array.sort not working on array of objects

Question

I got the following object array:

var arr = [{
    2: {
        1: { name: "test" },
        2: { name: "apple" }
    },
    3: {
        1: { name: "banana" },
        2: { name: "pear" }
    }
}];

Just some sample data. Now, I got 3 textareas:

<textarea id="first"></textarea>
<textarea id="second"></textarea>
<textarea id="third"></textarea>

And I have the following custom-made function:

function sort(alt)
{
    arr.sort(function (a,b) 
    {
        console.log(a);
        if (a[2].name < a[2].name)
            return (alt) ? 1 : -1;
        if (a[2].name > a[2].name)
            return (alt) ? -1 : 1;
        return 0;
    });
}

It should sort the array of objects by name, ascending or descending according to parameter. Now, I got 2 problems. This way I append all the values to the textareas:

for (var key in arr[0])
{
    var obj = arr[0][key];
    $(ID).append(obj[2].name + '\n');
}

The first time, that code will be executed without running sort. The second time, sort will be executed with false as parameter, than that code will be executed. The third time sort will be executed with true as parameter, than that code will be executed. However, the output of all textboxes is exactly the same.

This is a link to the jsfiddle:

http://jsfiddle.net/JoshB1997/gow4vzsc/

Also, the console.log(a) doesn't get printed in the console.

if (a[2].name < a[2].name): are you sure you want to compare a[2] with a[2]? — gefei
– gefei, Commented Nov 25, 2015 at 8:47
Oops... my bad! It's still early in the morning.. Still doesn't work though. Weird thing console.log(a) isn't printed either... — Joshua Bakker
– Joshua Bakker, Commented Nov 25, 2015 at 8:48
@JoshuaBakker: a[2] has no name property, its properties are 1 and 2. They refer to objects that have a name property. Also note that the arr in your question only has one entry in it. — T.J. Crowder
– T.J. Crowder, Commented Nov 25, 2015 at 8:51
arr.length is 1. I really wonder what you want to sort in it. — Bergi
– Bergi, Commented Nov 25, 2015 at 9:16

SidOfc · Accepted Answer · 2015-11-25 09:16:47Z

1

So variable arr is an array but as far as I can see it contains only one object.

You're trying to sort directly onto the array, since it only has one object it will simply never sort because there is nothing to sort.

You'll want to access arr[0] which is the object containing the actual objects you want to sort however the Object prototype doesn't contain any of the array functions so you cannot call sort on it even tho technically an Array is an Object an Array inherits from Object and not the other way around so the methods from Object are available to Array but not the other way around.

Also, you're trying to compare the same a[2].name with itself so it'll always be false since it's equal, not > or <.

In your case I extract all the name properties from every nested object you have like this (considering the usage of the original arr):

var compare = [];
var alt = false;
for (k in arr[0]) {
    if (arr[0].hasOwnProperty(k)) {
        for (l in arr[0][k])
            if (arr[0][k].hasOwnProperty(l))
                compare.push(arr[0][k][l].name);

compare.sort(function(a, b) {
    if (a == b)
        return 0;
    else if (a < b)
        return alt ? 1 : -1
    else
        return alt ? -1 : 1
});

Now you can use the compare array to output the sorted names correctly. Also - your construction seems overly complex? It has objects and within them are nested objects but you're only sorting and displaying names, is there any reason this structure has to be maintained?

If not I would highly recommend you simplify this to just be an array of names, the loop I made above is far from beautiful and I'd rather have you not use it since it assumes that the outmost object is an object filled with other objects that all have the name property. This code could still break without an extra arr[0][k][l].hasOwnProperty('name').

Either way, the compare array simply contains all the names and it easily sortable with the default sort if you don't make things to complex for yourself.

answered Nov 25, 2015 at 9:16

SidOfc

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2 Comments

Joshua Bakker Over a year ago

Cannot change it into an array of names as it could be possible it'll get data from a database and then it has to have other data. The a with a comparison is just a small mistake. Removing the object and the keys 2 and 3 fixed it though. Thanks for the clear explanation.

SidOfc Over a year ago

I hope it helps, removing code also makes the code less error prone, glad It helped anyways! Good luck with the project @JoshuaBakker

rakwaht · Accepted Answer · 2015-11-25 09:47:27Z

0

I suggest you to use http://underscorejs.org/ that contains quite really useful function to transform from object to arrays.

For example in this case you can use something like http://underscorejs.org/#values

let values = _.values(arr[0]);

Now values is an array that contains your two object (2,3) in the form

values = [
    {
      1: {
          name: "test"
      },
      2: {
          name: "apple"
      }
    },
    {
      1: {
          name: "banana"
      },
      2: {
          name: "pear"
      }
    }
]

and here you can call your sort function

There is my demo on your code with underscore.js http://jsfiddle.net/gow4vzsc/3/

EDIT: If you cant/wont to include an entire library you can write your code for get the values:

values = [];
for(key in arr[0]){
    values.push(arr[0][key]);   
}

Here a demo without underscore.js http://jsfiddle.net/3L7ttu2r/1/

edited Nov 25, 2015 at 9:47

answered Nov 25, 2015 at 9:14

rakwaht

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3 Comments

SidOfc Over a year ago

While I agree using a library is an easy way out it's definitely not the way to go to include an entire library for such a simple task - all it'll do is add overhead to the program.

rakwaht Over a year ago

@SidneyLiebrand updated my answer based on you comment

SidOfc Over a year ago

unfortunately the question has already been answered tho, I wouldn't put any more effort into it this time for your own sake :)

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