Re: [PHP-DB] Help creating tables and fields
- Original Message -
From: Karl James
To: 'bedul'
Sent: Thursday, March 29, 2007 10:40 AM
Subject: RE: [PHP-DB] Help creating tables and fields
bedul,
wow great information and so quick.
very greatfull for that.
Would you mine baby stepping for me?
[bedul]
i don't understand what u said.. but i read as something nice
my real name was gunawan (pronounce as Gun - A - One )
=
I am using dreamweaver 8 to build my pages.
[bedul]
i'm using not 1!! but 3!! text editor, html editor (dreamweaver) and text
based html editor..
fyi.. i build scratch in excel. that editor was crimson editor (this free) ,
Ace Html (trial).. since u already use DW.. don't try.. you might try to
download it..
plz remember.. u not superman.. we only able to create page using php. in
this era.. u probably able to use html, db, javascript and template
building..
owww man.. fyi.. i'm not that strong created by myself. even i already
create using both of i said earlier.
see attach
u might hard to read the languages i use. but i mean in the attach was..
this is my scratch design
===
But I can do in code as well.
I just created a connection to my database.
So, now I am just trying to create the username and password field
on my dwt template page? Would that be correct or what should be my first
step?
[bedul]
that's better.. but u supposed to create map link for the aplication
there were 3 who use this page.
let view the guest first..
1. guest enter the web (www.fotball.com)
2. in there what he/she able to see (for guest only) = what the scratch of
guest??
3. are there anything else that guest able to see??
then let view the user
1. user enter the web
2. user view opening page
3. login
4. if login failed
5. what he/she view after login
6. other page that he/she able to see
admin
i think.. for this.. self explain.
===
Should we just create the tables first through phpmyadmin page.
That is where it is easier to see everything, the tables that is.
[bedul]
the reason why you must to build the scratch (before goes online and able to
program).. is to figure what table we can use??
after u able to build the scratch page (html).. then u goes to build the
table..
===
Please continue with your thoughts.
Karl James (TheSaint)
[EMAIL PROTECTED]
[EMAIL PROTECTED]
www.theufl.com
[bedul]
last suggestion..
u can improve the skill if you
1. created as many page
2. complain by your user
3. lazy?? i too lazy type textbox and then i create a function that return
textbox
function inpText($name, $dValue, $size, $max){
$temp ='INPUT TYPE=text name='.$name.'';
$temp .='value='.$dValue.' size=';
$temp .=$size. maxlength=$max;
return $temp;
}
like above.. heheheehee
4. always ask to everyone (include me) about something u can't do like how
to make my page become ajax??
resepOnLine.xls
Description: MS-Excel spreadsheet
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[PHP-DB] echo
thanks to Chris and Dimiter,
I think I am close but still the problem is not solved, when I add
echo img src='/album/img/.$row[photoFileName].' /
to the code as it was sugested by Dimiter there is parse Error :
PHP Parse error: syntax error, unexpected $end in
C:\Inetpub\wwwroot\album\show.php on line 44
line 44 is end of the code just after /html
what does it mean?
I am storing just the name of photos in the database and the photos are
in /img folder ,and there is no permissions issue. My testing server
is IIS And the path would be something like this : Inetpub\wwwroot\album\img
as I am running out of time! could someone complete this code just with
one echo and img src so that I can retrive my photos ?
MySQL columns : photoID=seq number
photoFileName=name of my photo like 3sw.jpg
title=title
description=short description
-
body
?php
$link = mysql_connect('localhost', 'root', 'pw');
if (!$link) {
die('Not connected : ' . mysql_error());
}
echo 'connected!';
$db_selected = mysql_select_db('album', $link);
if (!$db_selected) {
die ('Can\'t use foo : ' . mysql_error());
}
$query = SELECT * FROM photo;
$result=mysql_query($query);
while ($row = mysql_fetch_array($result))
{
echo img src='/album/img/.$row[photoFileName].' /
}
mysql_free_result($result);
?
/body
/html
-
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Re: [PHP-DB] echo
elk dolk wrote:
thanks to Chris and Dimiter,
I think I am close but still the problem is not solved, when I add
echo img src='/album/img/.$row[photoFileName].' /
to the code as it was sugested by Dimiter there is parse Error :
PHP Parse error: syntax error, unexpected $end in
C:\Inetpub\wwwroot\album\show.php on line 44
line 44 is end of the code just after /html
what does it mean?
probably you don't have a semi colon at the end of your echo line
nor is there and end double quote.
I am storing just the name of photos in the database and the photos are
in /img folder ,and there is no permissions issue. My testing server
is IIS And the path would be something like this : Inetpub\wwwroot\album\img
as I am running out of time! could someone complete this code just with
one echo and img src so that I can retrive my photos ?
MySQL columns : photoID=seq number
photoFileName=name of my photo like 3sw.jpg
title=title
description=short description
-
body
?php
$link = mysql_connect('localhost', 'root', 'pw');
if (!$link) {
die('Not connected : ' . mysql_error());
}
echo 'connected!';
$db_selected = mysql_select_db('album', $link);
if (!$db_selected) {
die ('Can\'t use foo : ' . mysql_error());
}
$query = SELECT * FROM photo;
$result=mysql_query($query);
while ($row = mysql_fetch_array($result))
{
echo img src='/album/img/.$row[photoFileName].' /
}
mysql_free_result($result);
?
