Completed
Pull Request — master (#13)
by
unknown
02:30
created

ChangePassword   A

Complexity

Total Complexity 1

Size/Duplication

Total Lines 25
Duplicated Lines 0 %

Coupling/Cohesion

Components 1
Dependencies 0

Importance

Changes 1
Bugs 0 Features 0
Metric Value
wmc 1
c 1
b 0
f 0
lcom 1
cbo 0
dl 0
loc 25
rs 10

1 Method

Rating   Name   Duplication   Size   Complexity  
A forUser() 0 7 1
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<?php
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namespace SumoCoders\FrameworkMultiUserBundle\DataTransferObject\Form;
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use SumoCoders\FrameworkMultiUserBundle\User\UserInterface;
0 ignored issues
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Bug introduced by
This use statement conflicts with another class in this namespace, SumoCoders\FrameworkMult...ject\Form\UserInterface.

Let’s assume that you have a directory layout like this:

.
|-- OtherDir
|   |-- Bar.php
|   `-- Foo.php
`-- SomeDir
    `-- Foo.php

and let’s assume the following content of Bar.php:

// Bar.php
namespace OtherDir;

use SomeDir\Foo; // This now conflicts the class OtherDir\Foo

If both files OtherDir/Foo.php and SomeDir/Foo.php are loaded in the same runtime, you will see a PHP error such as the following:

PHP Fatal error:  Cannot use SomeDir\Foo as Foo because the name is already in use in OtherDir/Foo.php

However, as OtherDir/Foo.php does not necessarily have to be loaded and the error is only triggered if it is loaded before OtherDir/Bar.php, this problem might go unnoticed for a while. In order to prevent this error from surfacing, you must import the namespace with a different alias:

// Bar.php
namespace OtherDir;

use SomeDir\Foo as SomeDirFoo; // There is no conflict anymore.
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use Symfony\Component\Validator\Constraints as Assert;
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class ChangePassword
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{
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    /**
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     * @var string
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     *
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     * @Assert\Length(
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     *     min = 6,
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     *     minMessage = "sumocoders.multiuserbundle.form.length"
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     * )
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     */
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    public $newPassword;
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    /**
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     * @var UserInterface
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     */
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    public $user;
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    public static function forUser(UserInterface $user)
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    {
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        $dataTransferObject = new self();
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        $dataTransferObject->user = $user;
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        return $dataTransferObject;
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    }
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}
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