> 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.


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