Sort array of objects by single key with date value

Question

I have an array of objects with several key value pairs, and I need to sort them based on 'updated_at':

[
    {
        "updated_at" : "2012-01-01T06:25:24Z",
        "foo" : "bar"
    },
    {
        "updated_at" : "2012-01-09T11:25:13Z",
        "foo" : "bar"
    },
    {
        "updated_at" : "2012-01-05T04:13:24Z",
        "foo" : "bar"
    }
]

What's the most efficient way to do so?

score 476 · Accepted Answer · edited Dec 20 '20 at 20:01

476

You can use Array.sort.

Here's an example:

var arr = [{
    "updated_at": "2012-01-01T06:25:24Z",
    "foo": "bar"
  },
  {
    "updated_at": "2012-01-09T11:25:13Z",
    "foo": "bar"
  },
  {
    "updated_at": "2012-01-05T04:13:24Z",
    "foo": "bar"
  }
]

arr.sort(function(a, b) {
  var keyA = new Date(a.updated_at),
    keyB = new Date(b.updated_at);
  // Compare the 2 dates
  if (keyA < keyB) return -1;
  if (keyA > keyB) return 1;
  return 0;
});

console.log(arr);

edited Dec 20 '20 at 20:01

ssc

9,528
10
64
94

answered Jan 12 '12 at 15:23

gen_Eric

223,194
41
299
337

28

Couldn't you use `keyA - keyB` (or possibly `keyB - keyA`)? The date objects have a `valueOf()` method. – soktinpk Jul 17 '14 at 18:44
6

If you want to sort based on numeric values from your keys then this ES6 one-liner might be of help. `yourArr.sort((a, b) => a.yourNumericOrderingKey > b.yourNumericOrderingKey)`. In my case I had to sort based on an `order` key that was coming inside the objects of main array. – KeshavDulal Sep 20 '20 at 10:53
Not working in Chrome Mobile version >= 88.0.4324.152 – Rodrigo Vieira Feb 07 '21 at 02:52
1

Not really any need to convert to date objects because with ISO format alphabetical and chronological order are the same. So you can just sort as strings. As long as all values are in proper ISO format (including being in UTC) – Henry Woody Nov 25 '21 at 19:14
1

In 2012 this was a fine answer. In 2021, see my answer below https://stackoverflow.com/questions/8837454/sort-array-of-objects-by-single-key-with-date-value/70319634?noredirect=1#comment124367399_70319634 – donavon Dec 16 '21 at 00:20

score 196 · Answer 2 · edited May 23 '17 at 11:47

196

I already answered a really similar question here: Simple function to sort an array of objects

For that question I created this little function that might do what you want:

function sortByKey(array, key) {
    return array.sort(function(a, b) {
        var x = a[key]; var y = b[key];
        return ((x < y) ? -1 : ((x > y) ? 1 : 0));
    });
}

edited May 23 '17 at 11:47

Community

1
1

answered Jan 12 '12 at 15:23

David Brainer

6,223
3
18
16

Not working in Chrome Mobile version >= 88.0.4324.152 – Rodrigo Vieira Feb 07 '21 at 02:57

Patrick W. McMahon · Answer 3 · 2017-03-20T13:11:37.307

The Array.sort() method sorts the elements of an array in place and returns the array. Be careful with Array.sort() as it's not Immutable. For immutable sort use immutable-sort.

This method is to sort the array using your current updated_at in ISO format. We use new Data(iso_string).getTime() to convert ISO time to Unix timestamp. A Unix timestamp is a number that we can do simple math on. We subtract the first and second timestamp the result is; if the first timestamp is bigger than the second the return number will be positive. If the second number is bigger than the first the return value will be negative. If the two are the same the return will be zero. This alines perfectly with the required return values for the inline function.

For ES6:

arr.sort((a,b) => new Date(a.updated_at).getTime() - new Date(b.updated_at).getTime());

For ES5:

arr.sort(function(a,b){ 
 return new Date(a.updated_at).getTime() - new Date(b.updated_at).getTime();
});

If you change your updated_at to be unix timestamps you can do this:

For ES6:

arr.sort((a,b) => a.updated_at - b.updated_at);

For ES5:

arr.sort(function(a,b){ 
 return a.updated_at - b.updated_at;
});

At the time of this post, modern browsers do not support ES6. To use ES6 in modern browsers use babel to transpile the code to ES5. Expect browser support for ES6 in the near future.

