CODEWITHSUNDEEP
jqueryjquery-plugins
user198729
user198729

Reputation: 63636

How to implement a jQuery plugin with this feature?

So that it can be show or hide this way:

$(selector).pluginName('show')
$(selector).pluginName('hide')

The problem is how do I know which one to show or hide?

I'm now doing it this way:

$.fn.pluginName = function(opts) {  
    var conf = $.extend({},opts);
    return this.each(function() { 
        if(conf && 'conf' == conf.type)
        {
            //ClassName is defined elsewhere
            new ClassName(conf,$(this));
        }
        else
        {
            //**show or hide corresponding instance**
        }
    });
}

Upvotes: 3

Views: 123

Answers (2)

Doug Neiner
Doug Neiner

Reputation: 66191

You can use data to associate your object with the DOM element it belongs to:

$.fn.pluginName = function(opts) {
  if(typeof(opts) == "string"){
    this.each(function(){
      // Get the associated obj
      var obj = $(this).data('ClassName');
      if(obj){
        if(opts == "show") obj.myShowMethod();
        else if (opts == "hide") obj.myHideMethod();
      }
    })
  } else {
    var conf = $.extend({},opts);  
    this.each(function() { 
      var obj = new ClassName(conf,$(this));
      // Associate your object with this DOM element
      $(this).data('ClassName', obj);
    });
  }
  return this; // Don't break the chain
}

Also check out, Starter for jQuery, which uses the data pattern to associate the object with the DOM element.

Upvotes: 2

Emil Vikström
Emil Vikström

Reputation: 91922

.toggle() is probably what you are looking for. It will toggle the element between hidden and shown (depending on what state it is in at the moment).

Upvotes: -1

Related Questions