PHP7中的PHPSpec捕获TypeError [英] PHPSpec Catching TypeError in PHP7

问题描述

我想在PHP7中测试带有标量类型提示和严格类型的示例方法.当我不传递参数时，该方法应抛出TypeError. PHPSpec返回致命错误:

I want test an example method with scalar type hinting and strict types in PHP7. When I don't pass an argument, the method should throw a TypeError. PHPSpec return fatal error:

未捕获的TypeError:参数1传递给Example :: test

Uncaught TypeError: Argument 1 passed to Example::test

<?php

class Example
{
    public function test(string $name)
    {
        $this->name = $name;
    }
}


class ExampleSpec extends ObjectBehavior
{
    function it_is_initializable()
    {
        $this->shouldHaveType('Test\Example');
    }

    function it_check_test_method_when_not_pass_argument()
    {
        $this->shouldThrow('\TypeError')->during('test');
    }
}

一开始我声明:declare(strict_types=1);

怎么了?如何测试抛出TypeError?

What is wrong? How do I test throwing TypeError?

推荐答案

对我来说，如果我用以下注释单元测试，它就可以工作:

For me it works if I annotate the unit test with this:

/**
 * @expectedException \TypeError
 */

那我的测试是绿色的.

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