Array.sort() should receave a return value of one of 3 possible outcomes:

A positive number (first item > second item)
A negative number (first item < second item)
0 if the two items are equal

Note that the return value on the inline function can be any positive or negative number. Array.Sort() does not care what the return number is. It only cares if the return value is positive, negative or zero.

For Immutable sort: (example in ES6)

const sort = require('immutable-sort');
const array = [1, 5, 2, 4, 3];
const sortedArray = sort(array);

You can also write it this way:

import sort from 'immutable-sort';
const array = [1, 5, 2, 4, 3];
const sortedArray = sort(array);

The import-from you see is a new way to include javascript in ES6 and makes your code look very clean. My personal favorite.

Immutable sort doesn't mutate the source array rather it returns a new array. Using const is recommended on immutable data.

Another immutable alternative would be to use arr.slice().sort(...) — Simon, Nov 05 '20 at 15:17
as stated, you must use a transpiler like babel for ES6. You CAN NOT use ES6 in native browsers without a transpiler. — Patrick W. McMahon, Dec 18 '22 at 18:33

score 23 · Answer 4 · answered Apr 16 '16 at 02:40

Use underscore js or lodash,

var arrObj = [
    {
        "updated_at" : "2012-01-01T06:25:24Z",
        "foo" : "bar"
    },
    {
        "updated_at" : "2012-01-09T11:25:13Z",
        "foo" : "bar"
    },
    {
        "updated_at" : "2012-01-05T04:13:24Z",
        "foo" : "bar"
    }
];

arrObj = _.sortBy(arrObj,"updated_at");

_.sortBy() returns a new array

refer http://underscorejs.org/#sortBy and lodash docs https://lodash.com/docs#sortBy

Brad Parks · Answer 5 · 2018-02-09T13:05:20.340

21

Here's a slightly modified version of @David Brainer-Bankers answer that sorts alphabetically by string, or numerically by number, and ensures that words beginning with Capital letters don't sort above words starting with a lower case letter (e.g "apple,Early" would be displayed in that order).

function sortByKey(array, key) {
    return array.sort(function(a, b) {
        var x = a[key];
        var y = b[key];

        if (typeof x == "string")
        {
            x = (""+x).toLowerCase(); 
        }
        if (typeof y == "string")
        {
            y = (""+y).toLowerCase();
        }

        return ((x < y) ? -1 : ((x > y) ? 1 : 0));
    });
}

edited Feb 09 '18 at 13:05

answered Jan 22 '13 at 16:36

Brad Parks

66,836
64
257
336

3

The proposed solution might give an error if a[key] and b[key] are not both strings. I suggest to replace y=y.toLowerCase() with y = (""+y).toLowerCase() – user8074 Apr 05 '16 at 12:42
sort can accept any positive or negative number as a valid return. You're extra calculations to force it to be 1,0,-1 is not needed. You over complicated a simple return value. It's best to not add extra calculations that don't do anything. – Patrick W. McMahon Dec 30 '18 at 03:13

score 19 · Answer 6 · answered Feb 23 '17 at 14:46

19

With ES2015 support it can be done by:

foo.sort((a, b) => a.updated_at < b.updated_at ? -1 : 1)

answered Feb 23 '17 at 14:46

knowbody

8,106
6
45
70

1

no need for the inline if replace the < with - and remove '? -1 : 1" you will get a valid return. This example moves items that may be equal and so could give unexpected results. for equal items a 0 should be returned. – Patrick W. McMahon Mar 15 '17 at 16:53
if updated_at is an ISO time this will not work. This example assumes Unix timestamps but the OP posted data that was in ISO format. So you would have to convert to Unix timestamps to do a comparison. This can be done with `new Date(iso_str).getTime()` this will return a Unix timestamp. – Patrick W. McMahon Mar 15 '17 at 17:36

