PHP函数名称中的变量

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?

Answer 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.