assertIsA 的 PHP 单元测试版本 [英] PHPUnit Testing version of assertIsA

问题描述

PHPUnit 是否有一个检查值类型的断言

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

功能:

public function getTaxRate()
{
    return 21;  
}

我想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，我是 PHPUnit 测试的新手.

Sorry but i am new to PHPUnit testing.

我发现 SimpleTest 有 assertIsA();PHPUnit 有没有类似的东西.

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

问候

推荐答案

是一个数字"的概念在弱类型语言(如 php)中有点模糊.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() 可能会对您有所帮助:

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使用.所以你不能轻易表达断言 42.0 是整数或浮点数"的想法.这种更密集的断言可以归为一个私有的辅助断言方法:

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);
}

然后您只需在同一个测试用例类中的测试中使用它:

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断言，这可能是您最好的选择，除非我错过了什么.请参阅 Extending PHPUnit，它显示了这是如何完成的.

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.

相关:

• 在 PHP 中测试变量是否为数字的正确方法是什么?
• http://php.net/is_numeric

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