Undefined push when trying to insert to array

Question

I'm new to Javascript, I wanted to push all the data inside my text file to array, it works fine I do this:-

var pokemon_array = [];
$.get('pokemonlist.txt', function(data) {
        var lines = data.split('\n');

        for(var i=0;i<lines.length;i++) {
            var arr = lines[i].split('"');

            pokemon_array.push(arr[1]);
            pokemon_array.push(arr[3]);
            pokemon_array.push(arr[4]);
        }
        console.log(pokemon_array);
    });

I wanted it to have 2 dimensional array so I put this:-

var pokemon_array = [];
$.get('pokemonlist.txt', function(data) {
        var lines = data.split('\n');

        for(var i=0;i<lines.length;i++) {
            var arr = lines[i].split('"');

            pokemon_id = arr[1];
            pokemon_img = arr[3];
            pokemon_name = arr[4];

            pokemon_array[i].push(pokemon_id);
            pokemon_array[i].push(pokemon_img);
            pokemon_array[i].push(pokemon_name);
        }
        console.log(pokemon_array);
    });

Then I received this error:-

Uncaught TypeError: Cannot read property 'push' of undefined

Anything I did wrong here?

pokemon_array[i] is not an array, you'll need to define it as such first pokemon_array[i] = []; pokemon_array[i].push(pokemon_id); ..... — Titus
– Titus, Commented Aug 18, 2016 at 4:19

Nikolay Ermakov · Accepted Answer · 2016-08-18 04:23:28Z

3

Change

pokemon_array[i].push(pokemon_id);
pokemon_array[i].push(pokemon_img);
pokemon_array[i].push(pokemon_name);

to

pokemon_array.push([ pokemon_id, pokemon_img, pokemon_name ]);

otherwise pokemon_array[i] is not there yet.

Also you'd better either declare your pokemon_... variables with var or just do pokemon_array.push([ arr[1], arr[3], arr[4] ]); if the only use for them is to be added to array.

edited Aug 18, 2016 at 4:23

answered Aug 18, 2016 at 4:19

Nikolay Ermakov

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Comments

Meligy · Accepted Answer · 2016-08-18 04:29:46Z

0

The problem is that pokemon_array[i], which would be pokemon_array[0], pokemon_array[1] etc. was never assigned.

You can change the code to something like:

        pokemon_array[i] = [];
        pokemon_array[i].push(pokemon_id);
        pokemon_array[i].push(pokemon_img);
        pokemon_array[i].push(pokemon_name);

Or

pokemon_array[i] = [pokemon_id, pokemon_img, pokemon_name];

Or you can just write it shorter because you can just use the array you have:

pokemon_array[i] = arr;

But you don't need the [i] at all really, so you can go like:

pokemon_array.push([pokemon_id, pokemon_img, pokemon_name]);

which can be even shorter as mentioned above:

pokemon_array.push(arr);

This does not just append each element in arr. It adds the entire arr as one element of pokemon_array, as you want exactly.

edited Aug 18, 2016 at 4:29

answered Aug 18, 2016 at 4:23

Meligy

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4 Comments

MuthaFury Over a year ago

Thanks for the explaination!

Meligy Over a year ago

No problem. Please see the updated answer about multi-dimensional array in particular. And kindly accept the answer if it solves what you want, or leave another comment telling me what problem you still have. Cheers :)

MuthaFury Over a year ago

I got a question, is it possible to add a new value on based on the key? Example:- Instead of array([0]=>abc,[1]=>def); I want to make it based on my pokemon_id variable, so the output will be something like this array(['pikachu']=>abc, ['donkey']=>def) I can't declare the pokemon_array inside the function else it will only be usable inside the function, therefore I declaring the variable outside of the function

Meligy Over a year ago

If the IDs are unique, you are better off with an object instead of an array. var pokemons = {}; and pokemons[pokemon_id] = {id: pokemon_id, img: pokemon_img, name: pokemon_name};.

Văn Tuấn Phạm · Accepted Answer · 2016-08-18 04:25:52Z

0

You get the error because pokemon_array[i] is undefined

pokemon_array[i].push(pokemon_id);
pokemon_array[i].push(pokemon_img);
pokemon_array[i].push(pokemon_name);

to

pokemon_array.push(pokemon_id);
pokemon_array.push(pokemon_img);
pokemon_array.push(pokemon_name);

answered Aug 18, 2016 at 4:25

Văn Tuấn Phạm

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