PHPUnit的returnValueMap没有产生预期的结果

问题描述

我正在尝试使用PHPUnit的可对读取结果进行存根。它没有产生预期的结果，但是等效的returnCallback（）可以。如果您想自己检查，我已经提供了。

I'm trying to use PHPUnit's returnValueMap() to stub out the results of a read. It isn't yielding the expected results, but an equivalent returnCallback() does. I've made my test case available if you'd like to inspect it yourself.

returnValueMap（）

$enterprise = $this->getMock('Enterprise', array('field'));
$enterprise->expects($this->any())
    ->method('field')
    ->will($this->returnValueMap(array(
        array('subscription_id', null),
        array('name', 'Monday Farms')
    )));
$enterprise->subscribe('basic');

结果：

Subscription ID: NULL
Name: NULL

returnCallback（ ）

$enterprise = $this->getMock('Enterprise', array('field'));
$enterprise->expects($this->any())
    ->method('field')
    ->will($this->returnCallback(function ($arg) {
        $map = array(
            'subscription_id' => null,
            'name' => 'Monday Farms'
        );
        return $map[$arg];
    }));
$enterprise->subscribe('basic');

结果：

Subscription ID: NULL
Name: string(12) "Monday Farms"

企业：：订阅

public function subscribe() {
    echo 'Subscription ID: ';
    var_dump($this->field('subscription_id'));
    echo 'Name: ';
    var_dump($this->field('name'));
}

为什么returnValueMap（）不能按我预期的那样工作？我到底缺少什么？

Why doesn't returnValueMap() work as I expect it to? What exactly am I missing?

推荐答案

我遇到了同样的问题，最终发现returnValueMap（）必须映射所有参数。您的函数，包括可选的函数，然后是所需的返回值。

I had the same problem and eventually found out that returnValueMap() has to map all parameters of your function, including optional ones, then the desired return value.

Zend Framework中的示例函数：

Example function from Zend Framework:

public function getParam($key, $default = null)
{
    $key = (string) $key;
    if (isset($this->_params[$key])) {
        return $this->_params[$key];
    }

    return $default;
}

必须这样映射：

$request->expects($this->any())
        ->method('getParam')
        ->will($this->returnValueMap(array(array($param, null, $value))));

如果中间没有空值，它将不起作用。

Without the null in the middle, it won't work.

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