jQuery each() not working

Question

Am I doing something wrong here? I have an array of tags, and when I do a jQuery each() on the array it doesn't go into the each() I did have an alert in the each but nothing happens. I have checked my error log console and there are no errors. So, what am I doing wrong?

var tags = new Array();
tags["video-games"] = "Video Games";
tags["sports"] = "Sports";
tags["movies"] = "Movies";
tags["board-games"] = "Board Games";
tags["news"] = "News";
tags["television"] = "Television";
tags["computers"] = "Computers";
tags["opinions"] = "Opinions";
tags["reviews"] = "Reviews";

function updateTags(){
    console.log(tags);
    $("div.tags > div > span:first-child").nextAll().remove();
    $.each(tags, function(key, val){
        $("div.tags > div").append("<span><a class='tag' href='/tags/" + key + "'>" + val + "</a></span>");

    });
}

updateTags();

Arrays don't work like that, use an object.

rlemon
– rlemon

2013-10-08 15:15:23 +00:00
Commented Oct 8, 2013 at 15:15 — rlemon
– rlemon, Commented Oct 8, 2013 at 15:15
String keys are not supported javascript array

u_mulder
– u_mulder

2013-10-08 15:15:31 +00:00
Commented Oct 8, 2013 at 15:15 — u_mulder
– u_mulder, Commented Oct 8, 2013 at 15:15
You used Array[] like Object {}.

Murali Murugesan
– Murali Murugesan

2013-10-08 15:19:18 +00:00
Commented Oct 8, 2013 at 15:19 — Murali Murugesan
– Murali Murugesan, Commented Oct 8, 2013 at 15:19

SLaks · Accepted Answer · 2013-10-08 15:14:09Z

10

Arrays are expected to have numeric indexes.
You've created an empty array which happens to have some properties.

You should create an ordinary object instead:

var tags = {
    "video-games": "Video Games",
    ...
};

answered Oct 8, 2013 at 15:14

SLaks

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1 Comment

Akshay Khandelwal Over a year ago

I agree. As you have a key-value pair use th eobject and use the Javascript for(var key in obj){console.log(obj[key]);} to get the expected results

Akshay Khandelwal · Accepted Answer · 2013-10-08 15:23:34Z

0

So in this case

var tags = {};
tags["video-games"] = "Video Games";
tags["sports"] = "Sports";
tags["movies"] = "Movies";
tags["board-games"] = "Board Games";
tags["news"] = "News";
tags["television"] = "Television";
tags["computers"] = "Computers";
tags["opinions"] = "Opinions";
tags["reviews"] = "Reviews";

function updateTags(){
    //console.log(tags);
    $("div.tags > div > span:first-child").nextAll().remove();
    for(var key in tags){
        $("div.tags > div").append("<span><a class='tag' href='/tags/" + key + "'>" + tags[key] + "</a></span>");
    }
}

updateTags();

Should work.

edited Oct 8, 2013 at 15:23

answered Oct 8, 2013 at 15:18

Akshay Khandelwal

1,56811 silver badges19 bronze badges

1 Comment

rlemon Over a year ago

don't forget about Object.hasOwnProperty (In this case I don't see it making a difference), when iterating object keys it is a good thing to remember

Murali Murugesan · Accepted Answer · 2013-10-08 21:07:58Z

0

Your code is object and properties

var tags = {};
tags["video-games"] = "Video Games";

or

var tags = {
"video-games" : "Video Games";
};

Then

$.each(tags, function(key, val) {
    $("div.tags > div").append("<span><a class='tag' href='/tags/" + key + "'>" 
                                + val + "</a></span>");

});

edited Oct 8, 2013 at 21:07

answered Oct 8, 2013 at 15:21

Murali Murugesan

22.6k18 gold badges76 silver badges123 bronze badges

3 Comments

rlemon Over a year ago

in your second example you would get an invalid left hand side in assignment error because it sees the '-' as a operator and not part of the property name. Only bracket notation can be used for such names

Murali Murugesan Over a year ago

@rlemon, Thanks for informing.. I didnt noted - :( I will get change it

rlemon Over a year ago

your new example will also throw errors, you need to replace the '=' with ':' in object literal property assignments, and drop the semi-colon inside the object. var tags = { 'video-games' : 'Video Games' };

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