> Please take a look at the script below.
> If i want to show an image it doesnt work if it's under an if statement.
> Why in this case?
>
> <?
> $feedb=0;
> if($feedb==0){ echo "<img scr='somepic.jpg'>";}
> echo "<img src='somepic.jpg'>";
> ?>
> How can i solve this problem?By spelling src correctly. -Rasmus -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
