How to get list with of property from array of objects unless it contains another item with certain-CodePudding

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.

CodePudding user response：

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);

CodePudding user response：

You can use something like this

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