reload the page to continue
testing I get error Unable to execute query. Clearing the browsers cache
does not help.
Any ideas what's going on here?
?
$auth = false; // Assume user is not authenticated
if (isset( $PHP_AUTH_USER ) isset($PHP_AUTH_PW)) {
//Connect to MySQL
mysql_connect( 'localhost', 'autopilots', 'Xfm3dwj' )
or die ( 'Unable to connect to server.' );
//Select database on MySQL server
mysql_select_db( 'www2_simradusa_com' )
or die ( 'Unable to select database.' );
// Formulate the query
$sql = SELECT * FROM user WHERE
userid = '$PHP_AUTH_USER' AND
userpassword = '$PHP_AUTH_PW';
// Execute the query and put results in $result
$result = mysql_query( $sql );
if ( $result != false )
{ echo mysql_error(); }
// Get number of rows in $result.
$num = mysql_numrows( $result );
if ( $num != 0 ) {
// A matching row was found - the user is authenticated.
$auth = true;
}
}
if ( ! $auth ) {
header( 'WWW-Authenticate: Basic realm=Private' );
header( 'HTTP/1.0 401 Unauthorized' );
echo 'Authorization Required.';
exit;
} else {
header( 'Location:
http://www.simradusa.com/auth_dealers/dealers_page.php' );
}
?
--
Chip Wiegand
Computer Services
Simrad, Inc
www.simradusa.com
[EMAIL PROTECTED]
There is no reason anyone would want a computer in their home.
--Ken Olson, president, chairman and founder of Digital Equipment
Corporation, 1977
(They why do I have 9? Somebody help me!)
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[PHP-DB] odbc connect problem
I am attempting for the first time, to connect from freebsd/mysql/php box
to a NT box running MAS200. I am able
to connect from another NT workstation with Access and download data, so I
am trying from FreeBSD now.
When I run the code below I get error:
Fatal error: Call to undefined function: odbc_connect() in
/usr/local/apache/htdocs/odbctest.php on line 7
What am I doing wrong? There's not much info on this topic in the mysql
manual.
html
head
titleUntitled/title
/head
body bgcolor=white
?
$connect = odbc_connect(localhost, , -);
if (!$connect)
{
Error_handler( Error in odbc_connect , $connect );
}
$result = odbc_exec($connect, SELECT ProductLineDescription FROM
IMA_ProductLine, ItemNumber,
ItemDescription FROM IM1_InventoryMasterFile where DefaultWhse = 004);
if (!$result) {
Error_handler( Error in odbc_exec( no cursor returned ) ,
$connect );
}
echo table border=1trthBrand/ththPart
Number/ththDescription/th/tr\n;
// fetch the succesive result rows
while( odbc_fetch_row( $result ) ) {
$brand= odbc_result( $result, 1 ); // get the field
ProductLineDescription
$number= odbc_result( $result, 2 ); // get the field ItemNumber
$description= odbc_result( $result, 3 ); // get the field
ItemDescription
echo trtd$brand/tdtd$number/tdtd$description/td/tr
\n;
}
echo /table;
echo /body/html;
?
--
Chip Wiegand
Computer Services
Simrad, Inc
www.simradusa.com
[EMAIL PROTECTED]
There is no reason anyone would want a computer in their home.
--Ken Olson, president, chairman and founder of Digital Equipment
Corporation, 1977
(They why do I have 9? Somebody help me!)
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Re: [PHP-DB] odbc connect problem
Yeah, you're right, I see I need to add --with-iodbc. Now another question
- when I recompile to add
this support, do I have to add all the other options that I originally had
listed? Or can I add just this one
and everything else will be left the same? What about php.ini? Will it be
changed?
--
Chip W
Chip Atkinson [EMAIL PROTECTED] wrote on 08/09/2002 02:51:10 PM:
It looks like you didn't build php with odbc support.
Try putting in a call to
phpinfo ();
and see if it says that odbc support is compiled in.
(Another) Chip
On Fri, 9 Aug 2002 [EMAIL PROTECTED] wrote:
I am attempting for the first time, to connect from freebsd/mysql/php
box
to a NT box running MAS200. I am able
to connect from another NT workstation with Access and download data,
so I
am trying from FreeBSD now.
