Javascript functions in custom namespaces

Question

It is possible to declare 2 more functions in main function like this ?

var jquery4u = {
    init: function() {
        jquery4u.countdown.show();
    },
    countdown: function() {
        show: function() {
            console.log('show');
        },
        hide: function() {
            console.log('hide');
        }
    }
}
jquery4u.init();

and i receive the following error: Uncaught SyntaxError: Unexpected token ( on this line "show: function() {"

What happens when you try it? Did it work? If not, why not? (It won't work because of invalid syntax.. a function is not an object literal and so those 'keys' - show, hide - are really labels.) — user2864740
– user2864740, Commented Jul 16, 2015 at 23:28
show and hide need to be inside of an object {} not a function. — Mottie
– Mottie, Commented Jul 16, 2015 at 23:31

Mottie · Accepted Answer · 2015-07-16 23:33:26Z

2

Remove the function from the right of the countdown (demo)

var jquery4u = {
    init: function() {
        jquery4u.countdown.show();
    },
    countdown: {
        show: function() {
            console.log('show');
        },
        hide: function() {
            console.log('hide');
        }
    }
}
jquery4u.init();

Next time, use jsFiddle to make a demo and click the "JSHint" button.

answered Jul 16, 2015 at 23:33

Mottie

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1 Comment

Madalin Alexandru Over a year ago

ohhh thank you @Mottie !! also it's a new feature for me that "jshint".. haha thanks!

user5125586 · Accepted Answer · 2015-07-16 23:46:55Z

Actually, none of this will work. Unless you make countdown an object or you treat its sub-functions as proper functions. Why: Under countdown, you created an instance of object not a function.

var jquery4u = {
    countdown: function() {
        show = function() {
            console.log('show');
        }
        hide = function() {
            console.log('hide');
        }

        jquery4u.countdown.show();
    }
}

The above code is a valid code so it is possible. Unfortunately it will not return anything.

The proper way to do this is in this format:

var jquery4u = {
    countdown: {
        show: function() {
            console.log('show');
        },
        hide: function() {
            console.log('hide');
        }
    }
}

This will work. You can try it out by calling:

jquery4u.countdown.show();

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