score 7 · Answer 7 · answered Dec 20 '18 at 09:24

Data Imported

[
    {
        "gameStatus": "1",
        "userId": "c02cfb18-ae66-430b-9524-67d9dd8f6a50",
        "created_at": "2018-12-20 11:32:04"
    },
    {
        "gameStatus": "0",
        "userId": "c02cfb18-ae66-430b-9524-67d9dd8f6a50",
        "created_at": "2018-12-19 18:08:24"
    },
    {
        "gameStatus": "2",
        "userId": "c02cfb18-ae66-430b-9524-67d9dd8f6a50",
        "created_at": "2018-12-19 18:35:40"
    },
    {
        "gameStatus": "0",
        "userId": "c02cfb18-ae66-430b-9524-67d9dd8f6a50",
        "created_at": "2018-12-19 10:42:53"
    },
    {
        "gameStatus": "2",
        "userId": "c02cfb18-ae66-430b-9524-67d9dd8f6a50",
        "created_at": "2018-12-20 10:54:09"
    },
    {
        "gameStatus": "0",
        "userId": "1a2fefb0-5ae2-47eb-82ff-d1b2cc27875a",
        "created_at": "2018-12-19 18:46:22"
    },
    {
        "gameStatus": "1",
        "userId": "7118ed61-d8d9-4098-a81b-484158806d21",
        "created_at": "2018-12-20 10:50:48"
    }
]

FOR Ascending order

arr.sort(function(a, b){
    var keyA = new Date(a.updated_at),
        keyB = new Date(b.updated_at);
    // Compare the 2 dates
    if(keyA < keyB) return -1;
    if(keyA > keyB) return 1;
    return 0;
});

Example for Asc Order

[
    {
        "gameStatus": "0",
        "userId": "c02cfb18-ae66-430b-9524-67d9dd8f6a50",
        "created_at": "2018-12-19 10:42:53"
    },
    {
        "gameStatus": "0",
        "userId": "c02cfb18-ae66-430b-9524-67d9dd8f6a50",
        "created_at": "2018-12-19 18:08:24"
    },
    {
        "gameStatus": "2",
        "userId": "c02cfb18-ae66-430b-9524-67d9dd8f6a50",
        "created_at": "2018-12-19 18:35:40"
    },
    {
        "gameStatus": "0",
        "userId": "1a2fefb0-5ae2-47eb-82ff-d1b2cc27875a",
        "created_at": "2018-12-19 18:46:22"
    },
    {
        "gameStatus": "1",
        "userId": "7118ed61-d8d9-4098-a81b-484158806d21",
        "created_at": "2018-12-20 10:50:48"
    },
    {
        "gameStatus": "2",
        "userId": "c02cfb18-ae66-430b-9524-67d9dd8f6a50",
        "created_at": "2018-12-20 10:54:09"
    },
    {
        "gameStatus": "1",
        "userId": "c02cfb18-ae66-430b-9524-67d9dd8f6a50",
        "created_at": "2018-12-20 11:32:04"
    }
]

FOR Descending order

arr.sort(function(a, b){
    var keyA = new Date(a.updated_at),
        keyB = new Date(b.updated_at);
    // Compare the 2 dates
    if(keyA > keyB) return -1;
    if(keyA < keyB) return 1;
    return 0;
});

Example for Desc Order

[
    {
        "gameStatus": "1",
        "userId": "c02cfb18-ae66-430b-9524-67d9dd8f6a50",
        "created_at": "2018-12-20 11:32:04"
    },
    {
        "gameStatus": "2",
        "userId": "c02cfb18-ae66-430b-9524-67d9dd8f6a50",
        "created_at": "2018-12-20 10:54:09"
    },
    {
        "gameStatus": "1",
        "userId": "7118ed61-d8d9-4098-a81b-484158806d21",
        "created_at": "2018-12-20 10:50:48"
    },
    {
        "gameStatus": "0",
        "userId": "1a2fefb0-5ae2-47eb-82ff-d1b2cc27875a",
        "created_at": "2018-12-19 18:46:22"
    },
    {
        "gameStatus": "2",
        "userId": "c02cfb18-ae66-430b-9524-67d9dd8f6a50",
        "created_at": "2018-12-19 18:35:40"
    },
    {
        "gameStatus": "0",
        "userId": "c02cfb18-ae66-430b-9524-67d9dd8f6a50",
        "created_at": "2018-12-19 18:08:24"
    },
    {
        "gameStatus": "0",
        "userId": "c02cfb18-ae66-430b-9524-67d9dd8f6a50",
        "created_at": "2018-12-19 10:42:53"
    }
]

score 5 · Answer 8 · edited Jan 06 '23 at 08:56

5

As for today, answers of @knowbody (https://stackoverflow.com/a/42418963/6778546) and @Rocket Hazmat (https://stackoverflow.com/a/8837511/6778546) can be combined to provide for ES2015 support and correct date handling:

var arr = [{
    "updated_at": "2012-01-01T06:25:24Z",
    "foo": "bar"
  },
  {
    "updated_at": "2012-01-09T11:25:13Z",
    "foo": "bar"
  },
  {
    "updated_at": "2012-01-05T04:13:24Z",
    "foo": "bar"
  }
];
arr.sort((a, b) => {
   const dateA = new Date(a.updated_at);
   const dateB = new Date(b.updated_at);
   return dateA - dateB;
});

edited Jan 06 '23 at 08:56

Florian Feldhaus

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answered Nov 23 '17 at 04:24

