如何检查对象中所有级别的JavaScript中对象是否为空

Question

I wanted to find out if my object is empty or not for all its nested objects and key-value pairs.

for eg,

const x = {
  a:"",
  b:[],
  c:{
    x:[]
  },
  d:{
    x:{
      y:{
        z:""
      }
    }
  }
};

this should be an empty object and if any of this contains single value then it should be non empty.

Answer 1

Here is the way to do what using recursion

 const x = { a:"", b:[], c:{ x:[] }, d:{ x:{ y:{ z:'' } } } }; function checkEmpty(obj){ for(let key in obj){ //if the value is 'object' if(obj[key] instanceof Object === true){ if(checkEmpty(obj[key]) === false) return false; } //if value is string/number else{ //if array or string have length is not 0. if(obj[key].length !== 0) return false; } } return true; } console.log(checkEmpty(x)) xdxyz = 0; console.log(checkEmpty(x)); 

Answer 2

You can write a recursive function like following. Function creates a set with 2 possible values true and false. If the size of set is 1 and the value being false, which mean that the object is empty.

 const x = {a:"",b:[],c:{x:[]},d:{x:{y:{z:""}}}}; function isEmpty(o, r = new Set()) { for (let k in o) { if(typeof o[k] === "object") { if(Array.isArray(o[k])) r.add(!!o[k].length); else isEmpty(o[k],r); } else r.add(!(o[k] === "" || o[k] === undefined || o[k] === null)); } return r; } let result = isEmpty(x); console.log(result.has(false) && result.size == 1); 

Answer 3

I will use a recursive approach for this one, we iterate over the object.keys() and check is every value related to the key is empty, in the case the value is an object, we go one level deeper to check it.

 const x = { a:"", b:[], c:{x:[]}, d:{x:{y:{z:""}}} }; const x1 = [0,0,0]; const x2 = {0:0,1:0,2:0}; const isEmpty = (obj, empty=true) => { Object.keys(obj).forEach((key) => { if (typeof obj[key] === "object") empty = isEmpty(obj[key], empty); else empty = empty && (obj[key].length === 0); // Return early if we detect empty here. if (!empty) return empty; }); return empty; } console.log("original x: ", isEmpty(x)); xa = "I'm not empty"; console.log("x after edit: ", isEmpty(x)); console.log("x1: ", isEmpty(x1)); console.log("x2: ", isEmpty(x2)); 

Answer 4

try (we use here recursion , fat arrow , obj. keys , reduce , ternary operator and object checking )

let isEmpty = o => o.constructor.name === "Object" ? 
  Object.keys(o).reduce((y,z)=> y&&isEmpty(o[z]) ,true) : o.length == 0;

 const x = { a:"", b:[], c:{ x:[] }, d:{ x:{ y:{ z:"" } } } }; let isEmpty = o => o.constructor.name === "Object" ? Object.keys(o).reduce((y,z)=> y&&isEmpty(o[z]) ,true) : o.length == 0; // Test console.log(isEmpty(x)); xdxyz="Smile to life and life will smile to you"; console.log(isEmpty(x));