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javascriptjson
Vignesh
Vignesh

Reputation: 65

javascript - delete json elements that present in the given json

I know how to delete a JSON element using delete option. But I have a situation

I have two JSON objects like

var a = {
    key1 : 'v1',
    key2 : 'v2',
    key3 : 'v3',
    key4 : 'v4' 
}
var b = {
    key1 : 'v1',
    key3 : 'v3',
    key5 : 'v5'
}

now I want a minus b (i.e) delete elements in "var a" that present in "var b". I need my result as 

{
    key2 : 'v2',
    key4 : 'v4'
};

Is there any method available for this? Thanks in advance...

Upvotes: 0

Views: 41

Answers (4)

Ele
Ele

Reputation: 33736

If you don't want to mutate the source object, you can use the function Array.prototype.reduce as follow:

This approach compares both the key and value, however, if this is not the desired behavior, you can remove the value comparison

let a = {    key1 : 'v1',    key2 : 'v2',    key3 : 'v3',    key4 : 'v4' },
    b = {    key1 : 'v1',    key3 : 'v3',    key5 : 'v5'},
    bEntries = Object.entries(b);
    
let result = Object.entries(a).reduce((a, [k, v]) => {
  if (!bEntries.some(([bk, bv]) => k === bk && v === bv)) {
    return Object.assign(a, {[k]: v});
  }
  
  return a;
}, Object.create(null));

console.log(result);

Upvotes: 0

Carlos Menezes
Carlos Menezes

Reputation: 74

Going for an immutable approach (i.e. you don't modify neither a or b), all you need to check is if key k in a (as in a[k]) is not present in b (as in b[k] doesn't exist).

function objDifference(a, b) {
    let output = {}

    Object.keys(a).forEach((k) => {
        if (!(k in b)) { // If object b doesn't have the k key
            output[k] = a[k]
        }
    })

    return output;
}


function objDifference(a, b) {
    let output = {}

    Object.keys(a).forEach((k) => {
        if (!(k in b)) {
            output[k] = a[k]
        }
    })

    return output;
}

var a = {
    key1 : 'v1',
    key2 : 'v2',
    key3 : 'v3',
    key4 : 'v4' 
}
var b = {
    key1 : 'v1',
    key3 : 'v3',
    key5 : 'v5'
}
console.log(objDifference(a, b))

Upvotes: 0

Sebastian Kaczmarek
Sebastian Kaczmarek

Reputation: 8515

The simplest approach is to simply iterate over b keys and delete all of the keys in a that are present there:

var a = {
    key1 : 'v1',
    key2 : 'v2',
    key3 : 'v3',
    key4 : 'v4' 
}
var b = {
    key1 : 'v1',
    key3 : 'v3',
    key5 : 'v5'
}

for (let key in b) {
  if (a.hasOwnProperty(key)) {
    delete a[key];
  }
}

console.log(a);

Upvotes: 1

chakwok
chakwok

Reputation: 1040

for(let key of Object.keys(a)) {
    if(b.hasOwnProperty(key)) {
        delete a[key]
    }
}

Upvotes: 2

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