| Conditions | 1 |
| Paths | 1 |
| Total Lines | 21 |
| Code Lines | 13 |
| Lines | 0 |
| Ratio | 0 % |
| Changes | 2 | ||
| Bugs | 0 | Features | 1 |
| 1 | <?php |
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| 21 | public function testCreateUserGetsHandled() |
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| 22 | { |
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| 23 | $handler = new CreateUserHandler($this->userRepository); |
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| 24 | |||
| 25 | $user = new CreateUser('sumo', 'randomPassword', 'sumocoders'); |
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| 26 | |||
| 27 | $handler->handle($user); |
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| 28 | |||
| 29 | $this->assertEquals( |
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| 30 | 'sumo', |
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| 31 | $this->userRepository->findByUsername('sumo')->getUsername() |
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| 32 | ); |
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| 33 | $this->assertEquals( |
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| 34 | 'sumocoders', |
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| 35 | $this->userRepository->findByUsername('sumo')->getDisplayName() |
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| 36 | ); |
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| 37 | $this->assertEquals( |
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| 38 | 'randomPassword', |
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| 39 | $this->userRepository->findByUsername('sumo')->getPassword() |
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| 40 | ); |
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| 41 | } |
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| 42 | } |
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| 43 |
Let’s take a look at an example:
In the above example, the authenticate() method works fine as long as you just pass instances of MyUser. However, if you now also want to pass a different implementation of User which does not have a getDisplayName() method, the code will break.
Available Fixes
Change the type-hint for the parameter:
Add an additional type-check:
Add the method to the interface: