如果我只有文件名，如何在PHP中获得文件扩展名？

Question

I have the path to an image file /files/uploads/1 but as you can see, I don't have an extension so I can't display the image. How can I get the extension of the file so I can display it? Thanks!

EDIT: Here is come code I have tried. BMP, PNG, JPG, JPEG and GIF are the only possible extensions, but $path ends up never getting assigned a value.

$exts = array('bmp','png','jpg','jpeg','gif');
foreach ($exts as $ext) {
    if (file_exists("/files/uploads/" . $id . "." . $ext)) {
        $path = "/files/uploads/" . $id . "." . $ext;
    }
}

Answer 1

Personally, I had a problem where I had an unnecessary slash (/) before my path in the file_exists check, but the solution to the question is still the same.

$exts = array('bmp','png','jpg','jpeg','gif','swf','psd','tiff','jpc','jp2','jpx','jb2','swc','iff','wbmp','xbm','ico');
foreach ($exts as $ext) {
    if (file_exists("files/uploads/" . $id . "." . $ext)) {
        $path = "/files/uploads/" . $id . "." . $ext;
    }
}

Answer 2

You don't need a file extension to show an image.

<img src="/files/uploads/1">

Will show the image. It's the same way the placehold.it images work

<img src="http://placehold.it/350x150">

Answer 3

First of all , I think when you get the file name for the image that means no need to add extension , so you have the file name and image folder

it should simply look like this :

$fullpath = ('uploadword/'.$filename);
// then call it 
<img  src='".$fullpath."'> 

The normal scenario to upload all the images in specific folder then get the filename and save it in data base

move_uploaded_file($file_tmp,"upload/".$filename);

Some professional make it more sophisticated to make filename unique ,because they expected some duplicated values and many reasons , here is one simple technique :

$anynameorfunction = "";// random number etc...
$newname = $anynameorfunction.$filename;
move_uploaded_file($file_tmp,"upload/".$newname);

So, when they call it again it should be simple like this

  $fullpath = ('uploadword/'.$newname);
 // then call it 
  <img  src='".$newname."'> 

This is in very simple way and i wish you get what i mean