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Home/ Questions/Q 8424163
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Editorial Team
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Editorial Team
Asked: June 10, 20262026-06-10T03:54:53+00:00 2026-06-10T03:54:53+00:00

Consider the following scenario (this is not production code): class MyClass { public function

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Consider the following scenario (this is not production code):

 class MyClass {
    public function myMethod() {
        // create a directory
        $path = sys_get_temp_dir() . '/' . md5(rand());
        if(!mkdir($path)) {
            throw new Exception("mkdir() failed.");
        }

        // create a file in that folder
        $myFile = fopen("$path/myFile.txt", "w");
        if(!$myFile) {
            throw new Exception("Cannot open file handle.");
        }
    }
}

Right, so what’s the problem? Code coverage reports that this line is not covered:

throw new Exception("Cannot open file handle.");

Which is correct, but since I’m creating the folder above logically it would seem impossible for the fopen() to fail (except maybe in extreme circumstances, like disk at 100 %).

I could ignore the code from code coverage but thats kind of cheating. Is there any way I can mock the file system so that it can recognise myFile.txt and mock the file system unable to create the file?

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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-06-10T03:54:54+00:00Added an answer on June 10, 2026 at 3:54 am

    vfsStream is a stream wrapper for a virtual filesystem that is useful in unit tests to mock the real filesystem. You can install it from composer.

    More Info at:

    https://github.com/mikey179/vfsStream

    https://phpunit.de/manual/current/en/test-doubles.html

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