数组递归遍历，查询不到某一个元素，就删除的问题。

Question

需求：对后台返回的数据做处理。返回的数组中，第一个对象的value:[11,21,42],但是children中id不包含42，应该用数组中移除，最终value：[11,21]。同理，第二个数组中 value:[42]

后台返回格式：

[

{

"children": [

  {

    "id": 11,

    "typeName": "水果类"

  },

  {

    "id": 21,

    "typeName": "水果类"

  }

],

"title": "水果类",

"value": [11,21,42]

},

{

"children": [

  {

    "id": 42,

    "typeName": "展示类"

  }

],

"title": "展示类",

"value": [11,21,42]

}

Answer 1

let data = [{  "children": [{    "id": 11,    "typeName": "水果类"  }, {    "id": 21,    "typeName": "水果类"  }],  "title": "水果类",  "value": [11]}, {  "children": [{    "id": 42,    "typeName": "展示类"  }],  "title": "展示类",  "value": [11,21,42]}]data.forEach(item => {  let keys = item.children.map(a => a.id)  item.value = item.value.filter(i => keys.includes(i))})