Sorting Objects by value not working as expected

Question

I'm trying to sort 2 objects based on value.first_name. One ascending alphabetical order. The other descending alphabetical order. However, they are both sorting decending. What am I doing wrong? The goal is to arrange the objects in this array based on the value of first_name.

var participants = [
  {
    id: "992543",
    first_name: "",
    last_name: "",
    company: null,
    notes: "",
    registrationType: "",
    alerts: [ ],
    reg_scan: null
  },
  {
    id: "999070",
    first_name: "Tori",
    last_name: "Fullard",
    company: null,
    notes: "",
    registrationType: "Staff",
    alerts: [ ],
    reg_scan: null
  },
  {
    id: "99265",
    first_name: "Ronald",
    last_name: "Brown",
    company: null,
    notes: "",
    registrationType: "Dean's Guest",
    alerts: [ ],
    reg_scan: null
  },
  {
    id: "992279",
    first_name: "Laila",
    last_name: "Shetty",
    company: null,
    notes: "",
    registrationType: "Table Guest",
    alerts: [
      {
        alert_id: "1",
        dismissed: "0"
      }
    ],
    reg_scan: null
    },
    {
    id: "992248",
    first_name: "Paul",
    last_name: "Keenan",
    company: null,
    notes: "",
    registrationType: "Table Guest",
    alerts: [ ],
    reg_scan: null
    }
    ];

var az_part = participants;
var za_part = participants;
az_part.sort(function(a, b) {
    var nameA = a.first_name.toLowerCase();
    var nameB = b.first_name.toLowerCase();
    if (nameA > nameB) return 1;
    if (nameA < nameB) return -1;
    return 0;
});

za_part.sort(function(a, b) {
    var nameA = a.first_name.toLowerCase();
    var nameB = b.first_name.toLowerCase();
    if (nameA > nameB) return -1;
    if (nameA < nameB) return 1;
    return 0;
});

`az_part` and `za_part` are the SAME thing, `participants`. There is only one array, so you're sorting that same array twice. Check `participants` and you'll see that it's also sorted the same way even though your code doesn't explicitly sort it. You need to copy the array (`participants.slice(0)` will do it - you don't particularly need a deep-copy here) — Niet the Dark Absol, Apr 26 '17 at 18:03

score 3 · Accepted Answer · answered Apr 26 '17 at 18:03

All three variables az_part, za_part, and participants refer to the exact same array in memory. You are transforming that one array with an ascending sort and then immediately re-sorting it with a descending sort.

Use slice to create shallow copies of the array:

var az_part = participants.slice();
var za_part = participants.slice();

(By "shallow copy" I mean that this will create a new list, but it will not create new objects. Each resulting list will be different -- and can be sorted differently -- but if you mutate any of the objects inside, e.g., participants[2].notes = "foo", that will change the object as it exists in each list.)

score 0 · Answer 2 · edited May 23 '17 at 11:54

0

JavaScript objects (including arrays) are references to objects. So you're sorting the original array twice.

see this question for a good description of your problem.

Why does changing an Array in JavaScript affect copies of the array?

edited May 23 '17 at 11:54

Community

1
1

answered Apr 26 '17 at 18:14

Stuart H.

46
2

Sorting Objects by value not working as expected

2 Answers2