/body
/html
-
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[PHP-DB] DBX
Hello, Is there a DBX module for php5 on debian... I have just installed zCI on my debian box and I keep getting this error Fatal error: Call to undefined function dbx_query() in /var/www/zci/frontend/registered_device.php on line 79 I have been unable to locate the module... Please help. -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
[PHP-DB] echo
thank you all, it did it! [EMAIL PROTECTED] You need a trailing you have echo img src='/album/img/.$row[photoFileName].' / it needs to be echo img src='/album/img/.$row[photoFileName].' /; - Bored stiff? Loosen up... Download and play hundreds of games for free on Yahoo! Games.
RE: [PHP-DB] echo
try
echo img src='/album/img/{$row[photoFileName]}' / ;
warpping the array element in braces allows for proper evaluation
bastien
From: elk dolk [EMAIL PROTECTED]
To: php-db@lists.php.net
Subject: [PHP-DB] echo
Date: Thu, 29 Mar 2007 05:08:36 -0700 (PDT)
thanks to Chris and Dimiter,
I think I am close but still the problem is not solved, when I add
echo img src='/album/img/.$row[photoFileName].' /
to the code as it was sugested by Dimiter there is parse Error :
PHP Parse error: syntax error, unexpected $end in
C:\Inetpub\wwwroot\album\show.php on line 44
line 44 is end of the code just after /html
what does it mean?
I am storing just the name of photos in the database and the photos are
in /img folder ,and there is no permissions issue. My testing server
is IIS And the path would be something like this :
Inetpub\wwwroot\album\img
as I am running out of time! could someone complete this code just with
one echo and img src so that I can retrive my photos ?
MySQL columns : photoID=seq number
photoFileName=name of my photo like 3sw.jpg
title=title
description=short description
-
body
?php
$link = mysql_connect('localhost', 'root', 'pw');
if (!$link) {
die('Not connected : ' . mysql_error());
}
echo 'connected!';
$db_selected = mysql_select_db('album', $link);
if (!$db_selected) {
die ('Can\'t use foo : ' . mysql_error());
}
$query = SELECT * FROM photo;
$result=mysql_query($query);
while ($row = mysql_fetch_array($result))
{
echo img src='/album/img/.$row[photoFileName].' /
}
mysql_free_result($result);
?
/body
/html
-
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Browse Top Cars by Green Rating at Yahoo! Autos' Green Center.
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RE: [PHP-DB] DBX
Hve you tried the ODBC connections instead? http://ca3.php.net/manual/en/ref.dbx.php bastien From: Phares Kariuki [EMAIL PROTECTED] To: php-db@lists.php.net Subject: [PHP-DB] DBX Date: Thu, 29 Mar 2007 15:25:12 +0300 Hello, Is there a DBX module for php5 on debian... I have just installed zCI on my debian box and I keep getting this error Fatal error: Call to undefined function dbx_query() in /var/www/zci/frontend/registered_device.php on line 79 I have been unable to locate the module... Please help. -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php _ http://local.live.com/?mkt=en-ca/?v=2cid=A6D6BDB4586E357F!399 -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
[PHP-DB] array
Hi all,
I want to put result of query in a two dimensional array like this:
title[0]=x description [0]=y
title[1]=z description [1]=w
.
any idea would be appreciated
$query = SELECT * FROM photo;
$result=mysql_query($query);
while ($row = mysql_fetch_array($result))
{
$title=$row['title'];
$description=$row['description'];
}
-
Don't pick lemons.
See all the new 2007 cars at Yahoo! Autos.
Re: [PHP-DB] DBX
Phares Kariuki wrote: Hello, Is there a DBX module for php5 on debian... I have just installed zCI on my debian box and I keep getting this error Fatal error: Call to undefined function dbx_query() in /var/www/zci/frontend/registered_device.php on line 79 I have been unable to locate the module... Please help. apt-cache search dbx didn't return anything useful, so I guess you'll have to compile from source. -- Postgresql php tutorials http://www.designmagick.com/ -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
RE: [PHP-DB] array
That is not a two dimensional array. Its two one dimensional arrays and yo
do it like this
while ($row = mysql_fetch_array($result))
{
$title[] =$row['title'];
$description[] =$row['description'];
}
bastien
From: elk dolk [EMAIL PROTECTED]
To: php-db@lists.php.net php-db@lists.php.net
Subject: [PHP-DB] array
Date: Thu, 29 Mar 2007 13:29:45 -0700 (PDT)
Hi all,
I want to put result of query in a two dimensional array like this:
title[0]=x description [0]=y
title[1]=z description [1]=w
.
any idea would be appreciated
$query = SELECT * FROM photo;
$result=mysql_query($query);
while ($row = mysql_fetch_array($result))
{
$title=$row['title'];
$description=$row['description'];
}
-
Don't pick lemons.
See all the new 2007 cars at Yahoo! Autos.
_
Check Out Our List Of Trendy Restaurants. You'll Eat It Up!
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[PHP-DB] tables
Hi list, How can you show some pictures in separate tables in one page with echo? - Don't pick lemons. See all the new 2007 cars at Yahoo! Autos.
