我正在为现有代码编写单元测试,如下所示
class someClass {
public function __construct() { ... }
public function someFoo($var) {
...
$var = "something";
...
$model = new someClass();
model->someOtherFoo($var);
}
public someOtherFoo($var){
// some code which has to be mocked
}
}
在这里,我应该如何能够模拟函数"someOtherFoo
"的调用,这样它就不会在someOtherFoo
中执行"some code
"?
class someClassTest {
public function someFoo() {
$fixture = $this->getMock('someClass ', array('someOtherFoo'));
$var = "something";
....
// How to mock the call to someOtherFoo() here
}
}
是否可以模拟构造函数,以便它返回我自己构造的函数或变量?
谢谢
只要在测试的方法中有new XXX(...)
,您就注定要失败。将实例化提取到同一个类的新方法——createSomeClass(...)
。这允许您创建被测类的部分模拟,该模拟从新方法返回一个存根值或模拟值。
class someClass {
public function someFoo($var) {
$model = $this->createSomeClass(); // call method instead of using new
model->someOtherFoo($var);
}
public function createSomeClass() { // now you can mock this method in the test
return new someClass();
}
public function someOtherFoo($var){
// some code which has to be mocked
}
}
在测试中,在调用someFoo()
的实例中模拟createSomeClass()
,在从第一个模拟调用返回的实例中模拟someOtherFoo()
。
function testSomeFoo() {
// mock someOtherFoo() to ensure it gets the correct value for $arg
$created = $this->getMock('someClass', array('someOtherFoo'));
$created->expects($this->once())
->method('someOtherFoo')
->with('foo');
// mock createSomeClass() to return the mock above
$creator = $this->getMock('someClass', array('createSomeClass'));
$creator->expects($this->once())
->method('createSomeClass')
->will($this->returnValue($created));
// call someFoo() with the correct $arg
$creator->someFoo('foo');
}
请记住,因为实例正在创建同一个类的另一个实例,所以通常会涉及两个实例。如果更清楚的话,您可以在这里使用相同的模拟实例。
function testSomeFoo() {
$fixture = $this->getMock('someClass', array('createSomeClass', 'someOtherFoo'));
// mock createSomeClass() to return the mock
$fixture->expects($this->once())
->method('createSomeClass')
->will($this->returnValue($fixture));
// mock someOtherFoo() to ensure it gets the correct value for $arg
$fixture->expects($this->once())
->method('someOtherFoo')
->with('foo');
// call someFoo() with the correct $arg
$fixture->someFoo('foo');
}
可以用overload:
作为模拟类名的前缀
查看Mocking Hard Dependencies的文档。
你的例子应该是这样的:
/**
* @runTestsInSeparateProcesses
* @preserveGlobalState disabled
*/
class SomeClassTest extends 'PHPUnit'Framework'TestCase
{
public function test_some_foo()
{
$someOtherClassMock = 'Mockery::mock('overload:SomeOtherClass');
$someOtherClassMock->shouldReceive('someOtherFoo')
->once()
->with('something')
->andReturn();
$systemUnderTest = new SomeClass();
$systemUnderTest->someFoo('something');
}
}
我添加了@runTestsInSeparateProcesses
注释,因为通常模拟类也将在其他测试中使用。如果没有注释,自动加载程序就会因为class already exists
错误而崩溃。
如果这是您的测试套件中唯一使用mock类的地方,那么您应该删除注释。
我在这里找到了我的方法,试图白盒测试一个类__构造函数,以确保它调用自己的类方法,并将一些数据传递给__构造函数。
如果其他人出于同样的原因在这里,我想我会分享我最终使用的方法(没有在这个问题中使用的工厂式createSomeClass()方法)。
<?php
class someClass {
public function __constructor($param1) {
// here is the method in the constructor we want to call
$this->someOtherFoo($param1);
}
public function someOtherFoo($var){ }
}
现在PHPUnit测试:
<?php
$paramData = 'someData';
// set up the mock class here
$model = $this->getMock('someClass',
array('someOtherFoo'), // override the method we want to check
array($paramData) // we need to pass in a parameter to the __constructor
);
// test that someOtherFoo() is called once, with out test data
$model->expects($this->once())
->with($paramData)
->method('someOtherFoo');
// directly call the constructor, instead of doing "new someClass" like normal
$model->__construct($paramData);