Re: [PHP-DB] HTML tags in auth.php script
Check out the auto_prepend configuration option here: http://www.php.net/manual/en/configuration.directives.php I believe you can do this on a per-directory basis, but haven't tried it. -Steve On Friday, January 10, 2003, at 02:56 PM, [EMAIL PROTECTED] wrote: ThX John...One other question. I have an images folder and other 25 sub-folders underneath this folder. Each of this sub-folder is a web album with it own default index.html and thumbnails pics in it. My question is : Is there an easy way of having a folder level access control to the IMAGES folder and not having to get into each of those 25 index.html files and including my auth.php script in there. As of now any one can just type my whole URL upto the sub-folder and get to see my web pics. Thanks in advance. NT -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP-DB] SHOW PICTURE FROM DATABASE
On Tuesday, December 31, 2002, at 02:25 AM, nikos wrote: Steve b. the _type is the type of the binary data. From the PHP manual read: $userfile_type - The mime type of the file if the browser provided this information. An example would be image/gif Ok, but it didn't look like you saved the type information in the database, at least it wasn't in your query, so where is it coming from? I mean tacking _type on the variable doesn't magically tell you the mime type except in the case of file uploads (where the type is supplied by the browser), right? Ready to stand corrected... -Steve -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP-DB] SHOW PICTURE FROM DATABASE
Some suggestions...
On Monday, December 30, 2002, at 03:55 PM, Nikos Gatsis wrote:
where showpict.php:
$query=SELECT pict FROM pict WHERE pro_id= '$pro_id;
(You're missing an end ' there, but apparently that's not the problem)
$result=mysql_db_query($database, $query, $conn) or Die
(mysql_error());
list($photo)=mysql_fetch_row($result);
$type = $photo_type;
Where did the $photo_type variable come from? What's in it?
if (!empty($photo)) {
header(Content-Type: {$type});
I believe this header should be in the form header(Content-Type:
image/jpeg) or whatever image type you have.
You should probably include an additional header
header(Content-Length: . strlen($photo));
echo $photo;
}
THANX
Nikos
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Re: [PHP-DB] Insert path string into Mysql
On Monday, November 18, 2002, at 01:10 PM, Alan Kelly wrote:
$path = 'c:\\demo\\' ;
$query = insert into PH_PHOTO (PHOTO_PATH) VALUES ('$path');
$result=mysql_query($query);
I would guess that the backslashes are being interpreted once by php
when the variable $path is interpolated into the query string (to
become c:\demo\) and again by mysql at which point it tries to insert a
value for \d. You might try $path = 'c:demo' and see if that
works. Or maybe leave the path alone and do the query line like this:
$query = insert into PH_PHOTO (PHOTO_PATH) VALUES (' . $path . ');
-Steve
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Re: [PHP-DB] MySQL password protection?
You could put it anywhere. Stick it in a text file somewhere, fopen() and read the file for the password. Or keep it in a php script outside of the web root if that's the issue, then just include() it when you need to. Of course any file you put it in will have to be readable by whatever user the webserver is running as. -Steve On Wednesday, November 6, 2002, at 04:16 PM, 1LT John W. Holmes wrote: I was wondering if it is possible to protect my password to the MySQL-server from being in a PHP-script. Now I can't do that, so everybody who gets to see my php-sourcecode also can see my (not protected/not encrypted) password. How can I change this? You can't, unless you want to put it in php.ini or a my.conf file... ---John Holmes... -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP-DB] what's wrong with this ereg?
Well, one thing is you've got mismatched parentheses. You need another
another opening paren right after the if.
Also, don't you need to escape the last hyphen in your character sets?
As well as the . before the third character set?
-Steve
On Tuesday, August 20, 2002, at 11:11 AM, [EMAIL PROTECTED] wrote:
I am trying an email verification function, like this:
if (empty($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?