When I run the code below I get error:
Fatal error: Call to undefined function: odbc_connect() in
/usr/local/apache/htdocs/odbctest.php on line 7
What am I doing wrong? There's not much info on this topic in the mysql
manual.
html
head
titleUntitled/title
/head
body bgcolor=white
?
$connect = odbc_connect(localhost, , -);
if (!$connect)
{
Error_handler( Error in odbc_connect , $connect );
}
$result = odbc_exec($connect, SELECT ProductLineDescription FROM
IMA_ProductLine, ItemNumber,
ItemDescription FROM IM1_InventoryMasterFile where DefaultWhse = 004);
if (!$result) {
Error_handler( Error in odbc_exec( no cursor returned ) ,
$connect );
}
echo table border=1trthBrand/ththPart
Number/ththDescription/th/tr\n;
// fetch the succesive result rows
while( odbc_fetch_row( $result ) ) {
$brand= odbc_result( $result, 1 ); // get the field
ProductLineDescription
$number= odbc_result( $result, 2 ); // get the field
ItemNumber
$description= odbc_result( $result, 3 ); // get the field
ItemDescription
echo
trtd$brand/tdtd$number/tdtd$description/td/tr
\n;
}
echo /table;
echo /body/html;
?
--
Chip Wiegand
Computer Services
Simrad, Inc
www.simradusa.com
[EMAIL PROTECTED]
There is no reason anyone would want a computer in their home.
--Ken Olson, president, chairman and founder of Digital Equipment
Corporation, 1977
(They why do I have 9? Somebody help me!)
--
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RE: [PHP-DB] problem getting form input field to become part ofquery statement
On Sat, 2002-08-03 at 06:05, Rich Hutchins wrote:
Try referencing the $listbox variable in you SQL statement like this:
$sql = insert into .$listbox.
values(NULL,'$date','$exercise','$reps','$comments');
I'm guessing that it might also work like this:
$sql = insert into '$listbox'
values(NULL,'$date','$exercise','$reps','$comments');
I think your core problem is that the $listbox variable is not being
evaluated properly in the SQL statement. Once you solve that, you're good to
go.
Hope this helps.
Rich
I tried both suggestions and neither are working. I am using the get
format for the form so I can see what is being sent, and I am getting
this:
http://192.168.1.53/workout-absflexor.php?exerciselist=%24listboxexercise=80
reps=12comments=submit=Send+Data
Notice that the $listbox variable is still not being sent. Just above
this is an echo statement which shows what's in that variable, and it
displays the expected result (see code below). I am at a loss as to why
this is not working.
If I replace the $listbox variable with the table name shown in the echo
statement, a connection is made and the query is completed.
--
Chip W
www.wiegand.org
[EMAIL PROTECTED]
-Original Message-
From: Chip Wiegand [mailto:[EMAIL PROTECTED]]
Sent: Saturday, August 03, 2002 12:51 AM
To: phpdb
Subject: [PHP-DB] problem getting form input field to become part of
query statement
I have a form with a select list and a hidden field to save the selected
item. On submit another page is loaded with a few fields to be filled in
and submitted to a table. Using get I see the data is being passed from
the first page to the second properly, and the second page sends its
data properly. The hidden field from the first page is to be used by the
second page as the name of the table in the query. I have a place where
I echo the contents of the hidden field just to be sure it is correct,
and that does indeed show what I expect. I then make the query statement
point to the variable but it always responds that it cannot find the
table. The table does exist, the variable does contain the appropriate
table name, but is not being replaced by the name. What am I doing
wrong?
--
Chip W
www.wiegand.org
[EMAIL PROTECTED]
Below is the code for the first page --
html
head
title/title
/head
body
div align=center
?
$exercises=array(absflexor,absmachine,leglifts);
echo form action='workout-absflexor.php' method='get';
echo table width='70%' border='0' align='center';
echo trth align='center'h2Exercise Data Input/h2/th/tr;
echo trth align='center'select name='listbox';
echo option$exercises[0]/option;
echo option$exercises[1]/option;
echo option$exercises[2]/option;
echo /selectbrbr;
echo /th/tr/table;
echo input type='hidden' name='exerciselist' value='$listbox';
echo input type='submit';
echo /form;
?