Fyodor

247
3
14

Or as one liner `arr.sort((a, b) => { return new Date(a.updated_at) - new Date(b.updated_at)});`. – Florian Feldhaus Jan 06 '23 at 08:51

score 5 · Answer 9 · answered Dec 11 '21 at 22:49

I'm getting here a little late, but in 2021 the correct answer is to use Intl.Collator. updated_at is an ISO-8601 string and thus sortable as a string. Converting to a Date is a waste of time and so is performing if comparisons manually to return 0, 1, or -1.

const arr = [
  {
    "updated_at": "2012-01-01T06:25:24Z",
    "foo": "bar"
  },
  {
    "updated_at": "2012-01-09T11:25:13Z",
    "foo": "bar"
  },
  {
    "updated_at": "2012-01-05T04:13:24Z",
    "foo": "bar"
  }
];

const { compare } = Intl.Collator('en-US');
arr.sort((a, b) => compare(a.updated_at, b.updated_at));

Intl.Collator returns a function that can be used as the compareFunction for #Array.sort. Because we're sorting an object, we call compare with the values of the key that we want to sort by.

Note that if we were sorting an array of strings, we could simply do this:

arr.sort(compare);

Also note that as others have pointed out, sort will mutate the original array. If this is undesirable, you may want to clone it first. In 2021 you can do so as follows:

[...arr].sort((a, b) => compare(a.updated_at, b.updated_at));

This is the only one that works for me as of 12/14/2021! (for anyone else who's banging their heads against the wall) — SelfTaughtLooking4Job, Dec 14 '21 at 20:59

score 3 · Answer 10 · answered Feb 09 '18 at 06:46

As This answer's states, you can use Array.sort.

arr.sort(function(a,b){return new Date(a.updated_at) - new Date(b.updated_at)})

arr = [
    {
        "updated_at" : "2012-01-01T06:25:24Z",
        "foo" : "bar"
    },
    {
        "updated_at" : "2012-01-09T11:25:13Z",
        "foo" : "bar"
    },
    {
        "updated_at" : "2012-01-05T04:13:24Z",
        "foo" : "bar"
    }
];
arr.sort(function(a,b){return new Date(a.updated_at) - new Date(b.updated_at)});
console.log(arr);

Danny Mor · Answer 11 · 2017-07-12T14:17:57.620

2

Just another, more mathematical, way of doing the same thing but shorter:

arr.sort(function(a, b){
    var diff = new Date(a.updated_at) - new Date(b.updated_at);
    return diff/(Math.abs(diff)||1);
});

or in the slick lambda arrow style:

arr.sort((a, b) => {
    var diff = new Date(a.updated_at) - new Date(b.updated_at);
    return diff/(Math.abs(diff)||1);
});

This method can be done with any numeric input

edited Jul 12 '17 at 14:17

answered Jul 12 '17 at 13:57

Danny Mor

1,143
12
12

underated answer – two7s_clash Sep 22 '17 at 17:25
Why normalize the `diff`? JS sort only cares about the sign and zero status of the result, not the actual value. ([docs](https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/Array/sort)) – Henry Woody Nov 25 '21 at 19:25

score 2 · Answer 12 · edited Oct 24 '19 at 13:28

You could use the Lodash utility library to solve this problem (it's quite an efficient library):

const data = [{
    "updated_at": "2012-01-01T06:25:24Z",
    "foo": "bar"
  },
  {
    "updated_at": "2012-01-09T11:25:13Z",
    "foo": "bar"
  },
  {
    "updated_at": "2012-01-05T04:13:24Z",
    "foo": "bar"
  }
]

const ordered = _.orderBy(
  data,
  function(item) {
    return item.updated_at;
  }
);

console.log(ordered)

<script src="https://cdnjs.cloudflare.com/ajax/libs/lodash.js/4.17.15/lodash.min.js"></script>

