Sorting array by alphabet will not function correctly

Question

Code I am talking about: https://jsfiddle.net/sbe8yzv0/8/

I want to sort my array by the name with two buttons: First button sorts it by length and second buttons sorts it by alphabet.

When I sort by alphabet it doesn't want to function. It randomly puts the data in different positions. I have tried multiple ways of sorting it and they all act funky and doesn't sort it completely alphabetically. What can I do to make it actually sort the name in array by alphabet?

    function sortNameAlphabetically(a, b) {
        return a.name > b.name;
    }

The sort by length is working as intended except when it sorts alpabeticly after length it does it bakwards. How do I make sure it sorts the right way?

    function sortNameByLength(a, b) {
        return b.name.length - a.name.length;
        a.localeCompare(b);

[Your first comparison function is invalid](http://stackoverflow.com/q/24080785/1048572). — Bergi, Jun 14 '16 at 14:14

score 2 · Accepted Answer · answered Jun 14 '16 at 13:57

2

change the sortNameAlphabetically method to

  function sortNameAlphabetically(a, b) {
        return a.name.localeCompare( b.name );
    }

updated fiddle

Also, statement after return statement in sortNameByLength method is not reachable and not required anyways.

    function sortNameByLength(a, b) {
        return b.name.length - a.name.length;
    }

answered Jun 14 '16 at 13:57

gurvinder372

66,980
10
72
94

Awesome. Do you know how I can fix the second problem? – Thomasdc Jun 14 '16 at 14:00
1

@Thomasdc simply `return a.name.length - b.name.length;` – gurvinder372 Jun 14 '16 at 14:01
But how do I make it sort it alphabetically after being sorted by length then? So if I have ABA, BBBB, AAAA, AAA. It should be AAAA, BBBB, AAA, ABA – Thomasdc Jun 14 '16 at 14:05

Nina Scholz · Answer 2 · 2016-06-14T14:16:07.167

1

Part two of the question, to maintain the alphabetical sorting after length sorting, you can use a logical OR with the localeCompare method.

function sortNameByLength(a, b) {
    return b.name.length - a.name.length || a.name.localeCompare(b.name);
}

edited Jun 14 '16 at 14:16

answered Jun 14 '16 at 14:11

Nina Scholz

376,160
25
347
392

I have already tried this and it doesn't work. Nothing happens. https://jsfiddle.net/sbe8yzv0/10/ – Thomasdc Jun 14 '16 at 14:15
1

please try edit. you need to address the name property. – Nina Scholz Jun 14 '16 at 14:16
1

Very nice. Thanks a lot! :) – Thomasdc Jun 14 '16 at 14:18

Sorting array by alphabet will not function correctly

2 Answers2