I'm wanting to utilise PHP's built-in imagejpeg
and imagepng
functions. I have a variable $type
which defines the filetype jpeg or png.
I wanted to have one function call. Something like this:
image.$type($image, null, 100);
But that doesn't work. Any ideas?
It should works if you use the full function name in a var
$func = "image".$type;
$func($image,null,100);
Careful : imagepng
use a quality between 0-9 (not 0 and 100)
for completeness: you should use an approach more solid, using a OOP method with Factory method pattern . However I don't know your situation and your code. So it's just a note.
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