How to sort array using Jquery

Question

I have one array, which includes children. I have successfully sorted the array but unable to sort it, children.

Using the below code I am able to sort outer mail array elements but not its children.

var arry = [{
    'id': 301,
    'name': '2 Foo',
    'open': 'open',
    'children': [{
        'id': 1313,
        'name': '2.1 Foo ',
        'open': 'open'
      },
      {
        'id': 1143,
        'name': '2.3 Foo ',
        'open': 'open'
      },
      {
        'id': 1132,
        'name': '2.2 Foo ',
        'open': 'open'
      },
    ],
  },
  {
    'id': 30,
    'name': '1 Foo',
    'open': 'open',
    'children': [{
        'id': 1134,
        'name': '1.1 Foo ',
        'open': 'open'
      },
      {
        'id': 1130,
        'name': '1.3 Foo ',
        'open': 'open'
      },
      {
        'id': 1123,
        'name': '1.2 Foo ',
        'open': 'open'
      },
    ],
  },
];


function SortByName(a, b) {
  var aName = a.name.toLowerCase();
  var bName = b.name.toLowerCase();
  return ((aName < bName) ? -1 : ((aName > bName) ? 1 : 0));
}


$(document).ready(function() {
  var sorted_array = arry.sort(SortByName)
  console.log(sorted_array)
})

<script src="https://cdnjs.cloudflare.com/ajax/libs/jquery/3.3.1/jquery.min.js"></script>
<p id='data'>

</p>

@c.grey call .sort(SortByName) on the children of each element. — VLAZ
– VLAZ, Commented May 7, 2020 at 7:43

Nina Scholz · Accepted Answer · 2020-05-07 07:40:53Z

4

You need to iterate the array as well for sorting all children. An assignment is not necessary, because Array#sort mutates the array.

array.forEach(({ children }) => children.sort(sortByName));

answered May 7, 2020 at 7:40

Nina Scholz

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Comments

Deepu · Accepted Answer · 2020-05-07 07:47:00Z

1

Please modify your function as below

function SortByName(a, b){
	if(a.children){
  	a.children = a.children.sort(SortByName)
  }
  if(b.children){
  	b.children = b.children.sort(SortByName)
  }
  var aName = a.name.toLowerCase();
  var bName = b.name.toLowerCase();
  return ((aName < bName) ? -1 : ((aName > bName) ? 1 : 0));
}

answered May 7, 2020 at 7:47

Deepu

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6 Comments

Zuckerberg Over a year ago

Best solution so far!

Deepu Over a year ago

@c.grey i didn't do anything... just called your function itself........ :)

Nina Scholz Over a year ago

no, it's a worst solution, because the children are sorted for every parent element over and over.

VLAZ Over a year ago

@c.grey the complexity of the sorting witll be WAY higher by repeatedly sorting. Heck, based on the sorting algorithm implementation, you might get the worst complexity as algorithms like quicksort have worst performance on an already sorted dataset. Even if that's avoided, it's completely unnecessary to sort and re-sort the children.

rahul.m Over a year ago

@VLAZ do you have any optimal solution.

|

Aman · Accepted Answer · 2020-05-07 07:44:36Z

0

var updated_sorted_array = sorted_array.map(function(arr){
  arr.children = arr.children.sort(SortByName);
  return arr;
})

answered May 7, 2020 at 7:44

Aman

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1 Comment

riffnl Over a year ago

next time please add information and explanation on why and how

Makwana Prahlad · Accepted Answer · 2020-05-07 07:48:13Z

0

you need 2 functions to sort these objects, one for Desc and one for Asc sort like this:

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

And when you want to sort your objects you can pass in the original collection and the Column that you need to sort based on like this:

  function querySucceeded(data) {
     posts = [];
     posts = sortByKeyDesc(data, "TagName");
  }

answered May 7, 2020 at 7:48

Makwana Prahlad

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1 Comment

VLAZ Over a year ago

You never need to implement two functions from scratch for sorting. To sort in reverse order, all you need is to negate the result of the correct order, so a generic reverseSort = sortingFn => (...args) => -1 * sortingFn(...args) works. Or you can make it more concrete by just flipping the arguments reverseSort = fn => (a, b) => fn(a, b). If you have a functional utility library, you can just use a ready made solution sortAsc = { /* ... logic ... */ }; sortDesc = flip(sortAsc); This way you don't have to repeat yourself.

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