how to use getimagesize if I don't know extension of file?

marcnyc

Hello, I'd like to get the image size of a picture with getimagesize() but I don't know if it's a gif or jpg.

Is there a way I can do something like this (I know this can't be done but maybe there is some syntax I don't know about that does just this)?
getimagesize('path/file.gif|jpg');

I know the file name but I don't know if it's gif or jpeg...
Thanks

ereptur

If you know the actual filename then getimagesize() returns the mime-type as one of its elements.

list($filesize,$filetype) = getimagesize($imagefile);

Otherwise, you are going to have to test for the file to see if it exists

if(file_exists($imagefile)) {
  list($filesize) = getimagesize($imagefile);
} else if(file_exists($imagefile.jpg)) {
  list($filesize) = getimagesize($imagefile.jpg)

etc....etc...etc..

You get it

marcnyc

I know the name of the file but not the extension... any idea? any way?

Mordecai

Originally posted by ereptur -- edited slightly

if(file_exists($imagefile)) {
  list($filesize) = getimagesize($imagefile);
} else if(file_exists($imagefile.jpg)) {
  list($filesize) = getimagesize($imagefile.jpg);
}

[/B]

This is basically what you'd have to do. This just checks if the file exists without an extension, and then checks for .jpeg (there's a sytax error, however):
} else if(file_exists($imagefile.".jpg")) {

Then you can use something similar for .gif, .png, etc.

Weedpacket

I have to say I don't understand what the problem is. getimagesize doesn't even look at the filename when it's trying to figure out what type of file it is. I could have a PNG file named "foo.png", "foo", "foo.bar" or "foo.gif" and it recognises it as a PNG every time.

marcnyc

I am a little confused. Here is what I am doing now to find out if it is a gif or a jpg:

$size_logo_gif = @getimagesize('/'.$path_images.'/'.$name_file.'/'.$name_file.'_logo.gif');
$size_logo_jpg = @getimagesize('/'.$path_images.'/'.$name_file.'/'.$name_file.'_logo.jpg');
	if ( !$size_logo_gif && !$size_logo_jpg ) {
		// no pics at all
	} else if ( $size_logo_gif ) {
		// the pic is gif
	} else if ( $size_logo_jpg ) {
		// the pic is a jpg
	}

Of course I could use file_exists() instead of getimagesize(). I am not quite sure what the advantages are of using one over the other and if anybody knows if one of them is faster than the other let me know so I can reduce the load of the script and make it faster.

Anyway, this seems to work for me, I was just wondering if there was a faster way to do it or if there was a function that checks what type image it is.

feldon23

$typelist = array('jpg','JPG','jpeg','gif','GIF');

function findExtension ($filename) {
  global $typelist;
  foreach ($typelist as $type) {
    //strip extension if it exists
    if (strpos($filename, '.'.$type) >= 1) {
      $filename = substr($filename, 0, strpos($filename, '.'.$type));
    }
    //file found
    if (is_file($the_file.'.'.$type)) {
      return $the_file.'.'.$type;
    }
  }
}

//Search all extensions in $typelist
$logosize = getImageSize(findExtension("/{$path_images}/{$name_file}/{$name_file}_logo"));

Weedpacket

Originally posted by marcnyc
I am a little confused. Here is what I am doing now to find out if it is a gif or a jpg:

So you're not trying to see if a file is a GIF or a JPEG; you've got an incomplete filename and you're trying to reconstruct the full filename that the file is stored under? To put it another way, you're trying to see if a filename ends with .gif or .jpg, not trying to see if a file is a GIF or JPEG.

In that case you'd want to use file_exists to go through the possible filenames until you find a match, because you don't need to read a file if you just want to see if it's there. After all, what would you do if the files were images?

You can then use getImageSize() to check that the file extension is correct.