tags:

views:

21

answers:

2

For example:

$file = fopen("File.txt", "r");
$filename = $file->basename;

if there was a method like basename for file objects (file pointer resources).

+4  A: 

No, there is not.

By the way, in which scenario is this thing needed?

shamittomar
A: 

No, there is not a method to do. You should rather store the filename in a variable, like this:

<?php
$filename = "File.txt";
$file = fopen($filename, "r");
$basename = basename($filename);

Also, a little side note: a file pointer is not an object, it is a resource, which you can see by passing it to var_dump() (it would output something like resource(3) of type (stream)). This means that you cannot use it directly, you would have to use functions from the PHP core or a PHP extension to handle it. In the case of file pointers, you would use functions like fread(), fwrite() and fclose() to do so.

Frxstrem