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Home/ Questions/Q 1048589
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Editorial Team
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Editorial Team
Asked: May 16, 20262026-05-16T16:28:53+00:00 2026-05-16T16:28:53+00:00

For example: $file = fopen(File.txt, r); $filename = $file->basename; if there was a method

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For example:

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

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

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  1. Editorial Team
    Editorial Team
    2026-05-16T16:28:54+00:00Added an answer on May 16, 2026 at 4:28 pm

    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.

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