'PHP: remove filename from path
Say I have an path: images/alphabet/abc/23345.jpg
How do I remove the file at the end from the path? So I end up with: images/aphabet/abc/
Solution 1:[1]
You want dirname()
Solution 2:[2]
dirname()only gives you the parent folder's name, sodirname()will fail wherepathinfo()will not.
For that, you should use pathinfo():
$dirname = pathinfo('images/alphabet/abc/23345.jpg', PATHINFO_DIRNAME);
The PATHINFO_DIRNAME tells pathinfo to directly return the dirname.
See some examples:
For path
images/alphabet/abc/23345.jpg, both works:<?php $dirname = dirname('images/alphabet/abc/23345.jpg'); // $dirname === 'images/alphabet/abc/' $dirname = pathinfo('images/alphabet/abc/23345.jpg', PATHINFO_DIRNAME); // $dirname === 'images/alphabet/abc/'For path
images/alphabet/abc/, wheredirnamefails:<?php $dirname = dirname('images/alphabet/abc/'); // $dirname === 'images/alphabet/' $dirname = pathinfo('images/alphabet/abc/', PATHINFO_DIRNAME); // $dirname === 'images/alphabet/abc/'
Solution 3:[3]
<?php
$path = pathinfo('images/alphabet/abc/23345.jpg');
echo $path['dirname'];
?>
Solution 4:[4]
Note that when a string contains only a filename without a path (e.g. "test.txt"), the dirname() and pathinfo() functions return a single dot (".") as a directory, instead of an empty string. And if your string ends with "/", i.e. when a string contains only path without filename, these functions ignore this ending slash and return you a parent directory. In some cases this may be undesirable behavior and you need to use something else. For example, if your path may contain only forward slashes "/", i.e. only one variant (not both slash "/" and backslash "\") then you can use this function:
function stripFileName(string $path): string
{
if (($pos = strrpos($path, '/')) !== false) {
return substr($path, 0, $pos);
} else {
return '';
}
}
Or the same thing little shorter, but less clear:
function stripFileName(string $path): string
{
return substr($path, 0, (int) strrpos($path, '/'));
}
Sources
This article follows the attribution requirements of Stack Overflow and is licensed under CC BY-SA 3.0.
Source: Stack Overflow
| Solution | Source |
|---|---|
| Solution 1 | Byron Whitlock |
| Solution 2 | |
| Solution 3 | abney317 |
| Solution 4 |