You can find documentation here: https://lodash.com/docs/4.17.15#orderBy

aljgom · Answer 13 · 2018-12-14T06:23:39.727

With this we can pass a key function to use for the sorting

Array.prototype.sortBy = function(key_func, reverse=false){
    return this.sort( (a, b) => {
        var keyA = key_func(a),
            keyB = key_func(b);
        if(keyA < keyB) return reverse? 1: -1;
        if(keyA > keyB) return reverse? -1: 1;
        return 0;
    }); 
}

Then for example if we have

var arr = [ {date: "01/12/00", balls: {red: "a8",  blue: 10}},
            {date: "12/13/05", balls: {red: "d6" , blue: 11}},
            {date: "03/02/04", balls: {red: "c4" , blue: 15}} ]

We can do

arr.sortBy(el => el.balls.red)
/* would result in
[ {date: "01/12/00", balls: {red: "a8", blue: 10}},
  {date: "03/02/04", balls: {red: "c4", blue: 15}},
  {date: "12/13/05", balls: {red: "d6", blue: 11}} ]
*/

or

arr.sortBy(el => new Date(el.date), true)   // second argument to reverse it
/* would result in
[ {date: "12/13/05", balls: {red: "d6", blue:11}},
  {date: "03/02/04", balls: {red: "c4", blue:15}},
  {date: "01/12/00", balls: {red: "a8", blue:10}} ]
*/

or

arr.sortBy(el => el.balls.blue + parseInt(el.balls.red[1]))
/* would result in
[ {date: "12/13/05", balls: {red: "d6", blue:11}},    // red + blue= 17
  {date: "01/12/00", balls: {red: "a8", blue:10}},    // red + blue= 18
  {date: "03/02/04", balls: {red: "c4", blue:15}} ]   // red + blue= 19
*/

score 1 · Answer 14 · answered Jul 16 '19 at 06:51

Use Array.sort() to sort an array
Clone array using spread operator (…) to make the function pure
Sort by desired key (updated_at)
Convert date string to date object
Array.sort() works by subtracting two properties from current and next item if it is a number / object on which you can perform arrhythmic operations

const input = [
  {
    updated_at: '2012-01-01T06:25:24Z',
    foo: 'bar',
  },
  {
    updated_at: '2012-01-09T11:25:13Z',
    foo: 'bar',
  },
  {
    updated_at: '2012-01-05T04:13:24Z',
    foo: 'bar',
  }
];

const sortByUpdatedAt = (items) => [...items].sort((itemA, itemB) => new Date(itemA.updated_at) - new Date(itemB.updated_at));

const output = sortByUpdatedAt(input);

console.log(input);
/*
[ { updated_at: '2012-01-01T06:25:24Z', foo: 'bar' }, 
  { updated_at: '2012-01-09T11:25:13Z', foo: 'bar' }, 
  { updated_at: '2012-01-05T04:13:24Z', foo: 'bar' } ]
*/
console.log(output)
/*
[ { updated_at: '2012-01-01T06:25:24Z', foo: 'bar' }, 
  { updated_at: '2012-01-05T04:13:24Z', foo: 'bar' }, 
  { updated_at: '2012-01-09T11:25:13Z', foo: 'bar' } ]
*/

score 1 · Answer 15 · answered Aug 22 '19 at 08:36

I have created a sorting function in Typescript which we can use to search strings, dates and numbers in array of objects. It can also sort on multiple fields.

export type SortType = 'string' | 'number' | 'date';
export type SortingOrder = 'asc' | 'desc';

export interface SortOptions {
  sortByKey: string;
  sortType?: SortType;
  sortingOrder?: SortingOrder;
}


class CustomSorting {
    static sortArrayOfObjects(fields: SortOptions[] = [{sortByKey: 'value', sortType: 'string', sortingOrder: 'desc'}]) {
        return (a, b) => fields
          .map((field) => {
            if (!a[field.sortByKey] || !b[field.sortByKey]) {
              return 0;
            }

            const direction = field.sortingOrder === 'asc' ? 1 : -1;

            let firstValue;
            let secondValue;

            if (field.sortType === 'string') {
              firstValue = a[field.sortByKey].toUpperCase();
              secondValue = b[field.sortByKey].toUpperCase();
            } else if (field.sortType === 'number') {
              firstValue = parseInt(a[field.sortByKey], 10);
              secondValue = parseInt(b[field.sortByKey], 10);
            } else if (field.sortType === 'date') {
              firstValue = new Date(a[field.sortByKey]);
              secondValue = new Date(b[field.sortByKey]);
            }
            return firstValue > secondValue ? direction : firstValue < secondValue ? -(direction) : 0;

