How to check what image's format

There are serveral methods to check image format.

You can explode image filename by "." (dot) and take last element of result array.

Ex.

$image = 'abce.jpg';

$imarr = explode('.',$image);

$imgformat = $imarr[sizeof($imarr)-1];

I hope you may solved your problem using this.

getimagesize()

I am fully agree with the user archanapatel, and You can explode image filename by "." (dot) and take last element of result array.

Ex.

$image = 'abce.jpg';

$imarr = explode('.',$image);

$imgformat = $imarr[sizeof($imarr)-1];

I hope you may solved your problem using this.

and the user mtishetsky is totally wronge.

Maybe you should read up on getimagesize()

The "solution" you think is correct would only get the filename extension NOT the image format. So archanapatel and you are both wrong.

The "solution" you think is correct would only get the filename extension NOT the image format. So archanapatel and you are both wrong.

Absolutely!
Try to rename an jpg image to gif, and to open it in firefox. It will be displayed correctly.
A file is defined by it's content, not it's extension (except for window, but they surely will say that's a feature, and not a bug).

Linux/osx systems use what is called "the magic file", which gives content pattern and signatures to determine the mime type of a file based on it's content and not it's extension.

getimagesize() does exactly that; it parse the content of the file to determine it's mime type and dimensions.