Question

考虑一个存储用户信息的简单类：

<?php
class UserStore {
    private $users = array();

    function addUser($name, $mail, $pass) {
        if (isset($this->users['mail'])) {
            throw new Exception("User {$mail} already in system.\n");
        }

        if (strlen($pass) < 5) {
            throw new Exception("Password must have 5 or more letters.");
        }

        $this->users[$mail] = 
        array(
            'pass' => $pass,
            'mail' => $mail,
            'name' => $name,
        );

        return true;
    }   

    function notifyPasswordFailure($mail) {
        if(isset($this->users[$mail])) {
            $this->users[$mail]['failed'] = time();
        }
    }

    function getUser($mail) {
        return $this->users[$mail];
    }
}

这是我们的测试用例，以确保课程不会出现重复的电子邮件ID：

<?php
class UserStoreTest extends PHPUnit_Framework_TestCase {
    private $store;

    public function setUp() {
        $this->store = new UserStore();
    }

    public function tearDown() {

    }

    public function testAddUserDuplicate() {
        try {
            $ret = $this->store->addUser("Bob", "a@b.com", "123456");
            $ret = $this->store->addUser("Bill", "a@b.com", "123456");
            self::fail('Exception should\'ve been thrown.');
        } catch (Exception $e) {
            $const = $this->logicalAnd(
                    $this->logicalNot($this->contains("Bill")), 
                    $this->isType('array')
                );
            self::AssertThat($this->store->getUser("a@b.com"), $const);
        }
    }
}

这个例子来自一本书。逻辑似乎很简单：一旦在添加重复用户时抛出异常，我们确保getUser()不会给第二个用户。所以我运行此测试并得到以下错误：

有1次失败：

1) UserStoreTest::testAddUserDuplicate
Failed asserting that Array (
    'pass' => '123456'
    'mail' => 'a@b.com'
    'name' => 'Bill'
) does not contain 'Bill' and is of type "array".

WTF？测试失败了！怎么样？查看测试输出，我看到一个名为Bill的数组。这怎么可能？我看到它的方式，Bill从未添加到用户，因为抛出异常，那么为什么我们在输出中看到它？我在理解PHPUnit /这个例子时犯了错误，或者这本书的例子是错误的。请帮忙！

Answer 1

您的addUser方法中存在拼写错误 - 应该是

if (isset($this->users[$mail])) {
    throw new Exception("User {$mail} already in system.\n");
}

顺便说一句，我认为这是一个糟糕的测试，因为你甚至无法从第一个视图中得到错误：）

无法理解这一点并且拒绝重复的用户＆＃39;测试

1 个答案: