cannot print javascript array value

Question

var menu_items = 0;

$(".trueol").each(function(){

    menu_items = menu_items + 1;

});

var product_price = new Array(menu_items);

var product_name = new Array(menu_items);

        $(".price").each(function(){
        var i = 0;
            product_price[i] = $(this).data("p");
            document.write(product_price[i]); /*Inner document.write*/
            i++;
        });
         document.write(product_price[2]); /*Outer document.write*/

        $(".menu_product").each(function(){
            var j = 0;
            product_name[j] = $(this).text();
            document.write(product_name[j]); /*Inner document.write*/
            j++;
        });

           document.write(product_name[2]); /* Outer document.write */

Here, the inner document.write works just fine but outer ones print undefined. P.S menu_items has a value of 4 after iterating. What's the problem ?

It's a scope issue. product_name is correctly not defined inside that function. You need to read up on variable scope. :) — i-CONICA
– i-CONICA, Commented May 1, 2014 at 8:49
i think u should put var i=0 and var j=0 out of the loop. otherwise it will always save data in array at position 0 no increment — mayank chaturvedi
– mayank chaturvedi, Commented May 1, 2014 at 8:50
@i-Conica: the variable is defined, otherwise the op would get a reference error. — Felix Kling
– Felix Kling, Commented May 1, 2014 at 8:51
Yeah I looked again, it's either been edited or it's too early in the morning for me. :) — i-CONICA
– i-CONICA, Commented May 1, 2014 at 8:53

Reinstate Monica Cellio · Accepted Answer · 2014-05-01 08:55:24Z

1

Answers explaining using the each iterator's index are correct, but you're overcomplicating the issue. You can simply push the values onto the end of the arrays and miss out any counting...

var product_price = [];
var product_name = [];

$(".price").each(function(){
    var value = $(this).data("p");
    product_price.push(value);
    document.write(value); /*Inner document.write*/
});
document.write(product_price[2]); /*Outer document.write*/

$(".menu_product").each(function(){
    var value = $(this).text();
    product_name.push(value);
    document.write(value); /*Inner document.write*/
});

document.write(product_name[2]); /* Outer document.write */

https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/Array/push

Should you need to know the number of items in the arrays at any point, just use

product_price.length

or

product_name.length

One other hint. I assume you're using document.write as a method of outputting data for debugging purposes. Try using console.log("text"); instead, and then open the browser console (usually F12 and go to the console tab).

answered May 1, 2014 at 8:55

Reinstate Monica Cellio

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1 Comment

Reinstate Monica Cellio Over a year ago

You're welcome - it's much better to use that than pollute the DOM with something that doesn't need to be there.

martynas · Accepted Answer · 2014-05-01 09:02:51Z

1

The problem is you reset your counter at every single iteration (var i = 0 and var j = 0).

Just use .each() with an index.

// Arrays for your values
var product_price = [];
var product_name = [];

$(".price").each(function(index) {
    product_price.push($(this).data("p"));
    document.write(product_price[index]); /*Inner document.write*/
});

$(".menu_product").each(function(index) {
    product_name.push($(this).text());
    document.write(product_name[index]); /*Inner document.write*/
});

// Outer write
document.write(product_name[2]);

edited May 1, 2014 at 9:02

answered May 1, 2014 at 8:52

martynas

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Comments

Rohan Kumar · Accepted Answer · 2014-05-01 09:05:58Z

0

Try tro use push() as in every iteration your counter sets to 0,

var product_name = [],
    product_price = [];

$(".price").each(function(){
    t = $(this).data("p");
    document.write(t); /*Inner document.write*/
    product_price.push(t); // use push here
});
document.write(product_price[2]); /*Outer document.write*/

$(".menu_product").each(function(){
     t = $(this).text();
     document.write(t); /*Inner document.write*/
     product_name.push(t); // use push here
});
document.write(product_name[2]); /* Outer document.write */

edited May 1, 2014 at 9:05

answered May 1, 2014 at 8:52

Rohan Kumar

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