如何断言结果是PHPUnit中的整数？

I would like to be able test that a result is an integer (1,2,3...) where the function could return any number, e.g.:

我希望能够测试结果是一个整数（1,2,3 ...），其中函数可以返回任何数字，例如：

$new_id = generate_id();

I had thought it would be something like:

我以为它会是这样的：

$this->assertInstanceOf('int', $new_id);

But I get this error:

但我得到这个错误：

Argument #1 of PHPUnit_Framework_Assert::assertInstanceOf() must be a class or interface name

PHPUnit_Framework_Assert :: assertInstanceOf（）的参数＃1必须是类或接口名称

4 个解决方案

#1

$this->assertInternalType("int", $id);

#2

I prefer using the official PHPUnit class constants.

我更喜欢使用官方的PHPUnit类常量。

PHPUnit v5.2:

PHPUnit v5.2：

use PHPUnit_Framework_Constraint_IsType as PHPUnit_IsType;

// ...

$this->assertInternalType(PHPUnit_IsType::TYPE_INT, $new_id);

Or latest which at the moment of writing is v7.0:

或者最新的写作时刻是v7.0：

use PHPUnit\Framework\Constraint\IsType;

// ...

$this->assertInternalType(IsType::TYPE_INT, $new_id);

#3

Original answer is given below for posterity, but I would strongly recommend using assertInternalType() as suggested in other answers.

下面给出后代的原始答案，但我强烈建议使用其他答案中建议的assertInternalType（）。

Original answer:

原始答案：

Simply use assertTrue with is_int().

只需将assertTrue与is_int（）一起使用即可。

$this->assertTrue(is_int($new_id));

#4

I think better use this construction:

我认为最好使用这种结构：

$this->assertThat($new_id, $this->logicalAnd(
    $this->isType('int'), 
    $this->greaterThan(0)
));

because it will check not only the type of $new_id variable, but will check if this variable is greater than 0 (assume id cannot be negative or zero) and this is more strict and secure.

因为它不仅会检查$ new_id变量的类型，还会检查这个变量是否大于0（假设id不能为负或零），这更加严格和安全。

#1

$this->assertInternalType("int", $id);

#2