On Fri, 5 Jul 2002, Kelly Meeks wrote:
> If I had this information stored in a database field,
> <img src="<? echo $content_output["site_logo"]; ?>">
>
> how could I assign it to a variable and output it?
you have to ?> first to leave php mode so you can do
$content_output = "logo.gig";
$code = "<img src='<?php echo $content_output; ?>'>";
eval("?>$code");
which will echo the outut
or catch it with output buffering
$content_output = "logo.gig";
$code = "<img src='<?php echo $content_output; ?>'>";
ob_start();
eval("?>$code");
$evaled_code = ob_get_contents();
ob_end_clean();
I have used arrays in evals before you have to watch the quotes.
Paul Roberts
[EMAIL PROTECTED]
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