i have a record set filled with article records. however i have an image field in there, how can i get php to not show image if image field is empty. at present it shows as dead link ?
Well... you'll need to show us some code. What does the set look like? An array? How are you outputting the data? At present, I can do nothing more than show you a crude example:
if(!empty($myArray['image'])) { echo '<img src="' . $myArray['image'] . '" alt="image!" border=0 />'; }
i used the Dreamweaver wizard to create the record set articles. The data is ouputed in a dynamic table
As I said, you'll have to show us the part of the code that deals with outputting the image path and whatnot. I've shown you an example... look at the code and try to adapt it to suit your needs. As always, you have a very descriptive manual at your fingertips: here's a link to the man page on [man]empty/man.
Heres the code i got, thanks for ut help so far : )
<?php echo $row_rsArticle['article_imagelink']; ?>
how can ia dapt it, i nt good with this stuff ?
Take another look at my coding example of how to test whether a variable is empty before echo'ing it. Try to implement it in place of the code you posted.
If you honestly tried and can't manage it, post what you did attempt to use and I'll show you what went wrong.
okey dokey, soz for being lazy : (
