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);