php 如何断言结果是 PHPUnit 中的整数？
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How do I assert the result is an integer in PHPUnit?
提问by bnp887
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:
我原以为它会是这样的：
$this->assertInstanceOf('int', $new_id);
But I get this error:
但我收到此错误：
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
回答by Farid Movsumov
$this->assertInternalType("int", $id);
Edit: As of PHPUnit 8, the answer is:
编辑：从PHPUnit 8 开始，答案是：
$this->assertIsInt($id);
回答by Francesco Casula
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:
或者在撰写本文时最新的是v7.0：
use PHPUnit\Framework\Constraint\IsType;
// ...
$this->assertInternalType(IsType::TYPE_INT, $new_id);
回答by Mike Brant
Original answer is given below for posterity, but I would strongly recommend using assertInternalType()as suggested in other answers.
下面给出了原始答案供后代使用，但我强烈建议assertInternalType()按照其他答案中的建议使用。
Original answer:
原答案：
Simply use assertTrue with is_int().
只需将 assertTrue 与 is_int() 一起使用。
$this->assertTrue(is_int($new_id));
回答by Kudashev Serge
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 不能为负数或零），这更加严格和安全。
回答by ShadowBeast
As of PHPUnit 8, The other approaches are now deprecated, assertInternalType()will now throw this error and fail:
从 PHPUnit 8 开始，其他方法现已弃用，assertInternalType()现在将抛出此错误并失败：
assertInternalType() is deprecated and will be removed in PHPUnit 9. Refactor your test to use assertIsArray(), assertIsBool(), assertIsFloat(), assertIsInt(), assertIsNumeric(), assertIsObject(), assertIsResource(), assertIsString(), assertIsScalar(), assertIsCallable(), or assertIsIterable() instead.
assertInternalType() 已弃用，将在 PHPUnit 9 中移除。重构您的测试以使用 assertIsArray()、assertIsBool()、assertIsFloat()、assertIsInt()、assertIsNumeric()、assertIsObject()、assertIsResource()、assertIsString()、改为 assertIsScalar()、assertIsCallable() 或 assertIsIterable()。
It is now recommended to use assertIsNumeric()or assertIsInt()for this purpose.
现在建议使用assertIsNumeric()或assertIsInt()用于此目的。
回答by iateadonut
This answer is for people wondering how to make sure that the result is EITHER an integer or a string of integers. (The issue appears in some comments):
这个答案适用于想知道如何确保结果是整数或整数字符串的人们。（该问题出现在一些评论中）：
$this->assertEquals( 1 , preg_match( '/^[0-9]+$/', $new_id ),
$new_id . ' is not a set of digits' );
Here we make sure preg_match returns a '1', where preg_match is testing that all string characters are digits.
在这里，我们确保 preg_match 返回“1”，其中 preg_match 测试所有字符串字符都是数字。
