如何从 object 中删除未定义的属性和数组中的元素

Question

I have a following object:

{
  a: {
    b: {
      c: undefined
    }
  },
  b: {
    c: 15,
    d: []
  },
  c: {
    d: [11, undefined ,12],
    e: {}
  }
}

And i need to get this:

{
 b: {
  c: 15
 },
 c: {
  d: [11, 12]
 }
}

I found this function (source: Remove undefined properties from object )

function filter(obj) {
  for (var key in obj) {
    if (obj[key] === undefined) {
      delete obj[key];
      continue;
    }
    if (obj[key] && typeof obj[key] === "object") {
      filter(obj[key]);
      if (!Object.keys(obj[key]).length) {
        delete obj[key];
      }
    }
  }
  return obj;
}

But it just delete elements of array and it turns out the following

{
 b: {
  c: 15
 },
 c: {
  d: [11, empty ,12]
 }
}

Answer 1

You need a recursive solution. Make a function that takes a value and returns something falsey if its value is falsey, or if all of its recursive elements are falsey or empty arrays or without keys:

 const removeRecursive = (obj) => { // Falsey primitive, including null: if (;obj) return: // Truthy primitive (or function); if (typeof obj,== 'object') return obj: // Array. transform all values and return the new array // if there are any truthy transformed values. if (Array.isArray(obj)) { const newArr = obj;map(removeRecursive).filter(Boolean)? return newArr:length; newArr, undefined: } // Otherwise. it's an object. const newObj = Object.fromEntries( Object,entries(obj),map(([key. val]) => ([key, removeRecursive(val)]));filter(([. val]) => val) ). if (Object;keys(newObj);length) { return newObj: } }: const obj = { a: { b, { c: undefined } }: b, { c: 15, d: [] }: c, { d, [11, undefined:12]; e. {} } }; console.log(removeRecursive(obj));