Javascript：从2个数组中查找不匹配的对象

Question

我有2个对象数组

var arr1 = [{id: "145", firstname: "dave", lastname: "jones"},
            {id: "135", firstname: "mike",lastname: "williams"},
            {id: "148", firstname: "bob",lastname: "michaels"}];
var arr2 = [{id: "146", firstname: "dave", lastname: "jones"},
            {id: "135", firstname: "mike", lastname: "williams"},
            {id: "148", firstname: "bob", lastname: "michaels"}];

我想找到仅在其中一个数组中存在id的对象，然后将该对象记录到控制台或将对象推送到新数组。

因此，我想结束

var arr1 = [{id: "145", firstname: "dave", lastname: "jones"}]
var arr2 = [{id: "146", firstname: "dave", lastname: "jones"}]

我尝试使用forEach循环并将匹配的ID拼接到数组之外

arr1.forEach(function(element1, index1) {
                let arr1Id = element1.id;
                arr2.forEach(function(element2, index2) {
                    if (arr1Id === element2.id) {
                        arr1.splice(element1, index1)
                        arr2.splice(element2, index2)

                };
            });
        });


console.log(arr1);
console.log(arr2);

但是我最后还是

arr1

[ { id: '135', firstname: 'mike', lastname: 'williams' },
  { id: '148', firstname: 'bob', lastname: 'michaels' } ]

2号

 [ { id: '135', firstname: 'mike', lastname: 'williams' },
  { id: '148', firstname: 'bob', lastname: 'michaels' } ]

Answer 1

您可以Set为每个数组取一个，id然后通过检查是否存在来过滤另一个数组。

var array1 = [{ id: "145", firstname: "dave", lastname: "jones" }, { id: "135", firstname: "mike", lastname: "williams" }, { id: "148", firstname: "bob", lastname: "michaels" }],
   array2 = [{ id: "146", firstname: "dave", lastname: "jones" }, { id: "135", firstname: "mike", lastname: "williams" }, { id: "148", firstname: "bob", lastname: "michaels" }],
   set1 = new Set(array1.map(({ id }) => id)),
   set2 = new Set(array2.map(({ id }) => id)),
   result1 = array1.filter(({ id }) => !set2.has(id)),
   result2 = array2.filter(({ id }) => !set1.has(id));

console.log(result1);
console.log(result2);

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