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
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?
I've gone thru the info on eval() php.net, but it's not getting thru my skull very
well. I keep getting errors I can't correct.
Can anyone
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?
I'm not sure that you can eval() HTML.
Perhaps if your cell contained:
echo img
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