/div
/body
/html
And below is the code for the second page --
!DOCTYPE html PUBLIC -//W3C//DTD XHTML 1.0 Transitional//EN
http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd;
html xmlns=http://www.w3.org/1999/xhtml;
head
titleWorkout Data Entry Form/title
style type=text/css
body { background-color: aqua; }
div.c1 {text-align: center}
/style
/head
body
div class=c1
h2Work-Out Data Entry Screen/h2
form action=? PHP_SELF ? method=get
table summary= width=60% border=1 align=center
bgcolor=green
tr
thWeight/th
td align=leftinput type=text name=exercise
maxlength=4/td
/tr
tr
thReps/th
td align=leftinput type=text name=reps
maxlength=4/td
/tr
tr
thComments/th
td colspan=2textarea cols=50 rows=3
name=comments
/textarea/td
/tr
trtd? echo $listbox; ?/td/tr
/table
br /
input type=submit name=submit value=Send Data / input
type=reset /
/form
/div
?
if(isset($submit)):
$db = mysql_connect(localhost,root,carvin);
if(!$db) error_message(sql_error());
mysql_select_db(workout,$db) or die (Ack! Where's the database?);
$date = date(m-d);
$sql = insert into $listbox
values(NULL,'$date','$exercise','$reps','$comments');
mysql_query($sql) or die (Ack! No response when I queried the
server!);
endif;
?
/body
/html
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[PHP-DB] problem getting form input field to become part of query statement
I have a form with a select list and a hidden field to save the selected
item. On submit another page is loaded with a few fields to be filled in
and submitted to a table. Using get I see the data is being passed from
the first page to the second properly, and the second page sends its
data properly. The hidden field from the first page is to be used by the
second page as the name of the table in the query. I have a place where
I echo the contents of the hidden field just to be sure it is correct,
and that does indeed show what I expect. I then make the query statement
point to the variable but it always responds that it cannot find the
table. The table does exist, the variable does contain the appropriate
table name, but is not being replaced by the name. What am I doing
wrong?
--
Chip W
www.wiegand.org
[EMAIL PROTECTED]
Below is the code for the first page --
html
head
title/title
/head
body
div align=center
?
$exercises=array(absflexor,absmachine,leglifts);
echo form action='workout-absflexor.php' method='get';
echo table width='70%' border='0' align='center';
echo trth align='center'h2Exercise Data Input/h2/th/tr;
echo trth align='center'select name='listbox';
echo option$exercises[0]/option;
echo option$exercises[1]/option;
echo option$exercises[2]/option;
echo /selectbrbr;
echo /th/tr/table;
echo input type='hidden' name='exerciselist' value='$listbox';
echo input type='submit';
echo /form;
?
/div
/body
/html
And below is the code for the second page --
!DOCTYPE html PUBLIC -//W3C//DTD XHTML 1.0 Transitional//EN
http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd;
html xmlns=http://www.w3.org/1999/xhtml;
head
titleWorkout Data Entry Form/title
style type=text/css
body { background-color: aqua; }
div.c1 {text-align: center}
/style
/head
body
div class=c1
h2Work-Out Data Entry Screen/h2
form action=? PHP_SELF ? method=get
table summary= width=60% border=1 align=center
bgcolor=green
tr
thWeight/th
td align=leftinput type=text name=exercise
maxlength=4/td
/tr
tr
thReps/th
td align=leftinput type=text name=reps
maxlength=4/td
/tr
tr
thComments/th
td colspan=2textarea cols=50 rows=3
name=comments
/textarea/td
/tr
trtd? echo $listbox; ?/td/tr
/table
br /
input type=submit name=submit value=Send Data / input
type=reset /
/form
/div
?
if(isset($submit)):
$db = mysql_connect(localhost,root,carvin);
if(!$db) error_message(sql_error());
mysql_select_db(workout,$db) or die (Ack! Where's the database?);
$date = date(m-d);
$sql = insert into $listbox
values(NULL,'$date','$exercise','$reps','$comments');
mysql_query($sql) or die (Ack! No response when I queried the
server!);
endif;
?
