Javascript array + index

Question

Im still needing help with this, and have edited the jsfiddle post to show my problem. http://jsfiddle.net/7ztEf/6/

I want to return number to associated index value [0] =0 [1]=1 as you can see the index string returns all numbers. Thanks again Paul

I have a number generator script that returns values to DIV ID's. I need to hook into this somehow, to enable replacing color based upon the number value i.e. > 1 && <= 20 = red etc.

function myNumbers(numbers, type) {
    for (var x in numbers) {
       document.getElementById(type + x).innerHTML = numbers[x];
    } 
}

This script fills each of the DIVs named num0 ... num3 with a random number.

I have managed to query the first value of numbers[x] but need to set an index order to loop through the rest, or something.

Your code seems fine, Can you check this demo and tell what is the desired o/p jsfiddle.net/arunpjohny/7ztEf — Arun P Johny
– Arun P Johny, Commented Mar 4, 2013 at 3:42
You should not use for..in with arrays, it will iterate over all enumerable properties (including inherited properties), not just numeric indices. Also, order isn't guaranteed (but that probably isn't an issue here). Use a normal for loop with incrementing index, or forEach with a shim for older browsers. — RobG
– RobG, Commented Mar 4, 2013 at 4:02
Hi thanks Johny, using the jsfiddle you posted I am wanting to query the value of myNumbers, if value is 1 then div bg =blue, if value is two bg =red etc — tuvboy
– tuvboy, Commented Mar 4, 2013 at 10:28

pdoherty926 · Accepted Answer · 2013-03-04 03:36:31Z

0

Use Array.forEach.

numbers.forEach(function (number, index) {...})

answered Mar 4, 2013 at 3:36

pdoherty926

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slebetman · Accepted Answer · 2013-03-07 03:53:59Z

0

Don't use for..in for arrays. They're meant to be used on objects so using for..in on arrays will return such things as the length element.

Either use forEach as ethagnawl mentioned or use the traditional for loop:

for (var x=0; x < numbers.length; x++) {
   document.getElementById(type + x).innerHTML = numbers[x];
}

answered Mar 7, 2013 at 3:53

slebetman

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6 Comments

tuvboy Over a year ago

Thanks @slebetman! This still returns [0] = 0,1,2,3. I need to list index [0][1][2] and the numbers value. Cheers P

slebetman Over a year ago

Very close. There are still things you're misunderstanding. I just changed a few characters in you code to fix it: jsfiddle.net/h9AS6/1

slebetman Over a year ago

Read the differences between my jsfiddle and yours carefully.

tuvboy Over a year ago

Thanks again @slebetman that has helped me, now I can access the array index. Could you recommend how I should compare the returned index values? ie for each index[0][1][2][3] if myNumbers == 1 than do something else if myNumbers ==2 do something. JS is not my thing so appreciate any help!!! cheeers P

slebetman Over a year ago

numbers[x] is the value x is the key and numbers is the array. So just do if(numbers[x] == 1) ... etc

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