Solved

JavaScript example, whats wrong here?

Posted on 2014-04-08
3
199 Views
Last Modified: 2014-04-09
<div id="one"></div>
<div id="two"></div>
<div id="three"></div>

Open in new window


var customer = new function(name) {
    this.name = name;
    $('#one').html('Inside: ' + name);
    return {
        getName: function() {
            return name;
        },
        setName: function(newName) {
            name = newName;
        }
    };
}('Olaf');
customer.setName('Albert');
$('#two').html('Outside-1: ' + customer.getName());
$('#three').html('Outside-2: ' + customer.name)

Open in new window


Output:

Inside: Olaf
Outside-1: Albert
Outside-2: undefined

Why is the last output undefined? Didn't I set a property called name with a value?
0
Comment
Question by:itnifl
3 Comments
 
LVL 18

Expert Comment

by:Jerry Miller
ID: 39987259
I would use a different identifier from 'name', there may be some conflict using multiple instances of name since it is a keyword. I think that you were setting this.name to itself and that was producing your undefined error.

var customer = new function(n) {
    this.name = n;
    $('#one').html('Inside: ' + n);
    return {
        getName: function() {
            return n;
        },
        setName: function(newName) {
            n = newName;
        }
    };
}('Olaf');
customer.setName('Albert');
$('#two').html('Outside-1: ' + customer.getName());
$('#three').html('Outside-2: ' + customer.name)

Open in new window

0
 
LVL 82

Accepted Solution

by:
leakim971 earned 500 total points
ID: 39987296
Why is the last output undefined?

Open in new window


What do you return?
You're not returning the name

Check this page : http://jsfiddle.net/Q8q3h/

var Customer = function(name) {
    $('#one').html('Inside: ' + name);
    return {
        name:name,
        getName: function() {
            return this.name;
        },
        setName: function(newName) {
            this.name = newName;
        }
    };
}

var customer = new Customer("Olaf");
customer.setName('Albert');
$('#two').html('Outside-1: ' + customer.getName());
$('#three').html('Outside-2: ' + customer.name)

Open in new window

0
 
LVL 2

Author Comment

by:itnifl
ID: 39988026
If I do this customer.name also works:
var customer = new function(n) {
    this.name = n;
    $('#one').html('Inside: ' + n);
}('Olaf');

$('#three').html('Outside-2: ' + customer.name)

Open in new window


But if I choose to return what attributes are available like in the question, then this.name = n; will not set an attribute named name. Does it get overwritten by the return statement or what happens to it?
0

Featured Post

Is Your Active Directory as Secure as You Think?

More than 75% of all records are compromised because of the loss or theft of a privileged credential. Experts have been exploring Active Directory infrastructure to identify key threats and establish best practices for keeping data safe. Attend this month’s webinar to learn more.

Question has a verified solution.

If you are experiencing a similar issue, please ask a related question

Learn by example how to specify CSS selectors for Selenium WebDriver test automation software.
Get to know the ins and outs of building a web-based ERP system for your enterprise. Development timeline, technology, and costs outlined.
Explain concepts important to validation of email addresses with regular expressions. Applies to most languages/tools that uses regular expressions. Consider email address RFCs: Look at HTML5 form input element (with type=email) regex pattern: T…
This tutorial will teach you the core code needed to finalize the addition of a watermark to your image. The viewer will use a small PHP class to learn and create a watermark.

861 members asked questions and received personalized solutions in the past 7 days.

Join the community of 500,000 technology professionals and ask your questions.

Join & Ask a Question

Need Help in Real-Time?

Connect with top rated Experts

24 Experts available now in Live!

Get 1:1 Help Now