[PHP] need help with foreach

2006-10-31 Thread melinem 
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.

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Re: [PHP] need help with foreach

2006-10-31 Thread Dave Hamber 

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.

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Re: [PHP] need help with foreach

2006-10-31 Thread Richard Lynch 
On Tue, October 31, 2006 4:09 am, [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;

echo a href=\$row[url]\$row[url]/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.

I'm assuming you don't mean that you want the user to see the image
and then be auto-forwarded to the 'url' from the db...

That's a fish of a different color.

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