How to slice an input value? [duplicate] - php

Question

This question already has answers here:
How do I get a file name from a full path with PHP?
(14 answers)
Closed 8 years ago.
My input value is just like this: Oppa/upload/default.jpeg
I want to slice the value of an input according by / cause i want to get the image file name. Does anyone know some tricks to do this?
example: i want to get default.png
<input type="text" value="Oppa/upload/default.png" id="fileLink" name="fileLink" />

Answer 1

Use basename():
$path = "Oppa/upload/default.jpeg";
echo basename($path); //will output "default.jpeg"
echo basename($path, '.jpeg'); //will output "default"
The first parameter is the path of which the trailing component will be removed. If the first parameter ends in the optional second parameter, the second parameter will also be cut off.
On Windows, both slash (/) and backslash (\) are used as directory
separator character. In other environments, it is the forward slash
(/).
- PHP manual

Answer 2

You should use basename() PHP function.

Answer 3

This will work for you
$path = "Oppa/upload/default.jpeg";
echo basename($path);

Answer 4

Use pathinfo() php function
$path = "http://domain.tld/Oppa/upload/default.png";
$info = pathinfo ( $path, PATHINFO_BASENAME ); // returns default.png

Answer 5

Yet another solution, using preg_match()
<?php
$path = "http://domain.tld/Oppa/upload/default.png"; // or "C:\\domain.tld\Oppa\upload\default.jpg";
$pattern = '/[\/|\\\\]((?:.(?!\/|\\\\))+)$/';
if(preg_match($pattern, $path, $matches)){
echo $matches[1]; // default.png or default.jpg
}
?>
Note: People claim to have problems using basename() with asian characters

Answer 6

I am supposing that if your image path will not be change from Oppa/upload/ than this Should work using explode :
$str = "Oppa/upload/default.jpeg";
$s= explode("Oppa/upload/",$str);
echo $s[1];
Another Best thing you can do with defining the relative path as a constant, so :
const path = "Oppa/upload/";
$str = "Oppa/upload/default.jpeg";
$s= explode(path,$str);
echo $s[1];
will also work.