ev66 Posted September 1, 2006 Share Posted September 1, 2006 In the code below im trying to echo an image using a variable $a but cant get it to work.It works if i use the image name, but cant get it to work using the variable, any ideas please ?Thanks[color=green]THIS DOESNT WORK[/color][color=red]<?php $a = "ad_ink-spot.gif"; ?>[/color]<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd"><html xmlns="http://www.w3.org/1999/xhtml"><head><meta http-equiv="Content-Type" content="text/html; charset=iso-8859-1" /><title>Untitled Document</title></head><body><p></p><p> </p><p>[color=red]<?php echo '<a href=' . $_SERVER['PHP_SELF'] . '><img src=images/' . $a . 'width=200 height=121 border=0 /a>'; ?>[/color] </p></body></html>[color=green]THIS DOES WORK[/color][color=red]<?php $a = "ad_ink-spot.gif"; ?>[/color]<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd"><html xmlns="http://www.w3.org/1999/xhtml"><head><meta http-equiv="Content-Type" content="text/html; charset=iso-8859-1" /><title>Untitled Document</title></head><body><p></p><p> </p><p>[color=red]<?php echo '<a href=' . $_SERVER['PHP_SELF'] . '><img src=images/ad_ink-spot.gif width=200 height=121 border=0 /a>'; ?>[/color] </p></body></html> Quote Link to comment Share on other sites More sharing options...
trq Posted September 1, 2006 Share Posted September 1, 2006 [code=php:0]<?php echo '<a href="' . $_SERVER['PHP_SELF'] . '"><img src="images/' . $a . '" width="200" height="121" border="0"></a>'; ?>[/code] Quote Link to comment Share on other sites More sharing options...
ev66 Posted September 1, 2006 Author Share Posted September 1, 2006 Thanks. It works now. Quote Link to comment Share on other sites More sharing options...
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