How do I assert the result is an integer in PHPUnit?

Question

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.:

$new_id = generate_id();

I had thought it would be something like:

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

But I get this error:

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

Since generally ids shouldn't be zero, you'll probably also want to check that: 0 < $new_id in cases like this. — Kzqai, Jul 17 '14 at 17:21

score 126 · Accepted Answer · edited Apr 07 '19 at 16:13

126

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

Edit: As of PHPUnit 8, the answer is:

$this->assertIsInt($id);

edited Apr 07 '19 at 16:13

bnp887

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answered Feb 12 '14 at 09:38

Farid Movsumov

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@Andrew I don't have environment to test it. Could you please try and share results. – Farid Movsumov Sep 18 '18 at 11:04
2

@Andrew you probably figured it out by now. "34" doesn't work – Ela Buwa Oct 10 '18 at 23:49
Because type of "34" actually is string and it is normal – Farid Movsumov Oct 12 '18 at 10:39
`assertIsInt` method was added in version [7.5](https://phpunit.readthedocs.io/en/7.5/assertions.html#assertisint) – Raghavendra N Jan 07 '20 at 07:01

Francesco Casula · Answer 2 · 2018-02-13T18:46:50.153

28

I prefer using the official PHPUnit class constants.

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:

use PHPUnit\Framework\Constraint\IsType;

// ...

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

edited Feb 13 '18 at 18:46

answered Feb 11 '16 at 10:51

Francesco Casula

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Mike Brant · Answer 3 · 2017-06-27T13:57:54.763

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

Original answer:

Simply use assertTrue with is_int().

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

edited Jun 27 '17 at 13:57

answered Feb 09 '14 at 02:18

Mike Brant

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3

When creating tests, it's advised to go with the most specific test case. This will create better more descriptive errors. – Logan Bailey Aug 17 '16 at 19:27

score 5 · Answer 4 · answered Mar 22 '18 at 20:37

I think better use this construction:

$this->assertThat($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.

score 4 · Answer 5 · edited Apr 07 '19 at 19:18

4

As of PHPUnit 8, The other approaches are now deprecated, assertInternalType() will now throw this error and fail:

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.

It is now recommended to use assertIsNumeric() or assertIsInt() for this purpose.

edited Apr 07 '19 at 19:18

Robson

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answered Apr 07 '19 at 14:47

ShadowBeast

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Thank you for this answer! I included it in the top answer as well, so it's seen straight away. – bnp887 Apr 07 '19 at 16:13

score 0 · Answer 6 · answered Nov 20 '18 at 08:16

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):

$this->assertEquals( 1 , preg_match( '/^[0-9]+$/', $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.

How do I assert the result is an integer in PHPUnit?

6 Answers6