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Re: [PHP-DB] or statement in url
How about http://www.xyz.com/foo.php?FooID[]=1FooID[]=2 Then FooID will be an array of the values. -Steve On Thursday, July 18, 2002, at 03:20 PM, Matthew K. Gold wrote: can I use an OR operator in a variable that is passed through a url? ex. how can I combine the following two lines so that I end up with records that have an id of either 1 or 2? http://www.xyz.com/foo.php?FooID=1 http://www.xyz.com/foo.php?FooID=2 (I tried the following, but it didn't work: http://www.xyz.com/foo.php?FooID=1||FooID=2 ) thanks in advance for your help. Matt -- 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] header function
What's on line 3 of Auth_user.php ? If you have anything outside of the ?php ... ? tags it will start the html output. Even just an empty line. -steve On Monday, July 15, 2002, at 04:09 PM, Mohammad Forouhar-Fard wrote: Hi, I have a problem with function header(Location:xy.php?var=2). I have not any text (print echo or display any text) at all before I set a cookie If I try to execute this function of Apache server in my Company it is all auf OK. But if I try to run at home I have even the same error Cannot add header information - headers already sent by (output started at C:\httpd\HTDOCS\Auth_user.php on line 3 Can somebody help my I have at home this configuration: I have Winxp OmniHTTPd/2.09 Server. -- 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] what the heck? (elementary question)
On Monday, June 24, 2002, at 01:52 PM, Matthew Crouch wrote: this bit from my index page is giving me 2 headaches: 1. it isn't passing anything into the URL 2. the page that gets called (name.php) sits and tries to load forever. it looks like it's even filling up my hard drive with data (?!) Note: it does this (#2) even if I type in the URL with a variable and value, like name.php?lastname=smith ? printf (form method=\post\ action=\name.php?lastname=%s\, $lastname); ? div align=center input type=Text name=lastname size=75 input type=Submit name=submit value=Search for Last Name /div /form /div Can you do a get and a post at the same time? Tacking the lastname onto the url as in action=\name.php?lastname=%s\ is using a get method, but your form is supposed to set lastname using the post method. Is there some reason you want your script to set one version of lastname and let the user type in a different version of lastname? If so try using the post method with an input type=hidden name=otherlastname value=$lastname. -Steve -- 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] Date of Birth From Form
A combination of strtotime() and date() can make this easy. date(Y-m-d, strtotime($birthday)) Or see docs and comments at http://www.php.net/manual/en/function.strtotime.php -Steve On Friday, April 19, 2002, at 01:29 PM, Brandon Paul wrote: Hey all, I have a credit application form, and one of the required elements is the applicant's Date of Birth. I have creditapp table with a birthdate field and it is a DATE datatype. On the form, I want to be able to have them enter their Date of Birth as mm/dd/ and have it go into the database properly (-mm-dd). Is there a way to do it this way, or am I going about it wrong? Any help would be greatly appreciated. Thanks! Brandon -- 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] odbc versions
Hi. I'm looking at compiling one of the odbc function sets into php on my Mac OSX machine (I eventually want to be able to talk to MAS90 on a Windows machine -- if that's feasible). I'm finding the ODBC situation a bit confusing though. Should I use iodbc or unixodbc? Any suggestions? And will a database driver for one work with the other or are they incompatible? Thanks. -Steve -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP-DB] cron job and php
Where is your php binary? (try which php) The command path under crontab is not always the same as in a login shell, so you might need to spell it out like: */10 * * * * /usr/local/bin/php /home/harpreet/crontest.php -Steve On Tuesday, April 16, 2002, at 12:17 PM, Harpreet Kaur wrote: Thanks Lisi. The use of php in front of the path made /home/harpreet/crontest.php run one time when run from the telnet prompt. But it still doesnt run from the crontab. I created a mycron.txt file with the below statement. The put crontab mysql.txt to specify the cron. But it doesnt seem to run this way. What am i doing wrong. Please help. Help is appreciated. Regards, harpreet kaur From: Lisi [EMAIL PROTECTED] To: Harpreet Kaur [EMAIL PROTECTED], [EMAIL PROTECTED] Subject: Re: [PHP-DB] cron job and php Date: Tue, 16 Apr 2002 19:13:11 +0200 Try */1 * * * * php /home/harpreet/crontest.php or */1 * * * * php ~/crontest.php ~ means your home directory. I think this should work, every minute. -Lisi At 03:54 PM 4/16/02 +, Harpreet Kaur wrote: Can i run a php page using a crontab job. I am trying to do so but no luck. My cron job statement is : * * * * * lynx /home/harpreet/crontest.php Actually i want to run a script every 10 minutes to do some background job like we do in sql server. How can i do it with mysql and php. Help is appreciated, regards, Harpreet Kaur _ Get your FREE download of MSN Explorer at http://explorer.msn.com/intl.asp. -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php _ Send and receive Hotmail on your mobile device: http://mobile.msn.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
Re: [PHP-DB] Arrays
Your $eventdate variable is an array, not a string so you can't just
drop it into an sql query string like that.
