期望数组无效顺序

Question

使用Jasmine有一种方法可以测试2个数组是否包含相同的元素,但不一定是相同的顺序吗？即

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

expect(array1).toEqualIgnoreOrder(array2);//should be true

Answer 1

如果它只是整数或其他原始值,那么sort()在比较之前就可以使用它们.

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

如果是其对象,则将其与map()函数组合以提取将要比较的标识符

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());

Answer 2

您可以使用标准玩笑中的 expect.arrayContaining(array) ：

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

Answer 3

茉莉花 2.8 及更高版本有

jasmine.arrayWithExactContents()

它期望一个数组以任何顺序包含完全列出的元素。

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

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

Answer 4

// 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)

Answer 5

该笑话扩展包为我们提供了一些断言以便简化我们的测试中，它的更简洁和失败的测试误差更加明确。

对于这种情况，我们可以使用toIncludeSameMembers

expect([{foo: "bar"}, {baz: "qux"}]).toIncludeSameMembers([{baz: "qux"}, {foo: "bar"}]);

Answer 6

simple...

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

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

Answer 7

//Compare arrays without order
//Example
//a1 = [1, 2, 3, 4, 5]
//a2 = [3, 2, 1, 5, 4]
//isEqual(a1, a2) -> true
//a1 = [1, 2, 3, 4, 5];
//a2 = [3, 2, 1, 5, 4, 6];
//isEqual(a1, a2) -> false


function isInArray(a, e) {
  for ( var i = a.length; i--; ) {
    if ( a[i] === e ) return true;
  }
  return false;
}

function isEqArrays(a1, a2) {
  if ( a1.length !== a2.length ) {
    return false;
  }
  for ( var i = a1.length; i--; ) {
    if ( !isInArray( a2, a1[i] ) ) {
      return false;
    }
  }
  return true;
}