CODEWITHSUNDEEP
phpunit-testingphpunitphp-7.2
joe gates
joe gates

Reputation: 499

creating mock phpunit discrepency in mocked namespace and the original one

I have function like this:

class Bar{
    public function a():Foo{
    .
    .
    .
    }
}

now I am trying to create a mock for the class Bar with php unit test

    $mockedBar = $this->getMockBuilder(Bar::class)
     ->getMock()
     ->method('a')
     ->willReturn(new FakeFoo());

but when I am calling method a I am getting an error that method a return type must be instance of Foo not Mocked_blahblah.

unfortunately class Bar don't use any interface and the system is very big and I can't create an interface cause it make huge refactor in my codes; is there any way to disable return type of function a in mocked object? I am useing php7.2 and phpunit 6.0.13. Here is a real scenario:

class A
{
    public function b():B
    {
        echo "i am from class A function b";
    }
}
class B
{

}

class FakeB
{
}
class ATest extends TestCase
{
    public function testSayHi(){
        $mockedA = $this->getMockBuilder(A::class)
            ->getMock();
        $mockedA->method('b')->willReturn(new FakeB());
        $mockedA->b();
    }
}

Upvotes: 1

Views: 281

Answers (1)

Jakub Zalas
Jakub Zalas

Reputation: 36231

You can't disable return types. Perhaps you could try to do it with some kind of a hackish error handler, but it's a crazy thing to do.

Good news is that you're not trying to do anything unusual and your tests can be fixed.

Firstly, you need to assign the result of getMock to a variable. Next, you can define your test double:

class MyTest extends TestCase
{
    public function testIt()
    {
        $mockedBar = $this->getMockBuilder(Bar::class)->getMock();
        $mockedBar
             ->method('a')
             ->willReturn(new FakeFoo());

        $this->assertInstanceOf(Foo::class, $mockedBar->a());
    }
}

This will only work if FakeFoo is of type of Foo, for example extends it:

class FakeFoo extends Foo
{
    // override any Foo methods you'd like to fake
}

You don't need to create a Fake yourself, you can use PHPUnit to create a dummy:

class MyTest extends TestCase
{
    public function testIt()
    {
        $mockedBar = $this->createMock(Bar::class);
        $mockedBar
             ->method('a')
             ->willReturn($this->createMock(Foo::class));

        $this->assertInstanceOf(Foo::class, $mockedBar->a());
    }
}

To fix your second example:

class A
{
    public function b():B
    {
        echo "i am from class A function b";
    }
}

class B
{
}

class FakeB extends B
{
}

class ATest extends TestCase
{
    public function testSayHi(){
        $mockedA = $this->getMockBuilder(A::class)->getMock();
        $mockedA->method('b')->willReturn(new FakeB());
        $mockedA->b();
    }
}

Or, instead of using a fake let phpunit handle it:

class ATest extends TestCase
{
    public function testSayHi(){
        $mockedA = $this->getMockBuilder(A::class)->getMock();
        $dummyB = $this->createMock(B::class);

        $mockedA->method('b')->willReturn($dummyB);
        $mockedA->b();
    }
}

Upvotes: 1

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