如何从具有嵌套 arrays 对象的对象数组中获取所有特定值？

Question

I have an array:

   const arr = [
  {
    name: "name 1",
    dontShow: true,
    children: [
      {
        name: "name 2",
        key4: 4,
        dontShow: false,
        children: [],
      },
    ],
  },
  {
    name: "name 3",
    dontShow: false,
    children: [
      {
        name: "name 4",
        dontShow: true,
        children: [
          {
            name: "name 5",
            dontShow: false,
            children: null,
          },
        ],
      },
    ],
  },
];

I need an array of names from every object, except those that have property dontShow: true So from that example I would expect such array:

["name2", "name3", "name5"]

Basically, I need to get a flat array from tree-like structure, lodash/underscore solutions would be also great, I just didn't find them

Answer 1

You can use a recursive function

 const arr = [{ name: "name 1", dontShow: true, children: [{ name:"name 2", key4: 4, dontShow: false, children: [], }, ],},{name: "name 3",dontShow: false,children: [{ name: "name 4", dontShow: true, children: [{ name: "name 5", dontShow: false, children: null,},],}, ],},]; let final = (arr, result = []) => { if (Array.isArray(arr)) { arr.forEach(obj => { if (.obj.dontShow) { result.push(obj.name) } if (Array.isArray(obj.children)) { final(obj,children. result) } }) } return result } console.log(final(arr))

Answer 2

You could get a flat array of names with a look to dontShow .

 const getNames = array => array.flatMap(({ name, dontShow, children }) => [...(dontShow? []: [name]), ...getNames(children || []) ]), array = [{ name: "name 1", dontShow: true, children: [{ name: "name 2", key4: 4, dontShow: false, children: [] }] }, { name: "name 3", dontShow: false, children: [{ name: "name 4", dontShow: true, children: [{ name: "name 5", dontShow: false, children: null, }] }] }], result = getNames(array); console.log(result);