If you have separate fields in the db for year, month and day then you
probably want something like:
$query = insert into ... values ( ... , '$eventdate[0]',
'$eventdate[1]', '$eventdate[2]', ...);
Otherwise, if they're all supposed to go into the same field then do
something like:
$eventdatestring = join(-, $eventdate);
$query = insert into ... values ( ..., '$eventdatestring', ...);
This should insert the date in the form eventyear-eventmonth-eventday;
-Steve
On Monday, April 15, 2002, at 03:13 PM, Alex Francis wrote:
I am trying to insert event dats into a database and to make sure the
user
puts the date in the correct format. I am now trying to collect the
information from the three fields and insert it into one field.
code as follows: $eventdate = array (eventyear, eventmonth,
eventday);
when I insert it into the database like: $query = INSERT INTO
$tablename4
VALUES ('0', '$entername', '$date', '$eventdate', '$eventheading',
'$eventbody' );
for the $eventdate variable the word array is inserted.
Could someone please tell me where I am going wrong.
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Re: [PHP-DB] Using include file
I pasted your code into a file testinclude.php (changing only the server
name), then wrote another file test.php:
?
include('testinclude.php');
?
It parses okay for me. How are you doing the include?
-Steve
On Friday, April 12, 2002, at 03:02 PM, Alex Francis wrote:
I get a parse error on line 2. The code is as follows:
?
$server = mysql.xcalibre.co.uk;
$user = user;
$passwd = password;
$dbname = rschool;
$tablename1 = story;
$tablename2 = guestbook;
$tablename3 = drawing;
$tablename4=notices;
$link = mysql_connect ($server, $user, $passwd);
?
As I said, when I paste the code into each file I don't have a problem.
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Tel 01475 798106
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Steve Cayford [EMAIL PROTECTED] wrote in message
[EMAIL PROTECTED]">news:[EMAIL PROTECTED]...
Just a guess: Are you doing the include from a function? If so, make
sure you explicitly mark your global vars. Otherwise, what's the error
message?
-Steve
On Friday, April 12, 2002, at 12:28 PM, Alex Francis wrote:
I have one site which I am having problems connecting to my database.
If I
create my connections in an include config file I get an error on
the
server connection. When I cut and paste the code exactly as it is into
each
file I have no problems. I have a local server set up in my office
which I
use to test my databases and scripts and have a slightly different
config
file on that one. The include file works fine.
I have various other sites using include config files and have had
no
problems, but this one is a pain, I have to change each file when I
move it
mrom my test server to my hosting server.
Any help would be appreciated.
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Re: [PHP-DB] Images on MySQL
I keep hearing this from people (not to store images in mysql), but I would like to hear a bit more about why. Mysql has blob fields, so it seems perfectly reasonable to use them, doesn't it? I'm storing some images in a database and what's attractive to me about it is that I can put the images anywhere I like. I guess I could mount an image directory over NFS, but it seems easier and more consistent to use sql. Any thoughts on this? -Steve On Friday, March 29, 2002, at 05:43 AM, Jason Wong wrote: On Thursday 28 March 2002 19:35, Clever wrote: Hi, I'm designing a site and I have to store a lot of images. Which is the best for speed? 1) Store all images on a MySQL table? 2) Save them on disk like normal files and only have pointers to them on the database? 2) -- Jason Wong - Gremlins Associates - www.gremlins.com.hk /* The only difference in the game of love over the last few thousand years is that they've changed trumps from clubs to diamonds. -- The Indianapolis Star */ -- 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] elseif statement syntax
You probably have error reporting turned off in your php.ini file or you
may be directing errors into your log file, try checking there.
Why are you tacking or die(mysql_error()) on the end of each of your
query assignments? You're just assigning a string to a variable so
there's nothing to fail, or at least no mysql errors.
-Steve
On Wednesday, March 20, 2002, at 10:41 AM, Andrea Caldwell wrote:
Hi All, I'm pretty new at this, so go easy on me please ;-)
What is wrong w/ this syntax? If the search results are 0, it just
displays
a blank screen instead of echoing the error message if numresults ==0
or the
mysql_error message. If data is found, everything is fine. Thanks in
advance for your help!
if($searchterm){
$query = select directory.realname, directory.phone, directory.ext,
directory.phone2, directory.email, directory.location from directory
where
realname like '%.$searchterm.%' or die (mysql_error());
}
elseif($location){
$query = select directory.realname, directory.phone, directory.ext,
directory.phone2, directory.email, directory.location from directory
where
location like '%.$location.%' or die (mysql_error());
}
else{
$query = select directory.realname, directory.phone, directory.ext,
directory.phone2, directory.email, directory.location from directory
where
location like '%.$searchloc.%' or die (mysql_error());
}
$result = mysql_query($query) or die (mysql_error());
$num_results = mysql_num_rows($result)or die (mysql_error());;
if($num_results==0){
echo Sorry, nothing matched your search request. Please go back and
try
again.;
}
else {
echo pspan class=stdtext_boldNumber of Entries Found:
.$num_results./p/span;
}
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Re: [PHP-DB] A weird images database/web page problem
Hello again.
On Thursday, January 17, 2002, at 01:06 AM, chip wrote:
On Wednesday 16 January 2002 08:57 pm, Steven Cayford banged out on the
keys:
On 2002.01.16 19:59:37 -0600 chip wrote:
Let's tackle the first one first -
OK.
On Tuesday 15 January 2002 10:20 pm, Steven Cayford banged out on the
keys:
On 2002.01.15 23:30:25 -0600 chip wrote:
won't work)
$new_pic=$pic+12;
Is $pic already set by the HTTP_GET_VARS here?
I don't know. How do I set this? I was reading the online manual about
http_get_vars and it didn't help, and doesn't have any examples.
I was assuming that the $pic variable was set by the HTTP get request
since your link below is in the form:
a href=\index.php?pic=$new_pic\.
When the target script starts after being called like that the value
in the
global variable $HTTP_GET_VARS['pic'] will hold whatever $new_pic was
set
to in the reference. Assuming you have register_globals on,
Yes it is on.
then the global
variable $pic will also be set to the same value.
Makes sense
Otherwise $pic will be uninitialized and should evaluate to 0.
So, I guess I have to set the variable to 0 first when the first page is
loaded, then add 24 (or 12 or whatever) to for the link to the next
page. So,
with that in mind I added this
$pic=0;
before
$new_pic=$pic+12;
but that just caused it to reload the same images each time next was
pushed.
So, now that I appear to be a total dummy, and the manual doesn't
provide any
examples, and the two books I have don't seem to be able to help also,
would
you mind helping me a bit more? I'm I going about this next page stuff
all
wrong, is there a better way to do it?
--
Sorry, I think I got you off track. In this case you *do* want to set
$pic from the GET vars. I guess how I would do this would be something
like this:
if(! isset($pic)) {
$pic = 0;
}
This way, if $pic has been set by the GET vars, then the value will be
retained, otherwise it will be explicitly set to whatever default value
you want--in this case 0. This is probably anal of me, but I just don't
like using uninitialized variables. Also keep in mind that $pic can be
set to any value via the URL. So if someone types into their browser:
http://...your_site.../index.php?pic=hello; then when your script runs
$pic will hold the string hello instead of a number. Then if you stick
the value straight into your sql statement without verifying it you
could get all sorts of weird problems. If the site is just for your own
local use, then don't worry about it, but if its going to be available
to the wider world at some point then you may want to explicitly force
$pic to be an integer $pic = (int) $pic; or something like that.
In any case, this is all aside from the problem that you were running
into. I think if you change your sql statement below from select * from
ab limit $new_pic,12 to select * from ab limit $pic,12 you'll find
you can get the first page of images.
-Steve
(nothing new below here, just history)
$conn=mysql_connect(localhost, chip,carvin) or die (Could
not
get
the databse);
mysql_select_db(images, $conn) or die (Could not select the
database);
$sql=select * from ab limit $new_pic,12;
Assuming that $pic is 0, then $new_pic is 12, so you're selecting
with
LIMIT 12, 12. You probably want to use $pic here, right?
$result=mysql_query($sql);
while ($row=mysql_fetch_array($result))
{
printf(td align=\center\a href=\%s\img
src=\../thumbs/%s\/a/td\n, $row[name], $row[name]);
$i++;
if($i %6==0)
{
echo /tr\n;
}
Note that if the select statement gives you other than 6 or 12 images
(which it probably will on the last page), that /tr will never get
echoed.
}
echo tr\ntd colspan=\6\ align=\center\\na
href=\../index.html\Home/anbsp;\na
href=\index.php?pic=$new_pic\Next/a\n/td\n/tr\n;
?
To find out if $new_pic pointed to a valid image you would probably
need
to do a select count(*) from ab to get the total number of records.
If
$new_pic is less than the count, then show the link.
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Re: [PHP-DB] question
On Wednesday, December 12, 2001, at 09:03 AM, Robert Weeks wrote:
Hi,
I'm a little stumped here, maybe I haven't had enough coffee yet...
Anyway, I have a lot of SQL queries that return one row if there is a
match
and instead of having to write out something like this:
$firstname = $row[firstname];
$lastname = $row[lastname];
$midinitial = $row[midinitial];
$preferredname = $row[preferredname];
$address1 = $row[address1];
$address2 = $row[address2];
$city = $row[city];
$state = $row[state];
$zip = $row[zip];
I thought I'd loop through the result and and get the info out like
this:
$nrows = mysql_num_rows($result);
$nfields = mysql_num_fields($result);
$rowarray = mysql_fetch_row($result);
for ($i=0; $i$nfields; $i++){
$fname = mysql_fieldname($result,$i);
$val = $rowarray[$i];
$fname = $val;
Do you want $$fname = $val here? Taking the string in $fname as the name
of a variable to which you are assigning $val.
}
and then later in the page reference the values using ?=$address1?
Problem is this isn't working. If I echo the varibles to the page they
will
print out but they wont show up in the ?=$address1? tags.
I tried setting the varibles to global and that still doesn't seem to
work.
I know I'm missing something really simple and basic here. If someone
could
give me a push in the right direction I'd appreciate it.
Thanks,
Robert
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Re: [PHP-DB] losing first row of an array
On Friday, November 2, 2001, at 01:09 PM, Seany wrote:
having this code running PHP4.0b2
$result = mysql_query(SELECT clubac.id AS clubacid, clubac.club_id,
clubac.ac_year, clubac.ac_type, club.id
FROM clubac, club WHERE club.id = '$club_id' AND clubac.club_id =
club.id);
$row = mysql_fetch_array($result);
Here you've fetched the first row...
if ($row = mysql_fetch_array($result)) {
...and here you've overwritten it with the second row.
You probably just want: if ($row) { ...
do {
echo $row[club_id];
echo $row[id];
echo $row[ac_year];
echo $row[ac_type];
} while($row = mysql_fetch_array($result));
} else .blablabla
is there a fundamental miss on my part why the is printing the first or
not
even recognising it
if call direct in mysql with the above query it's fine with the results
even
if there is 1 row
duh???
Seany
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Re: [PHP-DB] Learning PHP Sessions
This is really off-topic for this list, but...
From my understanding of sessions, you really don't want session_start()
in an if{} block. Every time you hit this script, it will have no memory
of any session variables until you call session_start().
-Steve
On Tuesday, October 30, 2001, at 02:30 PM, Matthew Tedder wrote:
?php
/*
Hi,
I'm new to PHP and am having trouble understanding how to use PHP
sessions. My book tells in near the beginning how to start them and
register
session variables, but I can't figure out how to destroy a session or
later
read those session variables. I'm also trying to do this across
frames, but
can't even get it to work within a single page.
Here's what I've learned so far and what my problems are:
*/
/* To start a session */
session_start();
/* To register a session variable */
session_register(myvar);
$myvar = some value;
/*
PROBLEM #1: From the above commands, I get a $PHPSESSID that seems to
be
globally available for use, but I cannot seem to read my values back
out of
the registered session variable from anywhere... I tried:
*/
print $myvar\n; /* and absolutely nothing is printed */
/* To destroy a session */
session_destroy();
/*
PROBLEM #2: This says there is no session to destroy. It's rather
strange
because I can still print the $PHPSESSID value..
I've attached my code...
*/
?
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Re: [PHP-DB] Help! ¡Ayuda!
Source code- CÛdigo fuente:
?
$based=articulo.dbf;
if (($descriptor=dbase_open ($based, 0))==0){
printf (brError al abrir la base de datos);
}else{
printf (brBase de datos abierta);
$num_registros=dbase_numrecords($descriptor);
$num_campos=dbase_numfields($descriptor);
for ($i=1;$i=$num_registros;$i++){
$registro= dbase_get_record ($descriptor, $i);
for ($j=0;$j$num_campos;$j++){
printf (brFila %d,Campo %d vale %S, $i, $j, $registro[$j]);
Try using a lower-case '%s' here -^
I don't think the uppercase %S means anything to printf.
}
}
dbase_close($descriptor);
printf (brBase de datos cerrada);
}
?
Don't know much about dbase, but the rest looks okay.
-Steve
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Re: [PHP-DB] PHP and MySQL queries...
Yep, MySQL has DATE, DATETIME, and TIMESTAMP field types. You can order by them and everything. -Steve On Thursday, October 25, 2001, at 09:18 AM, Tim Foster wrote: I'm new to this list, to PHP and to MySQL (been doing VBScript on ASP for several years, tho). I'm curious... If you're going to store it as an integer, why not store 10/24/2001 as MMDD (20011024). This gives you the added benefit of being able to have the db sort your fields. This even works if you want to include the time with your date (provided all dates in the field consistently contain the same *amount* of info). For example, noon on Christmas will always be lower than noon of the following New Year ..as it should be: /MM/DD20011225 20020101 /MM/DD HH:MM 200112251200 200201011200 /MM/DD HH:MM:SS 2001122512 2002010112 I'm betting there's no easy way to sort it if you store it as MM/DD/YY MM/DD/10242001 12252001 (good) ..but NOT less than the following New Year's MM/DD/10242001 01012002 (bad) Granted, you might take up a bit more space in the DB, which would have a tiny impact on performance(??), but an extra $100 on the hard drive would effectively eliminate any reasonable space considerations and (IMHO) reduce the amount of programming/debugging to more than justify the overhead. FWIW, M$ likes to store their dates as two integers: one to hold the date portion, the other to hold the hours:minutes:seconds portion. If there's something about PHP/MySQL that makes this point moot, please let me know. TIM -He who always plows a straight furrow is in a rut. -Original Message- From: Mike Frazer [mailto:[EMAIL PROTECTED]] Sent: Wednesday, October 24, 2001 7:54 AM To: [EMAIL PROTECTED]; [EMAIL PROTECTED] Subject: Re: [PHP-DB] PHP and MySQL queries... Agreed. This is especially useful when you need to conserve every byte you can; a timestamp of 10/24/2001 or something similar is going to take 10 bytes as a string and an indeterminate number of bytes for an actual timestamp because of system variations, whereas an integer value of 10242001 will take you 2-4 bytes depending on the type of int you declare. Not a lot of space, but assume for a second you have 30 fields in your database and 5 million rows...suddenly those 6-8 bytes have multiplied on this one field alone. Space and speed are important in DBs :) Mike Frazer -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, e-mail: [EMAIL PROTECTED] For additional commands, e-mail: [EMAIL PROTECTED] To contact the list administrators, e-mail: [EMAIL PROTECTED] -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, e-mail: [EMAIL PROTECTED] For additional commands, e-mail: [EMAIL PROTECTED] To contact the list administrators, e-mail: [EMAIL PROTECTED]
Re: [PHP-DB] PHP and MySQL queries...
Oop. I guess I missed the point of that question. Still, the MySQL manual says a DATE takes 3 bytes, DATETIME 8 bytes, and TIMESTAMP 4 bytes. That seems fairly efficient. Using an INT for a date might actually take up more space. -Steve On Thursday, October 25, 2001, at 09:34 AM, Steve Cayford wrote: Yep, MySQL has DATE, DATETIME, and TIMESTAMP field types. You can order by them and everything. -Steve On Thursday, October 25, 2001, at 09:18 AM, Tim Foster wrote: I'm new to this list, to PHP and to MySQL (been doing VBScript on ASP for several years, tho). I'm curious... If you're going to store it as an integer, why not store 10/24/2001 as MMDD (20011024). This gives you the added benefit of being able to have the db sort your fields. This even works if you want to include the time with your date (provided all dates in the field consistently contain the same *amount* of info). For example, noon on Christmas will always be lower than noon of the following New Year ..as it should be: /MM/DD 20011225 20020101 /MM/DD HH:MM 200112251200 200201011200 /MM/DD HH:MM:SS 2001122512 2002010112 I'm betting there's no easy way to sort it if you store it as MM/DD/YY MM/DD/ 10242001 12252001 (good) ..but NOT less than the following New Year's MM/DD/ 10242001 01012002 (bad) Granted, you might take up a bit more space in the DB, which would have a tiny impact on performance(??), but an extra $100 on the hard drive would effectively eliminate any reasonable space considerations and (IMHO) reduce the amount of programming/debugging to more than justify the overhead. FWIW, M$ likes to store their dates as two integers: one to hold the date portion, the other to hold the hours:minutes:seconds portion. If there's something about PHP/MySQL that makes this point moot, please let me know. TIM -He who always plows a straight furrow is in a rut. -Original Message- From: Mike Frazer [mailto:[EMAIL PROTECTED]] Sent: Wednesday, October 24, 2001 7:54 AM To: [EMAIL PROTECTED]; [EMAIL PROTECTED] Subject: Re: [PHP-DB] PHP and MySQL queries... Agreed. This is especially useful when you need to conserve every byte you can; a timestamp of 10/24/2001 or something similar is going to take 10 bytes as a string and an indeterminate number of bytes for an actual timestamp because of system variations, whereas an integer value of 10242001 will take you 2-4 bytes depending on the type of int you declare. Not