Sorting data and its children using Javascript

Question

I have data. There are other data of the same type that have children in this data. I want to sort them all according to their 'id' fields, but I couldn't set up the algorithm scheme for it.

What I've tried:

const sortedData: [] = [];

function sortAscending(datas: Group[]) {
    sortedData.push(datas.sort((group1, group2) => group1.id - group2.id ));
    return sortedData;
}

With this code, I can only sort without children.

I tried to model the picture I wanted to describe:

Note: The returned data will be mapped and used later.

Sample array:

    0:
      children: [{…}]
      id: 1
      name: "name1"
      [[Prototype]]: Object
    1:
      children: []
      id: 7
      name: "name2"
      [[Prototype]]: Object

The 'children' field also has 'children', 'id' and 'name' fields just like themselves. What I want is to sort both the data itself and the data in the children field in it according to their 'id' fields.

Just as a note, shorten ur sort return by writing (group1.id - group2.id), there is no need to make it an if statement — Branchverse
– Branchverse, Commented Dec 17, 2021 at 13:01
Please provide a sample array and expected output after the sort array — Branchverse
– Branchverse, Commented Dec 17, 2021 at 13:03

Mr. A · Accepted Answer · 2021-12-17 13:13:56Z

1

first you need to know that array sort method sorts the array in-place it means it changes the array you pass to it. so

const rawData= [...] // your raw, un-sorted data

function sortAscending(data){...} // your function that does sorting

sortAscending(rawData) // after this, your array gets sorted 'in-place'

second, if your data has children and those children have their own children and you want them all be sorted ascending you need to write some recursive algorithm. for example:

function sortAscending(data){
  data.sort((group1, group2) => group1.id - group2.id));
  for(group of data){
     if (group.children.length) sortAscending(group.children);
  }
}

answered Dec 17, 2021 at 13:13

Mr. A

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