I need to mock a php function

In phpunit with Mockery:

I had the same problem a few months ago. This is what worked for me:

You could namespace your test with the namespace of the class under test. Then, before the test class, create the function in that namespace, like this:

<?php namespace The\Same\Like\The\Class\Under\Test

use \Mockery;

function shell_exec($cmd)
{
  return YourTest::$functions->shell_exec($cmd);
}

class YourTest extends \PHPUnit_Framework_TestCase {
  public static $functions;

  public function setUp()
  {
    self::$functions = Mockery::mock();
  }
}

Now, when you're testing your class, and want to mock the function, you could write it like this:

    self::$functions->shouldReceive('shell_exec')->with($cmd)->once();

Or in whatever way you want to mock it.

In phpspec:

This is exactly why I didn't use phpspec when I had this problem.

I think (but not sure) that with a phpspec approach in mind, you would wrap the shell command in a class, like this:

class Shell{
  public function exec($cmd)
  {
    return shell_exec($cmd);
  }
}

I guess, there's no need to test this class, since it's just a wrapper. Then, you would use this class instead of the function in your other classes, and when it comes to mocking, you could mock this class.

thanks just found this exact answer as well, great!

Instead of defining the namespaced function shell_exec() yourself you could do that with php-mock-phpunit . This has the advantage that you didn't create a side effect with your test, as the mock will be destroyed after a test:

<?php

namespace foo;

use phpmock\phpunit\PHPMock;

class ShellExecTest extends \PHPUnit_Framework_TestCase
{

    use PHPMock;
    
    public function testShellExec()
    {
        $exec = $this->getFunctionMock(__NAMESPACE__, "shell_exec");
        $exec->expects($this->once())->with("foo")->willReturn("bar");

        $this->assertEquals("bar", shell_exec("foo"));
    }
}

There are also php-mock integrations with other dialects:

• php-mock-mockery
• php-mock-prophecy