function doesn't work as another function parameter

Question

Can anyone tell me how to pass the draggable function as a parameter for clickit, which is an AJAX function? I want to pass draggable as a parameter to attach it to a dynamically added element. I can't get it to work defining the draggable function and pass it to clickit.

function clickit(fun){
  $.ajax({
        'url' : "url",
        'dataType' : 'json',
        'success' : function(data){
         var item_html ="";
            $.each(data.query.results.json,function(i,k){
                item_html += '<div class="dialog"><h3>'+k+'</h3></div>'
            });           
            $('.area').html(item_html);
            fun;         
          }
   });   
}
$('button').click(function(){
    var funpara = $('.dialog').draggable();
    clickit(funpara)
});

Here's the one that works:

function clickit(){
  $.ajax({
        'url' : "url",
        'dataType' : 'json',
        'success' : function(data){
         var item_html ="";
            $.each(data.query.results.json,function(i,k){
                item_html += '<div class="dialog"><h3>'+k+'</h3></div>'
            });           
            $('.area').html(item_html);
            $('.dialog').draggable();         
          }
   });   
}

$('button').click(function(){
    clickit()
});

It looks like the parameter fun cannot be passed to the success function.

Paul S. · Accepted Answer · 2014-09-11 13:53:21Z

3

It looks like you want to invoke fun, but currently you're just referencing to it without the invocation ()

function (data) {
    var item_html = "";
    $.each(data.query.results.json, function (i, k) {
        item_html += '<div class="dialog"><h3>' + k + '</h3></div>';
    });           
    $('.area').html(item_html);
    fun();         
}

In your context, it looks like you want to invoke $('.dialog').draggable(), so fun should look more like this

function fun(o) {
    return $('.dialog').draggable(o);
}

answered Sep 11, 2014 at 13:53

Paul S.

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