/body
/html
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Re: [PHP-DB] extract data from database into an array
Adam Alkins [EMAIL PROTECTED] wrote on 07/13/2002 07:56:46 PM: I have a database, all the data is numbers. I want to make a query that will extract the data and then make it available in an array, so the array is populated 'real-time'. I could just enter the number into an array manually, but I want to automate the job. I don't know what to start looking for, the 3 books I have either don't talk about this or I just am missing it. Could someone point me in the right direction? Well after you SELECT the data and run the query, you can use a few functions, it depends on your database type? For example, for MySQL, you can use mysql_fetch_array() which will take your data and store it in an array, both numeric and associative. i.e. $data = mysql_fetch_array($result) can be accessed like $data['column_name'] $data[column_number] You can also fetch an array only as an associative (mysql_fetch_assoc) or numeric (mysql_fetch_row). And simply if selecting multiple rows, use a loop while($data = mysql_fetch_array())... -- Adam Alkins I guess I wasn't very clear on exactly what results I expect. I can get the data and print all the numbers, that's no problem. What I want is to have all the numbers inserted into an array that would look like this: $datax = (50,50,50,60,60,60,70,70,70,65,65,70,75,75,80); Each number in the array is from a single row in the table (15 rows in this example). The database is weights, entered after a work-out, along with their reps. I only need the weights, as shown above. This array $datax is then processed by a php script to create a graph showing the progress of the person's wieght lifting over time. I can enter all the numbers into the array manually and the graph is drawn correctly, but prefer the web interface I made. This is okay for me, but not others. So, I made a web form that takes the data and inserts into the database. That works fine also. The table contains columns: squats, squats_reps, legpress, legpress_reps, etc etc. The first is the weight and the second is the number of reps. So after each workout the appropriate numbers would be entered via the form page, the graph would be generated with the updated numbers from another page. Hope this helps to explain it a bit better, -- Chip http://www.rasadam.com -- -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
[PHP-DB] A problem with insert to mysql from a web interface
I just set up a pc with mysql-3.23.51/php-4.2.1/apache-1.3.26. I have
created a database and tables and a web interface to insert data into
it. I can insert data from the command line just fine. From the web
interface there is nothing inserted and no error messages. I have tried
connection string with root and with user names, both of which can
insert data from a remote client as well as on the sql server itself.
When I enter data and press submit the page redraws, and I don't get the
message the data was entered, and checking the database, I see nothing
indeed was entered. I've used post and get, makes no differance. In the
database I have set the fields as int, small int, varchar (shows char in
describe table_name), makes no differance.
I have made other web interfaces to mysql databases before, on slightly
older versions of mysql/php/apache and have never had this problem.
If anyone has a few minute to review my code below, hopefully someone
will find whatever it is I am missing. I have pasted in the html for the
web page, and below that a .csv copy/pasted of the database.
--
Thanks
Chip W
www.wiegand.org
[EMAIL PROTECTED]
Below is the web page -
html
head/head
body bgcolor=aqua
div align=center
?
if(isset($submit)):
$db = mysql_connect(localhost, root, **) or die
(Could not connect to the server);
print brconnected to the server successfully;
mysql_select_db(workout, $db) or die (brCould not get the
databasebr
.mysql_error());
$sql = INSERT INTO legs
(squats,squats_reps,legpress,legpress_reps,legext,legext_reps,legcurls,
legcurls_reps,calfraise,calfraise_reps,hipabducter,hipabducter_reps,
hipadducter,hipadducter_reps,comments) VALUES('$squats','$squats_reps',
'$legpress','$legpress_reps','$legext','$legext_reps','$legcurls',
'$legcurls_reps','$calfraise','$calfraise_reps','$hipabducter',
'$hipabducter_reps','$hipadducter','$hipadducter_reps','$comments');
mysql_query($sql, $db) or die (Query failed .mysql_error());
print brThe data has been entered;
endif;
?
