I wanted to find out if my object is empty or not for all its nested objects and key-value pairs.
for e.g.,
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.
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))
x.d.x.y.z = 0;
console.log(checkEmpty(x));
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);
return true for 0 inside objectI 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));
x.a = "I'm not empty";
console.log("x after edit: ", isEmpty(x));
console.log("x1: ", isEmpty(x1));
console.log("x2: ", isEmpty(x2));true even if there will a value 0
{z:""} is considered empty. For me that is not empty. If he clarify with other cases I will update the answer, however the approach will be similar. Could you give me an example where my code fails, so I can analize it...x.a = 0; and then check it will return true[0,0,0] or not? If no, then {0:0, 1:0, 2:0} should not be considered empty also.[0,0,0] should be considered empty not {0:0, 1:0, 2:0}. In his question he showed only [] and ''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));
x.d.x.y.z="Smile to life and life will smile to you";
console.log(isEmpty(x));
deep check include null or undefined:
export const isEmptyObject = (obj: Object) => {
for (let key in obj) {
if (obj[key] instanceof Object === true) {
if (isEmptyObject(obj[key]) === false) return false;
}
else {
if (obj[key] !== null && obj[key] !== undefined && obj[key]?.length) return false;
}
}
return true;
};
