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Either the page from which you're calling this file isn't referencing it properly, which means the browser can't find it, or the image data returned by this file is broken/incomplete.
Real programmers don't comment their code. If it was hard to write, it should be hard to understand.
One possible suggestion, I found this worked for myself the other day, is to put a double-equals sign on the IF statements when trying to determine the file type :
if ($intype == 'gif')
{
}
Just on first glance it seems that may work, or it may not fit in with the rest of ur syntax, im not sure
if($intype = 'gif') will always return true, because a single '=' assigns a value to a variable. To compare values you have to use the '==' operator.
I also wouldn't use '<>' but instead '!='.