h2Work-Out Data Entry Screen/h2
h3Legs Work-out/h3
form action=? $PHP_SELF ? method=get
table summary= width=70% border=1 align=center bgcolor=green
tr
thExcercise/ththWeight/ththReps/th
/tr
tr
thSquats/th
td align=center
input type=text name=squats size=3 maxlength=3/td
td align=center
input type=text name=squats_reps size=3 maxlength=3/td
/tr
tr
thLeg Press/th
td align=center
input type=text name=legpress size=3 maxlength=3/td
td align=center
input type=text name=legpress_reps size=3 maxlength=3/td
/tr
tr
thLeg Extension/th
td align=center
input type=text name=legext size=3 maxlength=3/td
td align=center
input type=text name=legext_reps size=3 maxlength=3/td
/tr
tr
thLeg Curls/th
td align=center
input type=text name=legcurls size=3 maxlength=3/td
td align=center
input type=text name=legcurls_reps size=3 maxlength=3/td
/tr
tr
thCalf Raise/th
td align=center
input type=text name=calfraise size=3 maxlength=3/td
td align=center
input type=text name=calfraise_reps size=3 maxlength=3/td
/tr
tr
thHip Abducter/th
td align=center
input type=text name=hipabducter size=3 maxlength=3/td
td align=center
input type=text name=hipabducter_reps size=3 maxlength=3/td
/tr
tr
thHip Adducter/th
td align=center
input type=text name=hipadducter size=3 maxlength=3/td
td align=center
input type=text name=hipadducter_reps size=3 maxlength=3/td
/tr
tr
thComments/th
td colspan=2 align=center
textarea cols=50 rows=3 name=comments wrap=virtual
/textarea/td
/table
br /
input type=submit name=submit value=Send Data
/form
br /
/div
/body
/html
And here is the database (a mysqlnavigator .csv dump) -
id,int(3),,PRI,,auto_increment
squats,int(3),YES,,,
squats2,char(3)
squats_reps,char(3),YES,,,
legpress,char(3),YES,,,
legpress_reps,char(3),YES,,,
legext,char(3),YES,,,
legext_reps,char(3),YES,,,
legcurls,char(3),YES,,,
legcurls_reps,char(3),YES,,,
calfraise,char(3),YES,,,
calfraise_reps,char(3),YES,,,
hipabducter,char(3),YES,,,
hipabducter_reps,char(3),YES,,,
hipadducter,char(3),YES,,,
hipadducter_reps,char(3),YES,,,
comments,text,YES,,,
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Re: [PHP-DB] A problem with insert to mysql from a web interface
On Sat, 2002-07-13 at 00:52, Jason Wong wrote:
On Saturday 13 July 2002 14:49, Chip Wiegand wrote:
I just set up a pc with mysql-3.23.51/php-4.2.1/apache-1.3.26. I have
created a database and tables and a web interface to insert data into
it. I can insert data from the command line just fine. From the web
interface there is nothing inserted and no error messages. I have tried
connection string with root and with user names, both of which can
insert data from a remote client as well as on the sql server itself.
When I enter data and press submit the page redraws, and I don't get the
message the data was entered, and checking the database, I see nothing
indeed was entered. I've used post and get, makes no differance. In the
database I have set the fields as int, small int, varchar (shows char in
describe table_name), makes no differance.
I have made other web interfaces to mysql databases before, on slightly
older versions of mysql/php/apache and have never had this problem.
If anyone has a few minute to review my code below, hopefully someone
will find whatever it is I am missing. I have pasted in the html for the
web page, and below that a .csv copy/pasted of the database.
register_globals ?
That was it, Thanks, works fine now.
$sql = INSERT INTO legs
(squats,squats_reps,legpress,legpress_reps,legext,legext_reps,legcurls,
legcurls_reps,calfraise,calfraise_reps,hipabducter,hipabducter_reps,
hipadducter,hipadducter_reps,comments) VALUES('$squats','$squats_reps',
'$legpress','$legpress_reps','$legext','$legext_reps','$legcurls',
'$legcurls_reps','$calfraise','$calfraise_reps','$hipabducter',
'$hipabducter_reps','$hipadducter','$hipadducter_reps','$comments');
echo $sql; # ???
