variable in php function name

Question

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?

You might want to take a look at [variable variables].(php.net/manual/en/language.variables.variable.php) — EM-Creations
– EM-Creations, Commented Jan 7, 2013 at 10:36

Luca Rainone · Accepted Answer · 2013-01-07 10:23:32Z

1

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.

edited Jan 7, 2013 at 10:23

answered Jan 7, 2013 at 10:18

Luca Rainone

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