          })
          .reduce((pos, neg) => pos ? pos : neg, 0);
      }
    }
}

Usage:

const sortOptions = [{
      sortByKey: 'anyKey',
      sortType: 'string',
      sortingOrder: 'asc',
    }];

arrayOfObjects.sort(CustomSorting.sortArrayOfObjects(sortOptions));

score 1 · Answer 16 · edited Jan 12 '12 at 16:51

Sorting by an ISO formatted date can be expensive, unless you limit the clients to the latest and best browsers, which can create the correct timestamp by Date-parsing the string.

If you are sure of your input, and you know it will always be yyyy-mm-ddThh:mm:ss and GMT (Z) you can extract the digits from each member and compare them like integers

array.sort(function(a,b){
    return a.updated_at.replace(/\D+/g,'')-b.updated_at.replace(/\D+/g,'');
});

If the date could be formatted differently, you may need to add something for iso challenged folks:

Date.fromISO: function(s){
    var day, tz,
    rx=/^(\d{4}\-\d\d\-\d\d([tT ][\d:\.]*)?)([zZ]|([+\-])(\d\d):(\d\d))?$/,
    p= rx.exec(s) || [];
    if(p[1]){
        day= p[1].split(/\D/).map(function(itm){
            return parseInt(itm, 10) || 0;
        });
        day[1]-= 1;
        day= new Date(Date.UTC.apply(Date, day));
        if(!day.getDate()) return NaN;
        if(p[5]){
            tz= (parseInt(p[5], 10)*60);
            if(p[6]) tz+= parseInt(p[6], 10);
            if(p[4]== '+') tz*= -1;
            if(tz) day.setUTCMinutes(day.getUTCMinutes()+ tz);
        }
        return day;
    }
    return NaN;
}
if(!Array.prototype.map){
    Array.prototype.map= function(fun, scope){
        var T= this, L= T.length, A= Array(L), i= 0;
        if(typeof fun== 'function'){
            while(i< L){
                if(i in T){
                    A[i]= fun.call(scope, T[i], i, T);
                }
                ++i;
            }
            return A;
        }
    }
}
}

score 0 · Answer 17 · answered May 10 '18 at 06:22

For completeness here is a possible short generic implementation of sortBy:

function sortBy(list, keyFunc) {
  return list.sort((a,b) => keyFunc(a) - keyFunc(b));
}

sortBy([{"key": 2}, {"key": 1}], o => o["key"])

Note that this uses the arrays sort method that sorts in place. for a copy you can use arr.concat() or arr.slice(0) or similar method to create a copy.

score -1 · Answer 18 · edited Jun 01 '16 at 12:14

You can create a closure and pass it that way here is my example working

$.get('https://data.seattle.gov/resource/3k2p-39jp.json?$limit=10&$where=within_circle(incident_location, 47.594972, -122.331518, 1609.34)', 
  function(responce) {

    var filter = 'event_clearance_group', //sort by key group name
    data = responce; 

    var compare = function (filter) {
        return function (a,b) {
            var a = a[filter],
                b = b[filter];

            if (a < b) {
                return -1;
            } else if (a > b) {
                return 1;
            } else {
                return 0;
            }
        };
    };

    filter = compare(filter); //set filter

    console.log(data.sort(filter));
});

score -1 · Answer 19 · answered Jan 30 '20 at 11:10

I face with same thing, so i handle this with a generic why and i build a function for this:

//example:
//array: [{name: 'idan', workerType: '3'}, {name: 'stas', workerType: '5'}, {name: 'kirill', workerType: '2'}]
//keyField: 'workerType'
// keysArray: ['4', '3', '2', '5', '6']

sortByArrayOfKeys = (array, keyField, keysArray) => {
    array.sort((a, b) => {
        const aIndex = keysArray.indexOf(a[keyField])
        const bIndex = keysArray.indexOf(b[keyField])
        if (aIndex < bIndex) return -1;
        if (aIndex > bIndex) return 1;
        return 0;
    })
}

Sort array of objects by single key with date value

19 Answers19

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