What does that do? And would I put it somewhere in my web page code?
I've not seen this before.
Thanks for the help.
--
Chip
--
Jason Wong - Gremlins Associates - www.gremlins.com.hk
Open Source Software Systems Integrators
* Web Design Hosting * Internet Intranet Applications Development *
/*
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cheating on your income tax.
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[PHP-DB] extract data from database into an array
I have a database, all the data is numbers. I want to make a query that will extract the data and then make it available in an array, so the array is populated 'real-time'. I could just enter the number into an array manually, but I want to automate the job. I don't know what to start looking for, the 3 books I have either don't talk about this or I just am missing it. Could someone point me in the right direction? -- Chip -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
[PHP-DB] Graphing data
Is there any application available that can take data from my mysql database and draw graphs? I am using mysql/php on FreeBSD. -- Chip -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
[PHP-DB] need a solution for web page navigation
I have a database of the menu items used on my web site, there are more than 100 entries. How do I set up navigation arrows to point to the next appropriate page, and the previous page, from the database? I don't want to use javascript history function, navigation is too important to allow it to be turned off in someones browser. Example: Category 1 with it's own item1, item2, item3 Category 2 with it's own item1, item2, item3 Category 3 with it's own item1, item2, item3 all have next, previous, home links at the top of their pages, so category 1 links to item 1, then item 1 links to item 2 etc. Category 2 has item 1 linked to it's own item 2 and so on, and category 3 has it's own item 1 linked to it's own item 2 etc. Now if I want to make item 3 in category 3 available on the category 1 menu in addition to category 3, that's easy, but the links will point to category 3 pages, not category 1 pages. Hmm, I'm sure this is clear as mud. I know how to move forward and backward within the database, like in an image database, I've done that before. But this is a little different, I don't know if it is even possible. I am using php/mysql. If anyone has any idea what I am trying to do please attempt to help... -- Chip Wiegand Computer Services Simrad, Inc www.simradusa.com [EMAIL PROTECTED] There is no reason anyone would want a computer in their home. --Ken Olson, president, chairman and founder of Digital Equipment Corporation, 1977 (They why do I have 9? Somebody help me!) -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
[PHP-DB] sql query problem
I have a database layout similar to this- id order title namepagecat 0 0 title1 blueap 1 2 blue1 page1 ap 2 3 blue2 page2 ap 3 1 blue3 page3 ap I would like to get title1 (title column only) and blue2 (name column only) like this- title1 blue2 I am finding it difficult to write the proper select statement to get just those items in a web page using php/mysql. Suggestions? -- Chip Wiegand Computer Services Simrad, Inc www.simradusa.com [EMAIL PROTECTED] There is no reason anyone would want a computer in their home. --Ken Olson, president, chairman and founder of Digital Equipment Corporation, 1977 (They why do I have 9? Somebody help me!) -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
[PHP-DB] looking for a particular script, or how-to...
I'm looking for a script that will give me this - a web page that shows who is currently accessing my mysql database. If it could provide other details that would be great too. I'm not sure how to go about setting this up. I couldn't find any mysql command that shows the users accessing the database. Any suggestions? -- Chip Wiegand Computer Services Simrad, Inc www.simrad.com [EMAIL PROTECTED] There is no reason anyone would want a computer in their home. --Ken Olson, president, chairman and founder of Digital Equipment Corporation, 1977 (They why do I have 7? Somebody help me!)
[PHP-DB] Why the following error, yet it works anyway
I have the following short web page to delete dealer bulletins from a
database. My page lists
all the bulletins in the database - id and subject. There is a text input
field to enter the bulletin
id number and hit the delete button, and the bulletin is deleted. It works
except after the submit
delete I get an error -
-
Warning: Supplied argument is not a valid MySQL result resource in
/usr/local/apache/htdocs/bulletin_delete.php on line 35
-
line 35 is the while statement.
