PHPUnit Testing版本的assertIsA

本文介绍了PHPUnit Testing版本的assertIsA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

PHPUnit是否有一个断言来检查值的类型

Does PHPUnit have an assertion that checks the type of a value

功能:

public function getTaxRate()
{
    return 21;
}

我要测试返回的值是一个数字.

I want to test that the value returned is a number.

对不起，但是我是PHPUnit测试的新手.

Sorry but i am new to PHPUnit testing.

我发现SimpleTest具有assertIsA(); PHPUnit有类似的东西吗?

I found that SimpleTest has assertIsA(); is there something similar for PHPUnit.

致谢

推荐答案

在像php这样的弱类型语言中，是数字"的概念有些模糊.在php中，1 + "1"是2.字符串"1"是数字吗?

The notion that something "is a number" is a little fuzzy in weakly typed languages like php. In php, 1 + "1" is 2. Is the string "1" a number?

Phpunit断言assertInternalType() 可能会帮助您:

The Phpunit assertion assertInternalType() might help you:

$actual = $subject->getTaxRate();
$this->assertIternalType('int', $actual);

但是您不能将断言与逻辑运算符结合在一起.因此，您不能轻易表达断言42.0是整数还是浮点数"的想法.可以将此类更密集的断言分组为私有帮助程序断言方法:

But you can't combine assertions with logical operators. So you can't easily express the idea "assert that 42.0 is either an integer or a float". Such more intensive assertions can be grouped into a private helper assertion method:

private function assertNumber($actual, $message = "Is a number")  {
    $isScalar = is_scalar($actual);
    $isNumber = $isScalar && (is_int($actual) || is_float($actual));
    $this->assertTrue($isNumber, $message);
}

然后，您只需在同一测试用例类中的测试中使用它即可:

And then you just use it in your tests within the same testcase class:

$actual = $subject->getTaxRate();
$this->assertNumber($actual);

您也可以编写自己的自定义断言.如果您需要经常运行数字伪类型的断言，这可能是最好的选择，除非我错过了一些事情.请参见扩展PHPUnit ，其中显示了该操作的完成方式.

You can write your own custom assertion as well. This is probably your best bet if you need to run a number-pseudo-type assertion often, unless I've missed something. See Extending PHPUnit which shows how that is done.