Returning value from jquery array

Question

I created a simple html that enumerates names, with people's names on it as classes. I then get the class name and check it against the array. the result is undefined. here's my code:

var prof_name = $(this).attr('class');
var n = prof_name.split(' ');
var name_out = n[0];

var nameList = {'rom':'Rommel Santiago','holy':'Holiander Fields'};
var nameValue = nameList.nameout;

alert(nameValue); //this pops up as undefined

alert(nameList.rom) // this pops up as 'Rommel Santiago'

What did i missed?

thanks.

Axel · Accepted Answer · 2013-04-12 15:26:01Z

2

The issue is you're referencing your array incorrectly. I'm assuming your class name for $(this) is rom holy

You're retrieving rom like this: var name_out = n[0];

And you want to reference the rom key in nameList, which you can like this:

var nameValue = nameList[name_out];

After doing this, your updated code will look like this:

var prof_name = $(this).attr('class');
var n = prof_name.split(' ');
var name_out = n[0];

var nameList = {'rom':'Rommel Santiago','holy':'Holiander Fields'};
var nameValue = nameList[name_out];

alert(nameValue); //this pops up as Rommel Santiago

Working Example: http://jsfiddle.net/6GM7T/2/

Furthermore

I'd advise you to not use classes for storing data.

You can use the data attribute tag instead, like this:

<div id="myDiv" data-name="Rommel Santiago"></div>

And you can easily retrieve that data with jQuery using this:

$('#myDiv').data('name'); //returns Rommel Santiago

edited Apr 12, 2013 at 15:26

answered Apr 12, 2013 at 15:09

Axel

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3 Comments

Saint Dee Over a year ago

hi Axel, i still get the same result. if i alert nameList.name_out, i get undefined. If I alert nameList.rom then I get the value. It would appear that the name_out value is incorrectly formatted.

Axel Over a year ago

Can you post the HTML you're attempting to retrieve these values from?

Axel Over a year ago

I've updated my answer with the solution. I believe you're just referencing your array incorrectly.

jmoerdyk · Accepted Answer · 2013-04-12 15:21:38Z

1

I believe the OP wants to use the value of name_out as the index to nameList, in which case:

var nameValue = nameList.nameout;

should be:

var nameValue = nameList[name_out];

answered Apr 12, 2013 at 15:21

jmoerdyk

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Comments

Kartheek · Accepted Answer · 2013-04-12 16:02:09Z

1

As they said

It works with this

var prof_name = "rom holy";
var n = prof_name.split(' ');
var name_out = n[0];
alert("Parameter Selected = "+name_out);
var nameList = {'rom':'Rommel Santiago','holy':'Holiander Fields'};
var nameValue = nameList[name_out];

alert("From Object : "+nameValue); //this pops up as undefined

alert("Actual Value = "+nameList.rom) // this pops up as 'Rommel Santiago'

http://jsfiddle.net/Kartheek/nx3WQ/1/

if you call like nameList.nameout it actually searching for nameout as a parameter in namelist

if u use nameList[nameout] it searches for index nameout value.

answered Apr 12, 2013 at 16:02

Kartheek

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