Expect Arrays to be equal ignoring order

Question

If it's just integers or other primitive values, you can sort() them before comparing.

expect(array1.sort()).toEqual(array2.sort());

If its objects, combine it with the map() function to extract an identifier that will be compared

array1 = [{id:1}, {id:2}, {id:3}];
array2 = [{id:3}, {id:2}, {id:1}];

expect(array1.map(a => a.id).sort()).toEqual(array2.map(a => a.id).sort());

You could use expect.arrayContaining(array) from standard jest:

  const expected = ['Alice', 'Bob'];
  it('matches even if received contains additional elements', () => {
    expect(['Alice', 'Bob', 'Eve']).toEqual(expect.arrayContaining(expected));
  });

jasmine version 2.8 and later has

jasmine.arrayWithExactContents()

Which expects that an array contains exactly the elements listed, in any order.

array1 = [1,2,3];
array2 = [3,2,1];
expect(array1).toEqual(jasmine.arrayWithExactContents(array2))

See https://jasmine.github.io/api/3.4/jasmine.html

simple...

array1 = [1,2,3];
array2 = [3,2,1];

expect(array1).toEqual(jasmine.arrayContaining(array2));

// check if every element of array2 is element of array1
// to ensure [1, 1] !== [1, 2]
array2.forEach(x => expect(array1).toContain(x))

// check if every element of array1 is element of array2
// to ensure [1, 2] !== [1, 1]
array1.forEach(x => expect(array2).toContain(x))

// check if they have equal length to ensure [1] !== [1, 1]
expect(array1.length).toBe(array2.length)