[SOLVED] get image extension without knowing the extension (That's it !)

[Perfidus]

Hi there !

User's of my site can upload pictures to personal folders.

Pictures can be gif, jpg, png, whatever.

The point is: I know the folder and the name of the image but I haven't stored image extension in the DDBB because I thought it was possible to recover it later but now I find myself a little lost, because I was thinking on getimagesize(), but I can't using it without adding the extension...

Is there a way to get the extension of a file by it's name rather than storing the extension in the DDBB. And if it exists, is it better option than storing it it?

[wildteen88]

So you're uploading images, but when you save the image you remove the file extension! Why?

 

Also by DDBB do you mean Database?

[Perfidus]

No, sorry, I explained myself poorly.

The images on the server folder they have extensions, of course.

But when I upload them I bring then a name based in a numerical reference and that`s what I know, this number, but I'm not storing the extension on the DDB, only this reference.

I thought I was going to be easy to look at the folder, and check for file 7575757 and see what extension does it have, but now I find that I need to know the extension to access the file.

[Perfidus]

To make it simple, lets say:

1 - I have a folder where only images will exist (no other files)

2 - I know the names, not extensions, of this images and there wont be repeated names, not even with different extensions

3 - I want to know the extension that follows a given name in that folder

[Perfidus]

Mybe I can use an array with all allowed extensions and check if file_exists() till it returns true, but the question is:

Is it really worth? Isn't better to store the file extension with the name in the in the DDBB?

[Perfidus]

So, this is my own solution, if someone can bring some extra light over it, it will be wellcome...

$file = $path.$name.".";
$ext = array("jpg", "jpeg", "JPEG", "gif", "png", "bmp");
for($x = 0; $x < 6; $x++)
{
$image = $file.$ext[$x];
if(file_exists($image))
{
	$pic=$image;
}
}
if(isset($pic))
{
list($width, $height, $type, $attr) = getimagesize($pic);
echo "<img src=\"".$pic."\" width=\"".$width."\" height=\"".$height."\" alt=\"".$name."\" />";
}

[Daniel0]

if ($results = glob($path . $name . '.*')) {
$filename = $results[0];
}
else {
// doesn't exist...
}

 

This assumes that there will be no duplicates such that e.g. test.jpg and test.png exist though. Your code has the same limitation, however.

[Perfidus]

your code lokks clean, but it wont return the extension? Will it?

In my code finally I show the image, previously I need to get ext.

Anyway, this topic is solved.

[Daniel0]

glob returns the filenames of all files that match, so yes, it should return the extension as well.