CODEWITHSUNDEEP
arraysnode.jschai
Alissa
Alissa

Reputation: 714

Comparing arrays in chai

I'm writing some tests with chai and chai-as-promised (and more frameworks, but it doesn't matter in this case) and I need to check if array I get from a web-page is same as a predefined array. I tried to use expect(arrayFromPage).to.eventually.deep.equal(predefinedArray), but it won't work, because order of elements on page is sometimes different (which is OK, I don't need to check if they are in the same order).

I've found a way to workaround the issue by using expect(listFromPage).to.eventually.include.all.members(predefinedArray), but I'd like to know if there is a better solution.

What bothers me most in my workaround, is that I only assure that predefinedArray is subset of listFromPage, not that they are made of same elements.

So, I'd like to know if there is an assert that will pass for [1,2,3] and [3,2,1], but not for [1] and [1,2,3] or [1,2,3,4] and [1,2,3].

I know that I can use some second expectation (compare lengths, or something else), but I'd like to know if there is a one-line solution.

Upvotes: 25

Views: 22687

Answers (6)

user1108069
user1108069

Reputation: 513

https://medium.com/building-ibotta/testing-arrays-and-objects-with-chai-js-4b372310fe6d this explains well, short answer is to use eql

Upvotes: 0

CubanX
CubanX

Reputation: 5252

.members seems to be the way to do this now.

Specifically: have.members and have.deep.members

depending on your needs.

https://www.chaijs.com/api/bdd/#method_members

Upvotes: 2

Theophilus Omoregbee
Theophilus Omoregbee

Reputation: 2503

From the future, the way that worked for me, was to use .deepEqual which did the trick for me

assert.deepEqual(['name'], ['name'], 'this must be same to proceed');

Upvotes: 2

sindrenm
sindrenm

Reputation: 3482

Seeing as this was marked as resolved earlier, I tried doing the same thing as in the accepted answer. It probably worked back then, but doesn't seem to work anymore:

expect([1, 2, 3, 4]).to.have.all.members([2, 4, 3, 1]);

Gives the following error:

AssertionError: expected 1 to be an array

I did a little more research and found a pull request that added this functionality back in 2013:

https://github.com/chaijs/chai/pull/153

So the official way of doing this now is like this:

expect([1, 2, 3, 4]).to.have.same.members([2, 4, 3, 1]);

For completeness, here's the error that two different sets produces:

AssertionError: expected [ 1, 2, 3, 4 ] to have the same members as [ 4, 3, 1 ]

Hope this helps anyone searching for the same answer now. :-)

Upvotes: 47

pdenes
pdenes

Reputation: 802

It's not entirely clear from the documentation, but .to.have.all.members seems to work. I could only find a mention of this feature for .keys, but looks like it also works for .members with arrays.

Upvotes: 5

Magus
Magus

Reputation: 15104

You can do it with 2 lines :

expect(listFromPage).to.eventually.include.all.members(predefinedArray)
expect(predefinedArray).to.eventually.include.all.members(listFromPage)

With this, you'll check if both arrays contains the same values. But order does not matter.

Upvotes: 4

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