displaying array elements

Question

I am creating a JavaScript array is the following manner:

var selectedColors= { 'Orange' : $("#Orange").val(),
                         'Light Blue' : $("#LightBlue").val(),
                         'Dark Red' : $("#DarkRed").val(),
                         'Dark Blue' : $("#DarkBlue").val()};

Then loop through each item to see which color was not selected, and then store them in another array:

var colorsNotSelected = [];
$.each(selectedColors, function (key, value) {
    if (value.length == 0)
        colorsNotSelected.push({key:key});
});

Here I want to display the colors not selected, but doing it the following way display the keys: 0,1,2,3 instead of Orange, Light Blue, Dark Red, Dark Blue.

What am I doing wrong here?

if (colorsNotSelected.length > 0)
    $.each(colorsNotSelected, function (key) { alert(key) });
    return false;

Any help is much appreciated.

That is a Javascript Object, not a Javascript Array. FYI. Where is colorsNotSelected defined? — user400654
– user400654, Commented Nov 2, 2012 at 21:21
jsfiddle.net/6sp8s/2 <- Works fine as I see. Could you post what the value are — Selvakumar Arumugam
– Selvakumar Arumugam, Commented Nov 2, 2012 at 21:31
@KevinB miss type: added it above. selectedColors is a JS object, but why would that restrict me from accessing another array element? — user793468
– user793468, Commented Nov 2, 2012 at 21:55

Ian · Accepted Answer · 2012-11-02 22:04:15Z

1

The object and array would iterate the same in jQuery. It appears you need to use braces to keep that return false statement under check:

if (colorsNotSelected.length > 0) {
    $.each(colorsNotSelected, function (key) { alert(key) });
    return false;
}

This is unnecessary:

colorsNotSelected.push({key:key});

Just do this:

colorsNotSelected.push(key);

This is also assuming somewhere above your example code you have this:

var colorsNotSelected = [];

edited Nov 2, 2012 at 22:04

Ian

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answered Nov 2, 2012 at 21:24

SeanCannon

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1 Comment

user793468 Over a year ago

Still dont see the color names. 0,1,2,3 is displayed. Also, yes There is var colorsNotSelected = []; in my code

ContextSwitch · Accepted Answer · 2012-11-02 21:27:52Z

0

You might want to try a for / in loop instead:

for(var i in colorsNotSelected){
   alert(i);
}

answered Nov 2, 2012 at 21:27

ContextSwitch

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1 Comment

David Thomas Over a year ago

If you're going to use a for...in loop, use a if (colorsNotSelected.hasOwnProperty(i)) check to avoid iterating through the object's native/inherent properties.

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