Here is the complete page -
-
html
head
meta name=generator content=HTML Tidy, see www.w3.org /
titleDealer Bulletins, Simrad, Inc/title
/head
body
centerh2Simrad Dealer Bulletins - Delete Screen/h2
br /strongfont color=redNOTE: This is a permanent and
irreversible
delete! No second chances here!/font/strong/center
hr width=75% noshade=noshade /
form action=bulletin_delete.php method =POST
Enter the bulleting ID Number: input type=text name=newsid
input type=submit name=submit value=Delete/form
table summary= border=0 cellpadding=5 align=center width=90%
?
$db = mysql_connect(localhost, root)
or die (Can't get the database server);
mysql_select_db(bulletins, $db) or die (Can't get the database);
if (isset($submit)):
$sql = delete from dbulletins where news_id = '$newsid';
else:
$sql = select news_id, bulletin_subject from dbulletins;
endif;
$result = mysql_query($sql);
while ($row = mysql_fetch_array($result))
{
print tr\ntdDelete entry
strong.$row[news_id]./strong?nbsp;
.$row[bulletin_subject]./td\n/tr\n;
}
?
/table
center
a href=bulletin_admin.phpBack/a/center
/body
/html
--
Chip Wiegand
Computer Services
Simrad, Inc
www.simrad.com
[EMAIL PROTECTED]
There is no reason anyone would want a computer in their home.
--Ken Olson, president, chairman and founder of Digital Equipment
Corporation, 1977
(They why do I have 7? Somebody help me!)
RE: [PHP-DB] Why the following error, yet it works anyway
Rick Emery [EMAIL PROTECTED] wrote on 01/14/2002 10:35:37 AM:
First, if $submit is not set, then the delete statement will be
executed. This will not return an array upon which mysql_fetch_array()
will
act. Therefore $result will not be valid.
I made this change:
if (!isset($submit)):
$sql = select news_id, bulletin_subject from dbulletins;
elseif (isset($submit)):
$sql = delete from dbulletins where news_id = '$newsid';
endif;
but still get the same error. Shouldn't this fix the problem so now it
will
return the database list because the submit button is not pushed (on
returning
to the page)?
--
Chip
-Original Message-
From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED]]
Sent: Monday, January 14, 2002 11:18 AM
To: [EMAIL PROTECTED]
Subject: [PHP-DB] Why the following error, yet it works anyway
I have the following short web page to delete dealer bulletins from a
database. My page lists
all the bulletins in the database - id and subject. There is a text
input
field to enter the bulletin
id number and hit the delete button, and the bulletin is deleted. It
works
except after the submit
delete I get an error -
-
Warning: Supplied argument is not a valid MySQL result resource in
/usr/local/apache/htdocs/bulletin_delete.php on line 35
-
line 35 is the while statement.
Here is the complete page -
-
html
head
meta name=generator content=HTML Tidy, see www.w3.org /
titleDealer Bulletins, Simrad, Inc/title
/head
body
centerh2Simrad Dealer Bulletins - Delete Screen/h2
br /strongfont color=redNOTE: This is a permanent and
irreversible
delete! No second chances here!/font/strong/center
hr width=75% noshade=noshade /
form action=bulletin_delete.php method =POST
Enter the bulleting ID Number: input type=text name=newsid
input type=submit name=submit value=Delete/form
table summary= border=0 cellpadding=5 align=center
width=90%
?
$db = mysql_connect(localhost, root)
or die (Can't get the database server);
mysql_select_db(bulletins, $db) or die (Can't get the database);
if (isset($submit)):
$sql = delete from dbulletins where news_id = '$newsid';
else:
$sql = select news_id, bulletin_subject from dbulletins;
endif;
$result = mysql_query($sql);
while ($row = mysql_fetch_array($result))
{
print tr\ntdDelete entry
strong.$row[news_id]./strong?nbsp;
.$row[bulletin_subject]./td\n/tr\n;
}
?
/table
center
a href=bulletin_admin.phpBack/a/center
/body
/html
--
Chip Wiegand
Computer Services
Simrad, Inc
www.simrad.com
[EMAIL PROTECTED]
There is no reason anyone would want a computer in their home.
--Ken Olson, president, chairman and founder of Digital Equipment
Corporation, 1977
(They why do I have 7? Somebody help me!)
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For additional commands, e-mail: [EMAIL PROTECTED]
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