1 | <?php |
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8 | class User extends Model |
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9 | { |
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10 | |||
11 | /** @var int */ |
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12 | protected $id; |
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13 | |||
14 | /** @var string */ |
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15 | protected $username; |
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16 | |||
17 | /** |
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18 | * Create a new User given a username. |
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19 | * @param string $username |
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20 | */ |
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21 | public function __construct($username) |
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25 | |||
26 | /** |
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27 | * Get the username. |
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28 | * @return string |
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29 | */ |
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30 | public function getUsername() |
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34 | |||
35 | /** |
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36 | * Get the user's ID on the given project. |
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37 | * @param Project $project |
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38 | * @return int |
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39 | */ |
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40 | public function getId(Project $project) |
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44 | |||
45 | /** |
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46 | * Get a list of this user's groups on the given project. |
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47 | */ |
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48 | public function getGroups(Project $project) |
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51 | |||
52 | /** |
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53 | * Get the full URL to Special:UserRights for this user on the given project. |
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54 | * @param Project $project |
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55 | * @return string |
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56 | */ |
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57 | public function userRightsUrl(Project $project) |
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62 | } |
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63 |
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 sub-classes 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 parent class: