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))

})