HI,
I have a table in my database that has two fields, one called imgfile1 and the other called url

I can display the image from the imgfile1 field without a problem but what i want to do is create a hyperlink to another website using the address in the url field.

I am very much a novice at this and any help would be much appreciated

my code is below:

PHP Code:

<?php
$result = mysql_query("SELECT * FROM banners1") or die(mysql_error());  
while($row = mysql_fetch_array($result)){
  $imgfile1_txt=$row['imgfile1'];
  $url_txt=$row['url'];
 echo '<img src="http://www.webmaster-talk.com/images/'.$imgfile1_txt.'" width="161" border="0"vspace="3"></a>';
}
?>

The problem is with your html. You have an opening img tag and a closing a tag. If you want an image to function as a link the code is:

<a href="linkaddress"><img src="imageaddress" /></a>

In order to accomplish this in your code change:

PHP Code:

echo '<img src="images/'.$imgfile1_txt.'" width="161" border="0"vspace="3"></a>'; 


to

PHP Code:

echo '<a href="' . $url_txt . '"><img src="images/'.$imgfile1_txt.'" width="161" border="0"vspace="3" /></a>'; 


Thanks for the quick response and the solution to my problem.
Talkupation added!!