PHPunit模拟对象抽象和静态方法 [英] PHPunit mockobject abstract and static method

问题描述

我想测试一个抽象类中的方法.在此类中，有一个带有static的抽象方法.

I would like to test a method from an abstract class. In this class is there a abstract method with is static.

我使用PHPUnit.使用普通的抽象方法，它可以工作:

I use PHPUnit. With normal abstract methods it works:

<?php
abstract class AbstractClass
{
  public function concreteMethod()
  {
    return $this->abstractMethod();
  }

  public abstract function abstractMethod();
}

class AbstractClassTest extends PHPUnit_Framework_TestCase
{
  public function testConcreteMethod()
  {
    $stub = $this->getMockForAbstractClass('AbstractClass');
    $stub->expects($this->any())
         ->method('abstractMethod')
         ->will($this->returnValue(TRUE));

    $this->assertTrue($stub->concreteMethod());
  }
}
?>

phpunit file.php可以工作.

phpunit file.php works.

但是如果abstractMethod是静态的，它将显示:

But if the abstractMethod is static it displays:

PHP致命错误:类Mock_AbstractClass_6332ae11包含1个抽象方法，因此必须声明为抽象或实现/usr/local/apache2/php5.3/lib/php/PHPUnit/Framework/中的其余方法(AbstractClass :: abstractMethod) TestCase.php(1135):第33行上的eval()代码

PHP Fatal error: Class Mock_AbstractClass_6332ae11 contains 1 abstract method and must therefore be declared abstract or implement the remaining methods (AbstractClass::abstractMethod) in /usr/local/apache2/php5.3/lib/php/PHPUnit/Framework/TestCase.php(1135) : eval()'d code on line 33

推荐答案

您不能拥有抽象的静态方法.它将在PHP中生成一条E_STRICT消息.

You can't have abstract static methods. It will generate an E_STRICT message in PHP.

为您的类实现设计替代策略.

Devise an alternative strategy for your class implementation.

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