从 Object 数组中删除多个键

Question

Let's say we have an array of object:

var obj = [{a: 1, b: 2, c: 3, d: 4, e: 5 },{a: 6, b: 7, c: 
8, d: 9, e: 0 }];

and we want to delete key c,e from both of the objects.

How can it be done? One of the methods I found is:

['c', 'e'].forEach(e => delete obj[e]); //for object

Is there any other way so we don't have to use double for loop.

Answer 1

One way to do it is to use .map() together with object destructuring :

 var obj = [ { a: 1, b: 2, c: 3, d: 4, e: 5 }, { a: 6, b: 7, c: 8, d: 9, e: 0 }, ]; var newObj = obj.map(({ c, e, ...rest }) => rest); console.log(newObj)

This will create a new array with new objects which contain all of the other keys except for c and e .

Answer 2

You have 2 options to resolve it:

1. By using object destructuring : map(({ a,b,c,d,e }) => ({a,b,d})
2. Enhance option 1 by using using [Rest parameters] { c, e, ...rest }

• Object destructuring like below

  const obj = { a: 1, b: 2, c: 3, d: 4, e: 5 } var {c, e} = obj; // c = 3, e = 5

• With option 2, you will have c,e implicit name and the remaining items named rest . After that, you just need to get rest items.

Option 1

 var obj = [ { a: 1, b: 2, c: 3, d: 4, e: 5 }, { a: 6, b: 7, c: 8, d: 9, e: 0 }, ]; console.log(obj.map(({ a,b,c,d,e }) => ({a,b,d})));

Option 2

 var obj = [ { a: 1, b: 2, c: 3, d: 4, e: 5 }, { a: 6, b: 7, c: 8, d: 9, e: 0 }, ]; console.log(obj.map(({ c, e, ...rest }) => rest)); //...rest: the same as `a,b,d`