除非它包含具有特定值的另一个项目，否则如何从对象数组中获取具有属性的列表？

Question

I have an array of objects, and I need to get list with certain property from that array of objects. But i need that list to contain only those values where object was containing another property with certain element. This is very confusing so i made an example. Let's say i have an array with objects.

  employees = [
           {
            n: 'case 1',
            date: '2021-05-4',
            id: '123',
            user: [{name: 'Vlad', id: '1'}, {name: 'Misha', id: '2'}],
            isPresent : true,
           },
           {
            caseName: 'case 2',
            date: '2021-05-4',
            id: '124',
            user: [{name: 'Alina', id: '3'}, {name: 'Alex', id: '4'}],
            isPresent : true,
           },
           {
            caseName: 'case 3',
            date: '2021-05-4',
            id: '126',
            user: [],
            isPresent : false,
           },
        ]

And my task is to get a list of IDs from array of objects, but i need ID only from those objecrs which have isPresent as true . So i need ['123', '124'].

I could use a loops and conditions and so on. But i wondering is it possible to do with one line? Something like this:

employees.filter(item => { return item.isPresent === true }))

But i need only IDs not whole objects.

Answer 1

1) You can filter the elements with condition item.isPresent === true and then map over it to get the final result as:

employees
  .filter((item) => item.isPresent === true)
  .map((o) => o.id);

or you can also do as:

employees.filter((item) => item.isPresent).map((o) => o.id)

 const employees = [{ n: 'case 1', date: '2021-05-4', id: '123', user: [{ name: 'Vlad', id: '1' }, { name: 'Misha', id: '2' }], isPresent: true, }, { caseName: 'case 2', date: '2021-05-4', id: '124', user: [{ name: 'Alina', id: '3' }, { name: 'Alex', id: '4' }], isPresent: true, }, { caseName: 'case 3', date: '2021-05-4', id: '126', user: [], isPresent: false, }, ] const result = employees.filter((item) => item.isPresent === true).map((o) => o.id); console.log(result);

2) You can also achieve the same result using reduce as:

employees.reduce((acc, curr) => {
  curr.isPresent && acc.push(curr.id);
  return acc;
}, []);

 const employees = [ { n: "case 1", date: "2021-05-4", id: "123", user: [ { name: "Vlad", id: "1" }, { name: "Misha", id: "2" }, ], isPresent: true, }, { caseName: "case 2", date: "2021-05-4", id: "124", user: [ { name: "Alina", id: "3" }, { name: "Alex", id: "4" }, ], isPresent: true, }, { caseName: "case 3", date: "2021-05-4", id: "126", user: [], isPresent: false, }, ]; const result = employees.reduce((acc, curr) => { curr.isPresent && acc.push(curr.id); return acc; }, []); console.log(result);

Answer 2

You can use something like this

employees.filter((item) => item.isPresent).map((obj) => obj.id);