php 在 PHPUnit Mocks 的 returnCallback() 中修改对象
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Modifying objects in returnCallback() of PHPUnit Mocks
提问by Alex Lawrence
I want to mock a method of a class and execute a callback which modifies the object given as parameter (using PHP 5.3 with PHPUnit 3.5.5).
我想模拟一个类的方法并执行一个回调来修改作为参数给出的对象（使用 PHP 5.3 和 PHPUnit 3.5.5）。
Let′s say I have the following class:
假设我有以下课程：
class A
{
function foobar($object)
{
doSomething();
}
}
And this setup code:
这个设置代码：
$mock = $this->getMockBuilder('A')->getMock();
$mock->expects($this->any())->method('foobar')->will(
$this->returnCallback(function($object) {
$object->property = something;
}));
For some reason the object does not get modified. On var_dumping $objectI see it is the right object. Could it be that the object gets passed by value? How can I configure the mock to receive a reference?
由于某种原因，该对象没有被修改。在var_dumping$object我看到它是正确的对象。难道对象是按值传递的？如何配置模拟以接收参考？
回答by edorian
He Alex,
他亚历克斯，
i talked to Sebastian (the phpunit creator) about that problem and yes: The argument gets cloneed before it is passed to the callback.
我和 Sebastian（phpunit 创建者）讨论了这个问题，是的：参数在clone传递给回调之前被ed。
From the top of my head i can't offer you any workaround but i choose to answer anyway to at least tell you that you are doing nothing wrong and that this is expected behavior.
在我的脑海中，我无法为您提供任何解决方法，但无论如何我选择回答以至少告诉您您没有做错任何事情，这是预期的行为。
To quote Sebastians comment on IRC on why it clones the argument:
引用 Sebastians 对 IRC 的评论，解释为什么它会克隆这个论点：
It's a long-going debate between me, myself, and users of PHPUnit whether or not this is right ;-)
这是我、我自己和 PHPUnit 用户之间的长期争论，这是否正确；-)
To have something copy/pasteable:
要复制/粘贴某些内容：
Assertion 3 in this codesample will fail. (The variable is only changed in the returned object)
此代码示例中的断言 3 将失败。（变量只在返回的对象中改变）
<?php
class A
{
function foobar($o)
{
$o->x = mt_rand(5, 100);
}
}
class Test extends PHPUnit_Framework_TestCase
{
public function testFoo()
{
$mock = $this->getMock('A');
$mock->expects($this->any())
->method('foobar')
->will($this->returnCallback(function($o) { $o->x = 2; return $o; }));
$o = new StdClass;
$o->x = 1;
$this->assertEquals(1, $o->x);
$return = $mock->foobar($o);
$this->assertEquals(2, $return->x);
$this->assertEquals(2, $o->x);
}
}
Update:
更新：
Starting with PHPUnit 3.7 the cloning can be turned off. See the last argument off:
从 PHPUnit 3.7 开始，可以关闭克隆。查看最后一个参数：
public function getMock(
$originalClassName,
$methods = array(),
array $arguments = array(),
$mockClassName = '',
$callOriginalConstructor = TRUE,
$callOriginalClone = TRUE,
$callAutoload = TRUE,
$cloneArguments = FALSE
);
It might even be off by default :)
它甚至可能默认关闭:)
回答by David Harkness
The class that performs the cloning of the parameters passed to the mocked method is PHPUnit_Framework_MockObject_Invocation_Static. Looking at cloneObject()you can see that it will return the original object if the parameter's class's __clone()method is not public.
执行克隆传递给模拟方法的参数的类是PHPUnit_Framework_MockObject_Invocation_Static. 查看cloneObject()您可以看到，如果参数的类的__clone()方法不是公共的，它将返回原始对象。
If you have control over the class of the parameters objects andyou don't need to ever clone them yourself, you can add a private empty __clone()method.
如果您可以控制参数对象的类并且您不需要自己克隆它们，您可以添加一个私有的空__clone()方法。
回答by iisisrael
This is really old, but it came up at the top of a search and pointed me in the right direction, so worth updating. Since PHPUnit 6.0, use disableArgumentCloning(), like so:
这真的很旧，但它出现在搜索的顶部并为我指明了正确的方向，因此值得更新。从 PHPUnit 6.0 开始，使用disableArgumentCloning()，像这样：
return $this->getMockBuilder('A')
->disableOriginalConstructor()
->disableArgumentCloning()
->setMethods(array('foobar'))
->getMock()
;
