我的类中有一个generate()
方法,它只是创建
类实例的简写方式。它接受一个 request
,该请求在方法上有类型提示。我正在尝试对此进行单元测试,我知道的唯一方法是 make
一个 answer
并通过它。这不起作用,因为它不是 request
。有解决办法吗?方法如下:
public static function generate(Question $question, Request $request): self
{
return self::create([
'user_id' => Auth::user()->getKey(),
'question_id' => $question->getKey(),
'answer_body' => $request->answer_body,
]);
}
这里是测试
/** @test */
public function it_can_generate_a_new_instance()
{
$user = factory(User::class)->create();
$this->actingAs($user);
$question = factory(Question::class)->create();
$answer = factory(Answer::class)->make();
Answer::generate($question, $answer);
$this->assertEquals($user->getKey(), Answer::first()->user_id);
$this->assertEquals($question->getKey(), Answer::first()->question_id);
$this->assertEquals($answer->answer_body, Answer::first()->answer_body);
}
测试通过,直到我在方法中键入提示 Request
。
最佳答案
您可以使用给定的属性创建一个新的请求对象。它可能有点脆弱,但它应该可以工作:
public function it_can_generate_a_new_instance()
{
$user = factory(User::class)->create();
$this->actingAs($user);
$question = factory(Question::class)->create();
$answer = factory(Answer::class)->make();
$request = new Request([ 'answer_body' => $answer->answer_body ]);
Answer::generate($question, $request);
$this->assertEquals($user->getKey(), Answer::first()->user_id);
$this->assertEquals($question->getKey(), Answer::first()->question_id);
$this->assertEquals($answer->answer_body, Answer::first()->answer_body);
}
关于php - 通过测试中的方法传递请求,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/57461937/