a lot of space, but assume for a second you have 30 fields in your database and 5 million rows...suddenly those 6-8 bytes have multiplied on this one field alone. Space and speed are important in DBs :) Mike Frazer -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, e-mail: [EMAIL PROTECTED] For additional commands, e-mail: [EMAIL PROTECTED] To contact the list administrators, e-mail: php-list- [EMAIL PROTECTED] -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, e-mail: [EMAIL PROTECTED] For additional commands, e-mail: [EMAIL PROTECTED] To contact the list administrators, e-mail: [EMAIL PROTECTED] -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, e-mail: [EMAIL PROTECTED] For additional commands, e-mail: [EMAIL PROTECTED] To contact the list administrators, e-mail: [EMAIL PROTECTED]
Re: [PHP-DB] PROBLEM ACCESSING DBASE WITH PHP
You only need to post your question once, and you really shouldn't cross-post to multiple lists...please, I just got six copies of your question. Did you compile php with the --enable-dbase configure flag? You can check with phpinfo() for what's enabled and also what configuration flags were used. -Steve On Thursday, October 25, 2001, at 12:31 PM, Sebas wrote: I need to open a dbase file, so I use the dbase_open function and then I compile it but it tells me Fatal error: Call to undefined function: dbase_open() it seems like this function does not exists!.Does anybody could help me? Thanks [EMAIL PROTECTED] --- Necesito abrir una base de datos dbase, por lo que uso la funciÛn dbase_open y al compilar, me sale el error: Fatal error: Call to undefined function: dbase_open() Parece como si la funciÛn no existiera. øøPuede alguien ayudarme?? °°Es muy urgente!!Gracias [EMAIL PROTECTED] -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, e-mail: [EMAIL PROTECTED] For additional commands, e-mail: [EMAIL PROTECTED] To contact the list administrators, e-mail: [EMAIL PROTECTED] -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, e-mail: [EMAIL PROTECTED] For additional commands, e-mail: [EMAIL PROTECTED] To contact the list administrators, e-mail: [EMAIL PROTECTED]
Re: [PHP-DB] MySQL Query Weirdness
When you use mysql_fetch_array you're actually retrieving an array with
two copies of the value you're looking for. One is indexed by the column
number and one indexed by the column name. So it looks like you're
looping through them both and printing each out. You might want
mysql_fetch_assoc (to use just the column name) or mysql_fetch_row (to
use just the column number)
-Steve
On Thursday, September 20, 2001, at 05:02 PM, Chris S. wrote:
Hello,
I'm new at this php/mysql stuff, so go easy on my guys.
I've got my query script up and working, but the problem is I'm getting
the
same column printed twice on the html output. Here is the output:
Connected successfully
Chris Chris
Mark Mark
Mike Mike
Dee Dee
etc...
Here is my .php script:
?php
$link = mysql_connect(localhost, dbuser,dbpassword)
or die (Could not connect);
print (Connected successfully);
mysql_select_db (dapscores) or die (Could not select database);
$query = SELECT first_name FROM overall_results;
$result = mysql_query ($query) or die (Query failed);
// printing HTML result
print table\n;
while ($line = mysql_fetch_array($result)) {
print \ttr\n;
while(list($col_name, $col_value) = each($line)) {
print \t\ttd$col_value/td\n;
}
print \t/tr\n;
}
print /table\n;
mysql_close($link);
?
Here is my database layout:
mysql describe overall_results;
+---+-+--+-+-+---+
| Field | Type| Null | Key | Default | Extra |
+---+-+--+-+-+---+
| Match_Date| date| YES | | NULL| |
| Place | varchar(10) | YES | | NULL| |
| Last_Name | varchar(20) | YES | | NULL| |
| First_Name| varchar(20) | YES | | NULL| |
| USPSA | varchar(10) | YES | | NULL| |
| Class | char(3) | YES | | NULL| |
| Division | varchar(20) | YES | | NULL| |
| PF| varchar(7) | YES | | NULL| |
| Lady | char(3) | YES | | NULL| |
| Mil | varchar(4) | YES | | NULL| |
| Law | varchar(4) | YES | | NULL| |
| F0r | char(3) | YES | | NULL| |
| Age | varchar(20) | YES | | NULL| |
| Match_Pts | float | YES | | NULL| |
| Match_percent | float | YES | | NULL| |
+---+-+--+-+-+---+
If I run the query from mysql, it works fine, just the HTML output
shows
the double column thing. Is this a database problem? I've tried
different
variations of my script and I get the same output each time.
Thanks
--
Chris S.
[EMAIL PROTECTED]
PGP 0xDA39672B
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Re: [PHP-DB] Query construction
Usually what I do when I'm having this type of problem is put the whole
sql statement into a string variable then print the variable out just
before running the query so you know exactly what's being sent to mysql.
Then I log into mysql from the command line and enter the query manually
to see what sort of response or error message I get.
-Steve
On Wednesday, September 19, 2001, at 09:01 AM, Russ Michell wrote:
Hi there:
Further to a previous submission (which can be ignored), I'd like the
following query to 'search' a
table field of different team names (stored as a single string), but it
does't seem to work!
SELECT * FROM $table_users WHERE usrName='$username' AND
usrPswd=password('$password') AND affil_team RLIKE '$team+';
(I've also tried: RLIKE '$team?' )
This should match the string found in the variable: '$team' with the
some of contents (string)
found in 'affil_team'.
For example my test has been, trying to find 'footballSat', so $team =
'footballSat'. 'footballSat'
exists as part of the string in the 'affil_team' field but the above
query refuses to find
'footballSat'!! (No error is received though)
I'm using MySQL 3.22.32 + php4.0.3pl1
Cheers
Russ
#---#
Believe nothing - consider everything
Russ Michell
Anglia Polytechnic University Webteam
Room 1C 'The Eastings' East Road, Cambridge
e: [EMAIL PROTECTED]
w: www.apu.ac.uk/webteam
t: +44 (0)1223 363271 x 2331
www.theruss.com
#---#
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Re: [PHP-DB] Table Search ... HELP
All the info you want should be in $result after the query. But to get it
out of $result you need to access one row at a time with
mysql_fetch_array($result), which returns a numerically keyed array (if I
remember correctly), or mysql_fetch_assoc($result), which returns a string
keyed array or hash. So, like this (more or less)...
if ($TechContact == ){
$TechContact = '%';
}
$result = mysql_query (SELECT * FROM enet
WHERE TechContact LIKE '$TechContact%');
// this part added...
while($row = mysql_fetch_assoc($result)){
print $row[TechContact];
print $row[SomeOtherColumnName];
print $row[YetAnotherColumnName];
print br\n;
}
On 2001.09.06 19:41:55 -0500 Devon wrote:
Below is an example of my code which searches a table and prints the
result,
the problem is that it only displays the TechContact where I want it to
display all the fields that associated with it in that row off the colum
eg.
Mobile, AdminContact etc etc Any suggestions?
if ($TechContact == )
{$TechContact = '%';}
$result = mysql_query (SELECT * FROM enet
WHERE TechContact LIKE '$TechContact%');
print $row[TechContact];
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Re: [PHP-DB] Forms Question
On Wednesday, September 5, 2001, at 04:50 PM, Jeff Grossman wrote:
Hello,
Here is the code I have:
while ($row=mysql_fetch_array($result)) {
$store=$row[store];
$jobdesc=$row[jobdesc];
echo FORM METHOD=post ACTION=update.php;
echo PStore:
Select NAME=\$store\
option VALUE=\Signal Hill\Signal Hill
option VALUE=\Reseda\Reseda
option VALUE=\Orange\Orange
option VALUE=\West Covina\West Covina
option VALUE=\Riverside\Riverside
option VALUE=\Norwalk\Norwalk
option VALUE=\Fountain Valley\Fountain Valley
option VALUE=\Pasadena\Pasadena
option VALUE=\Redondo Beach\Redondo Beach
option VALUE=\San Bernardino\San Bernardino
option VALUE=\Kearny Mesa\Kearny Mesa
option VALUE=\San Marcos\San Marcos
option VALUE=\Chino\Chino
option VALUE=\Coporate Office\Corporate Office
/select/P;
echo PINPUT TYPE=text SIZE=35 NAME=\Jobdesc\
VALUE=\$jobdesc\/P;
echo pINPUT TYPE=submit VALUE=\submit\ LABEL=\Save
Changes\/P;
}
Is want I am trying to do possible? I want the value which is stored in
$store to automatically fill in on the drop down list. But, for some
reason it is defaulting to the first option, and not using the value
that is in the database.
Can I use a drop down menu, or should I just go to radio buttons?
Thanks,
Jeff
If I understand your question...
In order to have your value preset in the drop down list you need
indicate that with
option value=\blahblah\ selectedblahblah
What I've been using for this is a hash like this:
while ($row=mysql_fetch_array($result)) {
$selected = array();
$selected[$row[store]] = selected;
$store=$row[store];
$jobdesc=$row[jobdesc];
echo FORM METHOD=post ACTION=update.php;
echo PStore:;
echo Select NAME=NameOfVariableToBePosted
echo option VALUE=\Signal Hill\ . $selected[Signal
Hill] . Signal Hill;
echo option VALUE=\Reseda\ . $selected[Reseda] .
Reseda;
...etc, etc., etc.
something along those lines, anyway. So, if $row[store] == Signal
Hill, then $selected[Signal Hill] will be set to selected, while
$selected[Reseda] and all the others will be blank.
This is a very keen thing about php.
-Steve
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