2011-05-16 10:04
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Trying to do a simple mock of Zend_Config for a method that requires a Zend_Config object type, but the mock returns a type of Mock_Zend_Config.

Surely I missed something at this late hour and I am obviously wrong in the function call but I fail to spot my error.

$config = $this->getMock("Zend_Config"); 

Returns Mock_Zend_Config, and my object needs to be of type Zend_Config. Looked up the function signature in a cheatsheet and changed the method call to:

$config = $this->getMock("Zend_Config", array(), array($confArray),"Zend_Config",true); 

This version generates a fatal error with message "Zend_Config already exists".

On a sidenote and probably not related to phpunit as such but the typehint does not generate a fatal error as it should , and does so when run without tests.

Any idea of what I'm missing in la mock?

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尝试为需要Zend_Config对象类型的方法执行Zend_Config的简单模拟,但mock返回一个类型 Mock_Zend_Config。


   $ config = $ this-> getMock(“Zend_Config”);  

返回Mock_Zend_Config,我的对象需要是Zend_Config类型。 在备忘单中查找函数签名并将方法调用更改为:

  $ config = $ this-> getMock(“Zend_Config”,array(),array($  confArray), “Zend_Config的”,真正的);  


在旁注中,可能与 phpunit就是这样,但是typehint不会产生致命错误,并且在没有测试的情况下运行。

知道我在la mock中缺少什么吗?

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  • douzi0609 2011-05-16 10:13

    Mock objects extend the mocked object. A type hint for Zend_Config will be satisfied by any class extending Zend_Config because by definition a Mock_Zend_Config extends Zend_Config and therefor is a Zend_Config. Consequently, you will get a Fatal Error when trying to name the mock like the class it extends and none if you dont.

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