Check if file is ANY image (general check)- PHP

Question

I could of course check mime types through the exif_imagetype or getimagesize and check mime types one by one... But I just want ANY image - I dont care what type... So I was wondering - can I do something like this: ?

// PHP manual says: Determine the type of an image
// and that Imagetype Constants are 1-17 so :
$tmp_imagetype = exif_imagetype('image.gif'); 
if ( ($tmp_imagetype>=1) && ($tmp_imagetype<=17) ) {
    echo "It is an image!";
} else{
    echo "It isn't an image.";
}

Can I rely on that?

What happens when the file is not an image? Will it just return non-image constant value or will it throw a warning or error

Thanks

This looks like a duplicate of http://stackoverflow.com/questions/15408125/php-check-if-file-is-an-image. — Ashley, Feb 12 '14 at 15:29
@Ashley But it is not as I stated in the begining of the question (but the question name could be confusing, I repaired it) — jave.web, Feb 12 '14 at 15:30
@PeeHaa I am looking for an automatic solution and asking about the particular exif_imagetype - possible "duplicates" are all solving how to detect image type or how to check it one by one, or search for it in an array... And I dont know if I can rely on this my current solution — jave.web, Feb 12 '14 at 15:35

score 1 · Accepted Answer · answered Feb 12 '14 at 15:35

1

You really don't need all that you just need the following as exif_imagetype returns false if its not an image

if($imagetype = exif_imagetype('image.gif')){
    // its an image
} else {
    // its not an image
}

answered Feb 12 '14 at 15:35

cmorrissey

8,493
2
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Ah you are right I missread the manual it really says "When a correct signature is found, the appropriate constant value will be returned otherwise the return value is **FALSE**. ". Do you think I should also add `@` before the exif_imagetype() call? I dont want any warnings when checking for the too small or no-a-file... – jave.web Feb 12 '14 at 15:38
@jave.web you can put a `@` if you want but i usually avoid them for debugging purposes – cmorrissey Feb 12 '14 at 15:40
Yea me too, I just wanted to know if it is a completely stupid idea or not :) – jave.web Feb 12 '14 at 15:44

score 1 · Answer 2 · answered Feb 12 '14 at 15:42

1

You could check if the function returns (exactly) FALSE:

$file="SOMEFILE";
if(exif_imagetype($file)===FALSE){
    print("NOT IMAGE");
}else{
    print("IMAGE");
}

answered Feb 12 '14 at 15:42

zaf

22,776
12
65
95

score 0 · Answer 3 · answered Feb 12 '14 at 15:34

0

Why don't you just put all types of images in an array and then check it with this code:

if(in_array($value, $array)) {
   //it's an image! Hooray!
}else {
  //it's not.
}

answered Feb 12 '14 at 15:34

annascott

50
1
5

Hint: use `in_array` in your if – Hanky Panky Feb 12 '14 at 15:36
annascott because I want to do a general check not an "in_array" check – jave.web Feb 12 '14 at 15:37
well then use what I used before, an == in your if. – annascott Feb 12 '14 at 15:38
1

You dont understand me. I want to do GENERAL CHECK - like some general condition if file is an image (see upvoted answers), I dont want to have >any< list of image types. – jave.web Feb 12 '14 at 15:48

Check if file is ANY image (general check)- PHP

3 Answers3