Question

我使用Yii2和内置代码，并希望编写单元测试。在代码的文档中我发现了这个例子test：

function testUserCanBeActivatedWithValidKey()
{
    // lookup for user with Eloquent API
    $user = User::find($this->user_id);
    // executing action
    $isActivated = $user->activate('123456'));
    // performing assertion
    $this->assertTrue($isActivated);
    // checking that data was actually saved into database
    $this->tester->seeRecord('users', [
        'id' => $this->user_id,
        'is_active' => 1
     ]);
}

但是当我自己尝试编写测试时，没有$this->tester对象。我如何从seeInDatabase()等方法中受益？

Answer 1

试试这个：

function testUserCanBeActivatedWithValidKey(WebGuy $I)
{
    // lookup for user with Eloquent API
    $user = User::find($this->user_id);
    // executing action
    $isActivated = $user->activate('123456'));
    // performing assertion
    $this->assertTrue($isActivated);
    // checking that data was actually saved into database
    $I->seeRecord('users', [
        'id' => $this->user_id,
        'is_active' => 1
     ]);
}

您将测试对象作为测试函数的参数。如果它是TestGuy，WebGuy或其他东西，你必须查看你的codeception.yml配置。

Answer 2

class UserTest extends \yii\codeception\TestCase
{
   /**
    * @var \CodeGuy
    */
    protected $tester;

function testUserCanBeActivatedWithValidKey()
{
    // lookup for user with Eloquent API
    $user = User::find($this->user_id);
    // executing action
    $isActivated = $user->activate('123456'));
    // performing assertion
    $this->assertTrue($isActivated);
    // checking that data was actually saved into database
    $this->tester->seeRecord('users', [
        'id' => $this->user_id,
        'is_active' => 1
     ]);
}

覆盖$ tester并开始使用它。

Yii2＆amp; Codeception - 在单元测试中启用测试器模块

2 个答案: