I search a lots but what I got is how to merge objects and keeps properties of both. How about keep only the same props only? For example:
const obj1 = {a: 1, b:2, c:3} const obj2 = {a: 3, b:3, d:5, e:7} Is there any way to create an obj3 which is {a:3, b:3} (only keep props in both objects)?
One option would be to reduce by obj2's entries, assigning them to the accumulator object if the property exists in obj1:
const obj1 = {a: 1, b:2, c:3} const obj2 = {a: 3, b:3, d:5, e:7} console.log( Object.entries(obj2).reduce((a, [key, val]) => { if (key in obj1) a[key] = val; return a; }, {}) );
Accepted answer is a good way to go but the following might turn out to be more efficient;
var obj1 = {a: 1, b:2, c:3}, obj2 = {a: 3, b:3, d:5, e:7}, res = {}; for (k in obj1) k in obj2 && (res[k] = obj2[k]); console.log(res);You could convert to an array (with the help of Object.entries) to filter – then convert back (with the help of reduce)
Object.entries(obj2) .filter(([key]) => (key in obj1)) .reduce((obj, [key, val]) => ({...obj, [key]: val}), {}) Here's a way that tries to improve efficiency by looping over the object with fewer keys (less iterations):
const obj1 = {a: 1, b:2, c:3} const obj2 = {a: 3, b:3, d:5, e:7} // find sm/lg object by key length let [sm,lg]=[obj1,obj2].sort((a,b)=>( Object.keys(a).length-Object.keys(b).length )); const merged = {} // make small object outer loop for (let key in sm){ if (key in lg) merged[key]=obj2[key] // only store if key in both objects } console.log(merged);