如何 PHPUnit 断言函数-php问题

HowTo PHPUnit assertFunction(如何 PHPUnit 断言函数)

本文介绍了如何 PHPUnit 断言函数的处理方法，对大家解决问题具有一定的参考价值，需要的朋友们下面随着小编来一起学习吧！

问题描述

我想知道如何验证类"是否具有函数.assertClassHasAttribute 不起作用，这是正常的，因为 Function 不是 Attribute.

I was wondering if how I can verify if a 'class' has a Function. assertClassHasAttribute does not work, it's normal since a Function is not an Attribute.

推荐答案

当 PHPUnit 没有提供断言方法时，我要么创建它，要么使用带有详细消息的低级断言之一:

When there's not an assertion method provided by PHPUnit I either create it or use one of the lower-level assertions with a verbose message:

$this->assertTrue(
  method_exists($myClass, 'myFunction'), 
  'Class does not have method myFunction'
);

assertTrue() 是最基础的.它提供了很大的灵活性，因为您可以使用任何内置的 php 函数来为您的测试返回 bool 值.因此，当测试失败时，错误/失败消息根本没有帮助.类似 Failed asserting that <FALSE>为真.这就是为什么将第二个参数传递给 assertTrue() 来详细说明测试失败的原因很重要.

assertTrue() is as basic as you can get. It allows a great deal of flexibility because you can use any built-in php function that returns a bool value for your test. Consequently, when the test fails the error/failure message isn't helpful at all. Something like Failed asserting that <FALSE> is TRUE. That's why it's important to pass the second param to assertTrue() detailing why the test failed.

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