i have a function like this:
const getKeysAs = (key1, key2) => {
return {
[key1]: state.key1,
[key2]: state.key2
}
}
So if state.key1 is 'a' and state.key2 is 'b', calling getKyesAs('one', 'two') would return
{
one: 'a',
two: 'b'
}
Now, if one of the argument is undefined, is there a way to not include it in the returned object ?
You can Conditionally add properties to an Object with object destructuring
const obj = {
...(key1 && { [key1]: state[key1] }),
...(key2 && { [key2]: state[key2] })
};
If some of the args function is undefined, null or 0 (falsy values) then it will no be added to the object.
There is a very scalable way to do it:
const state= {
a: "hello",
}
function getKeysAs (keys) {
return [...arguments].reduce((acc, cur)=> {
const newValue = state[cur] && {[cur]: state[cur]}
return {...acc, ...newValue}
}, {})
}
console.log(getKeysAs("a", "b"))This way, you can pass as much keys as you need without worrying about scalability & undefined values.
state[key1], not state.key1 as in their example.
Use Object.assign().
const getKeysAs = (key1, key2) => {
return Object.assign({}, key1 && {[key1]: state[key1]}, key2 && {[key2]: state[key2]});
}
Assuming you actually mean to do state[key1], not state.key1, here is a solution that doesn't create superfluous objects:
const getKeysAs = (...keys) => {
const result = {};
for (const key of keys) {
if (key != null) {
result[key] = state[key];
}
}
return result;
}
const o = {}; if (key1) o[key1] = ...;…?state[key1]?