If you want to use two different functions in the chain then you need
to define two plugin functions:
// wrap it all in a closure so you can share private data easily
(function() {
$.fn.myPlugin1 = function(options) {
var myOptions = $.extend(defaults, options);
// do stuff
};
$.fn.myPlugin2 = function(options) {
var myOptions = $.extend(defaults, options);
// do stuff
};
var defaults = {
def1: 'default1',
def2: 'default2',
...
defx: 'defaultx'
};
})();
On Tue, Mar 4, 2008 at 2:09 PM, Leanan <[EMAIL PROTECTED]> wrote:
>
> Ok, I'm really trying to wrap my head around this, and it's irritating
> me. I've checked out the following pages:
>
> http://docs.jquery.com/Plugins/Authoring
> http://www.learningjquery.com/2007/10/a-plugin-development-pattern
>
> And they help some but there's one problem I'm running into.
>
> I have:
>
> $.fn.myPlugin = function(options) {
> $.myPlugin.defaults = $.extend($.myPlugin.defaults, options);
> return(this);
> };
>
> $.myPlugin = {
> defaults: {
> def1: 'default1',
> def2: 'default2',
> ...
> defx: 'defaultx'
> },
>
> myFunc: function() {
> var options = $.myPlugin.defaults;
> //do something here, anything really
> },
>
> myFunc2: function() {
> var options = $.myPlugin.defaults;
> //do something here, again, anything
> }
> };
>
> Now, I want to do something like this:
>
> jQuery().myPlugin({def1: 'other_def1', def2: 'other_def2'}).myFunc();
>
> Can I?
>
> Because it's not working. It tells me that jQuery().myPlugin({def1:
> 'other_def1', def2: 'other_def2'}) is not a function.
>
> However, if I change it so that myFunc is $.fn.myFunc = function()
> { }, then I can, but I have to do it as:
>
> jQuery().myPlugin({def1: 'other_def1', def2:
> 'other_def2'}).myPlugin.myFunc();
>
> Which I don't like.
>
> None of the tutorials or documentation I can find is clear on this
> point, though they do refer to the functions in $.myPlugin as being
> user-accessable... and they are, but only if I do $.myPlugin.myFunc.
> I can't chain. And I want to be able to do that.
>
