Funny you should ask - just done it myself:
Why you use mysql_result function?
This can be done a bit simpler way (and more effective):
?php
mysql_connect(db, $user, $password);
$result = mysql_query( SELECT eventName FROM $table);
while(list($event) = mysql_fetch_row($result))
Thanks for that! I was just using the same structure as I found in a code
sample somewhere.
- seb
-Original Message-
From: Lenar [mailto:[EMAIL PROTECTED]]
Sent: 26 July 2001 13:16
To: [EMAIL PROTECTED]
Subject: Re: [PHP] html form question
Funny you should ask - just done it myself
To: [EMAIL PROTECTED]
Subject: Re: [PHP] html form question
Funny you should ask - just done it myself:
Why you use mysql_result function?
This can be done a bit simpler way (and more effective):
?php
mysql_connect(db, $user, $password);
$result = mysql_query( SELECT eventName FROM $table
: Lenar [mailto:[EMAIL PROTECTED]]
Sent: 26 July 2001 13:16
To: [EMAIL PROTECTED]
Subject: Re: [PHP] html form question
Funny you should ask - just done it myself:
Why you use mysql_result function?
This can be done a bit simpler way (and more effective):
?php
mysql_connect(db
What's the method for populating any number of html
form option.../option tags with query results?
I've seen lots of php-embedded examples for CHECKBOX,
RADIO and even SELECT, but I can't seem to figure out
how to create a simple drop-down menu. HELP!!!
Wouldn't it just be:
select
. /select;
return ($tmpOut);
}
-Original Message-
From: David Robley [mailto:[EMAIL PROTECTED]]
Sent: July 26, 2001 9:02 AM
To: CGI GUY; [EMAIL PROTECTED]
Subject: Re: [PHP] html form question
On Thu, 26 Jul 2001 10:27, CGI GUY wrote:
What's the method for populating any number
Funny you should ask - just done it myself:
?php
$table=shoots;
require (connect.php4);
$result=MYSQL_QUERY( SELECT eventName FROM $table);
$num_rows = mysql_num_rows($result);
for ($i=0;$i$num_rows;$i++)
{
echo option;
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