我有一个 PHPUnit 模拟对象,它返回 'return value'
无论它的参数是什么:
I've got a PHPUnit mock object that returns 'return value'
no matter what its arguments:
// From inside a test...
$mock = $this->getMock('myObject', 'methodToMock');
$mock->expects($this->any))
->method('methodToMock')
->will($this->returnValue('return value'));
我希望能够根据传递给模拟方法的参数返回不同的值.我尝试过类似的方法:
What I want to be able to do is return a different value based on the arguments passed to the mock method. I've tried something like:
$mock = $this->getMock('myObject', 'methodToMock');
// methodToMock('one')
$mock->expects($this->any))
->method('methodToMock')
->with($this->equalTo('one'))
->will($this->returnValue('method called with argument "one"'));
// methodToMock('two')
$mock->expects($this->any))
->method('methodToMock')
->with($this->equalTo('two'))
->will($this->returnValue('method called with argument "two"'));
但是如果没有使用参数 'two'
调用 mock,这会导致 PHPUnit 抱怨,所以我假设 methodToMock('two')
的定义覆盖第一个的定义.
But this causes PHPUnit to complain if the mock isn't called with the argument 'two'
, so I assume that the definition of methodToMock('two')
overwrites the definition of the first.
所以我的问题是:有没有办法让 PHPUnit 模拟对象根据其参数返回不同的值?如果有,怎么做?
So my question is: Is there any way to get a PHPUnit mock object to return a different value based on its arguments? And if so, how?
使用回调.例如(直接来自 PHPUnit 文档):
Use a callback. e.g. (straight from PHPUnit documentation):
<?php
class StubTest extends PHPUnit_Framework_TestCase
{
public function testReturnCallbackStub()
{
$stub = $this->getMock(
'SomeClass', array('doSomething')
);
$stub->expects($this->any())
->method('doSomething')
->will($this->returnCallback('callback'));
// $stub->doSomething() returns callback(...)
}
}
function callback() {
$args = func_get_args();
// ...
}
?>
在 callback() 中做任何你想做的处理,并根据你的 $args 适当地返回结果.
Do whatever processing you want in the callback() and return the result based on your $args as appropriate.
这篇关于如何让 PHPUnit MockObjects 根据参数返回不同的值?的文章就介绍到这了,希望我们推荐的答案对大家有所帮助,也希望大家多多支持跟版网!