How one can write a function, which takes only few attributes in most-compact way in ES6?
I’ve came up with solution using destructuring + simplified object literal, but I don’t like that list of fields is repeated in the code.
Is there an even slimmer solution?
(v) => { let { id, title } = v; return { id, title }; }
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Answer
Here’s something slimmer, although it doesn’t avoid repeating the list of fields. It uses “parameter destructuring” to avoid the need for the v
parameter.
({id, title}) => ({id, title})
(See a runnable example in this other answer).
@EthanBrown’s solution is more general. Here is a more idiomatic version of it which uses Object.assign
, and computed properties (the [p]
part):
function pick(o, ...props) { return Object.assign({}, ...props.map(prop => ({[prop]: o[prop]}))); }
If we want to preserve the properties’ attributes, such as configurable
and getters and setters, while also omitting non-enumerable properties, then:
function pick(o, ...props) { var has = p => o.propertyIsEnumerable(p), get = p => Object.getOwnPropertyDescriptor(o, p); return Object.defineProperties({}, Object.assign({}, ...props .filter(prop => has(prop)) .map(prop => ({prop: get(props)}))) ); }