asked on

jquery plugin: call to method within plugin

Hi,

I have this plugin:

(function( $ ){

    var methods = {
        init : function( options ) {
            var settings = $.extend( {
                'location' : 'top',
                'background-color' : 'blue'
            }, options);

            alert ( settings.location );
        },
        show : function( ) {
            // IS
        },
        hide : function( ) {
            // GOOD
        },
        update : function( content ) {
            // !!!
        },
        display : function ( test1, test2 ) {
            var test1 = test1;
            alert ( "test1: " + test1 + "\ntest2: " + test2 );
            self.display2( test2 );
        },
        display2 : function ( test3 ) {
            alert ( "test3: " + test3 );
        }
    };

    $.fn.tooltip = function( method ) {

        // Method calling logic
        if ( methods[method] ) {
            return methods[ method ].apply( this, Array.prototype.slice.call( arguments, 1 ));
        } else if ( typeof method === 'object' || ! method ) {
            return methods.init.apply( this, arguments );
        } else {
            $.error( 'Method ' +  method + ' does not exist on jQuery.tooltip' );
        }

    };

})( jQuery );

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There is this method display() that is calling the other method display2() but display2 is never executed. How can I make it work?

I have the following in my as my init.js:

$(document).ready(function() {

// calls the init method
    $('div').tooltip();

// calls the init method
    $('div').tooltip({
        'location' : 'left'
    });

    $().tooltip('display', 'John', '316');
});

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All the alerts are working except display2().

thanks,
Victor

Michel Plungjan

Can you try this. instead of self.

Michel Plungjan

And define display2 before you call it

Victor Kimura

ASKER

this.display2 ( test2 );

The error is Uncaught TypeError: Object [object Object] has no method 'display2'

self.display2 ( test2 );

gives the same error.

tried this but it doesn't work:
$('').tooltip.display2( test2 );

No certain what you mean by "defining" display2 before it's called. How do I do that?

Thanks,

ASKER CERTIFIED SOLUTION

Jon Norman

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See answer

Victor Kimura

ASKER

Thank you. =)