javascript - 从列表中删除列表 - 逻辑

Question

我有 2 个列表对象，需要从 ListA 中删除包含 ListB 的所有项目并返回剩余的项目。我的方法如下:

列表A

[{
    "id": 1,
    "name": "ant"
}, {
    "id": 2,
    "name": "ant2"
}, {
    "id": 3,
    "name": "ant3"
}, {
    "id": 4,
    "name": "ant3"
}, {
    "id": 5,
    "name": "ant3"
}]

列表B

[{
    "id": 1,
    "name": "ant"
}, {
    "id": 4,
    "name": "ant4"
}, {
    "id": 3,
    "name": "ant3"
} ]

我尝试过的:

const xxx = this.listA.filter(x => !listB.includes(x) != null);

注意(替代场景):当有 2 个相同的 List 时，预期结果是 []。但是，就我而言，它与单个列表相同。

Answer 1

给你:)

const listBSerialized = listB.map(x => JSON.stringify(x)) 
const xxx = listA.filter(x => !listBSerialized.includes(JSON.stringify(x)));

您将得到以下结果:

[ { id: 2, name: 'ant2' },
  { id: 4, name: 'ant3' },
  { id: 5, name: 'ant3' } ]

您还可以在序列化时需要属性顺序保证时使用模型，有时对象可以具有相同的属性但顺序不同(这取决于对象来自哪里):

function Ant(id, name) {
  this.id = id
  this.name = name
}

listA = [
  new Ant(1, "ant"),
  new Ant(2, "ant2"),
  new Ant(3, "ant3"),
  new Ant(4, "ant3"),
  new Ant(5, "ant3")
]

listB = [
  new Ant(1, "ant"),
  new Ant(4, "ant4"),
  new Ant(3, "ant3")
]

const listBSerialized = listB.map(x => JSON.stringify(x)) 
const xxx = listA.filter(x => !listBSerialized.includes(JSON.stringify(x)));