This package integrates the function mock library PHP-Mock with PHPUnit.
Use Composer:
composer require --dev php-mock/php-mock-phpunit
PHP-Mock integrates with the trait
PHPMock
into your PHPUnit test case. This trait extends the framework
by the method
getFunctionMock()
.
With this method you can build a mock in the way you are used to build a
PHPUnit mock:
namespace foo;
class BuiltinTest extends \PHPUnit\Framework\TestCase
{
use \phpmock\phpunit\PHPMock;
public function testTime()
{
$time = $this->getFunctionMock(__NAMESPACE__, "time");
$time->expects($this->once())->willReturn(3);
$this->assertEquals(3, time());
}
public function testExec()
{
$exec = $this->getFunctionMock(__NAMESPACE__, "exec");
$exec->expects($this->once())->willReturnCallback(
function ($command, &$output, &$return_var) {
$this->assertEquals("foo", $command);
$output = ["failure"];
$return_var = 1;
}
);
exec("foo", $output, $return_var);
$this->assertEquals(["failure"], $output);
$this->assertEquals(1, $return_var);
}
}
There's no need to disable the mocked function. The PHPUnit integration does that for you.
This library comes with the same restrictions as the underlying
php-mock
:
-
Only unqualified function calls in a namespace context can be mocked. E.g. a call for
time()
in the namespacefoo
is mockable, a call for\time()
is not. -
The mock has to be defined before the first call to the unqualified function in the tested class. This is documented in Bug #68541. In most cases you can ignore this restriction. But if you happen to run into this issue you can call
PHPMock::defineFunctionMock()
before that first call (e.g. with@beforeClass
). This would define a side effectless namespaced function. Another effective approach is running your test in an isolated process (e.g. with@runInSeparateProcess
).
This project is free and under the WTFPL. Responsable for this project is Markus Malkusch markus@malkusch.de.
If you like this project and feel generous donate a few Bitcoins here: 1335STSwu9hST4vcMRppEPgENMHD2r1REK