Its not very clear what you want to achieve with that code snippet.
If you are pulling database rows out of your database with
mysql_fetch_array(), then you do not need a foreach loop.
I think you may want something like this, but without knowing your
database table structure and the query you have used it is only a guess.
while($row = mysql_fetch_array($result2,MYSQL_ASSOC)){
echo "<a href=client.php?art=".$row['pix']."
border='0'>{$row['jobType']}</a>\n";
if($row['url']!='')echo "html to use with ".$row['url'];
}
On 31/10/2006 10:09 [EMAIL PROTECTED] wrote:
I want to tell the server to do 2 things when I click on the "jobType"
link.
This is the code that I have now which displays the "art" when I click
on "jobType"
foreach($row as $jobType)
{
$row = mysql_fetch_array($result2,MYSQL_ASSOC);
echo "<a href=client.php?art=".$row['pix']."
border='0'>{$row['jobType']}</a>\n";
}
I now want to, with the same click, also show the url if I have a url
for that "jobType" in my database column called "url".
Thank you for any help.
--PHP General Mailing List (http://www.php.net/)
To unsubscribe, visit: http://www.php.net/unsub.php
--
PHP General Mailing List (http://www.php.net/)
To unsubscribe, visit: http://www.php.net/unsub.php
