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Imageduckspeaker wrote in Imagephp

creating variable

Stupid question I’m sure…but I just can’t see the solution.

I have the follow lines...that create a variable ($uploadfile) that I use later to load a file onto the site.

$uploaddir = '../../../mem_hoa/';
$uploadfile = $uploaddir . $filename;

It works great!

BUT…I want to allow for user selectable variable to determine directory location.

The following DOESN’T work…and I don’t know why

$uploaddir = '../../../$memselection/';
$uploadfile = $uploaddir . $filename;

Help?

(NOTE: I have verified the variable ($memselection) is working correctly, and is actually used for some table insert and manipulation on the same php page.)