Question

我有2个测试类，所有测试运行良好。但问题是我不知道如何重用其他测试类方法。我很抱歉是Phpunit的新手，我已经在搜索了，我找不到解决方案。

class A_Test {

    public function create_something_1()
    {
        $this->visit('/a/create')
             ->type('Name','name')
             ->press('Submit')
             ->seeInDatabase('something_1', ['name' => 'Name']);
        $data = \App\Models\Something1::first();

        return $data;
    }

}


class B_Test {

    public function create_something_2()
    {

        // The problem is here, I result of new A_Test always null
        $a = (new A_Test)->create_something_1();

        $this->visit('/b/create')
             ->select($a->id, 'something_1')
             ->type('New Name','name')
             ->press('Submit')
             ->seeInDatabase('something_2', ['name' => 'New Name']);

    }

}

因此，为了使B_Test工作，我总是像这样

复制A_Test类中的方法

class B_Test {

    public function create_something_1()
    {
        $this->visit('/a/create')
             ->type('Name','name')
             ->press('Submit')
             ->seeInDatabase('something_1', ['name' => 'Name']);
        $data = \App\Models\Something1::first();

        return $data;
    }

    public function create_something_2()
    {
        // This is work
        $a = $this->create_something_1();

        $this->visit('/b/create')
             ->select($a->id, 'something_1')
             ->type('New Name','name')
             ->press('Submit')
             ->seeInDatabase('something_2', ['name' => 'New Name']);

    }

}

Answer 1

在tests/TestCase.php文件中，您可以执行此操作：

class TestCase extends Illuminate\Foundation\Testing\TestCase
{
    ....
    protected function create_something_resusable()
    {
        $this->visit('/a/create')
             ->type('Name','name')
             ->press('Submit')
             ->seeInDatabase('something_1', ['name' => 'Name']);
        $data = \App\Models\Something1::first();

        return $data;
    }
    ....
}

然后所有测试都可以调用$this->create_something_resusable()，但请记住在所有测试中扩展TestCase类。

如何在Laravel中重用phpunit测试方法

1 个答案: