You could also bypass the num_rows entirely...
<?php
$query="select field from table";
$result=mysql_query($query) or exit(mysql_error()."<BR>".$query);
echo '<select name="categories">';
for ($i=0; $row = mysql_fetch_assoc($result); $i++) {
echo '<option>'.($row["field"]).'';
}
echo '</select>';
?>
# Nathan
----- Original Message -----
From: "Gurhan Ozen" <[EMAIL PROTECTED]>
To: "mailing list" <[EMAIL PROTECTED]>; <[EMAIL PROTECTED]>
Sent: Friday, March 15, 2002 4:17 PM
Subject: RE: drop down list using php and mysql
A correction... for line has the greater sign wrong:) It has to be :
for ($i=0; $numresult > $i; $i++)
Sorry about that:)
Gurhan
-----Original Message-----
From: Gurhan Ozen [mailto:[EMAIL PROTECTED]]
Sent: Friday, March 15, 2002 6:01 PM
To: mailing list; [EMAIL PROTECTED]
Subject: RE: drop down list using php and mysql
Hi,
This type of qustion should be forwarded to PHP-DB list. Its adres is:
[EMAIL PROTECTED] .
As for the answer to your question.. IF you'd like to create a dynamic
drop-down menu.. You can just just query the table for the options to be
displayed and loop through the result set and print them on html .. For
example:
$query="select field from table";
$result=mysql_query($query);
$numrows=mysql_num_rows($result);
echo '<select name="categories">';
for ($i=0; $numresult < $i; $i++)
{
$row=mysql_fetch_array($result);
echo '<option>'.($row["field"]).'';
}
Gurhan
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