Fails to call a plugin - jquery-plugins

im trying to develop a plugin that render a datagrid to html element. For example in:
<div id="datagrid">
<!-- Renderizado via Backbone.js -->
</div>
I have this plugin definition:
// csDatagrid.js
(function($) {
$.fn.csDatagrid = function(options) {
// Code ...
};
}, (jQuery));
And i call the function in this way:
// csAuditarSesiones.js
$("#datagrid").csDatagrid({
columns: cols,
url: $("#sfAction").val(),
filterBy: 'user_str'
});
Chrome says:
Uncaught TypeError: Object [object Object] has no method 'csDatagrid'
Load library queue (Chrome Developer Tools):
JQuery
csDatagrid (my plugin)
csAuditarSesiones (script with code for the current page, have the plugin call)
Thanks !
EDIT 1
Apparently the plugin not load, follow code always print "Not Loaded !":
if(jQuery().csDatagrid) {
console.log("Loaded !");
}else{
console.log("Not Loaded !");
}

The reason here is you are defining your plugin in the document ready event. Therefore the elements calling this plugin must call the plugin after the plugin has been loaded.
Try defining your new plugin outside of a load\ready event such as.
(function($) {
$.fn.csDatagrid = function(options) {
// Code ...
};
}, (jQuery));
$(document).ready(function(){
$("#datagrid").csDatagrid({
columns: cols,
url: $("#sfAction").val(),
filterBy: 'user_str'
});
});
Just for refrence (surely reading it now). http://learn.jquery.com/plugins/basic-plugin-creation/
EDIT:
Is it possible you have multiple jQuery versions being loaded? This can happen when there are conflicts. Also ensure that your plugin is loaded after the javascript file and before the document ready function.
There is one other (i have no idea why I believe it is when multiple versions are loaded) phenomenom that happens and you have to add the $ call back to your ready function. Such as.
$(document).ready(function($){
//TODO: Code here
});

There is error in your immediate invoked function's last line.
}, (jQuery)); should be :
})(jQuery);

Related

Applying JQuery plugin to new elements inserted into DOM through AJAX using each()

I'm using the JQuery plugin customSelect() http://adam.co/lab/jquery/customselect/ to style dropdown boxes.
If I initialize the script more than once on the same elements, they cease to work. I need to figure out a way to only initialize the plugin on newly created elements inserted into the DOM through AJAX.
So, I initialize it on all drop downs (the desired effect) -
$('select').customSelect();
and elements that I select this plugin work for are given a class of "hasCustomSelect", like so
<select class="hasCustomSelect"></select>
but if I initialize twice -
$('select').customSelect();
$('select').customSelect();
it will literally stop working.
So, in my AJAX request, I have it initializing on elements WITHOUT that class, but it doesn't seem to be working. Here is the AJAX -
$.ajax({
type: 'POST',
url: 'wp-content/themes/themetitle/functions/get-child.php',
data: { childID : theID, URL : theURL, classlevel : classlevel },
beforeSend:function() {
},
success:function(data) {
$('#makesort').append(data);
makesort();
$.each('select', function() {
if( $(this).hasClass('hasCustomSelect') ) {
//DO NOTHING!
} else {
$(this).customSelect();
}
});
},
error:function() {
}
});
So, the thought I had was after the data is appended to my div, I'll go through each <select> element, determine if it already contains the class hasCustomSelect, and if not, then initialize the plugin for this element.
This doesn't seem to have the desired effect, however, and in fact, it's producing an error. There are probably several ways to see if a plugin is already applied, but I thought checking to see if the default class that's added is there would be sufficient.
Try this for your each:
$('select:not(.hasCustomSelect)').each(function() {
$(this).customSelect();
});
or...
$.each($('select:not(.hasCustomSelect)'), function() {
$(this).customSelect();
});
The reason you're getting an error is because the first parameter to $.each(... is a collection of objects, not a selector.

simple modal loader

I am new in using jquery. I am trying add in the simplemodal.js (Eirc Martin's simple modal) a function called 'jBox' that will take the data (ie link) and options and using ajax will load the content into the modal container. This way I want on my pages in several places easy call this function
jBox = function(address, options){
$.get(address, function(data) {
$.modal(data);
});
};
});
The code is working fine, but i would like to add a loading image before the content is fully loaded. There is a lots of similar posts about loader/ spinner in simplebox but none of the works for me.
I was trying following code
$('#test').load('<img src="loader.gif">').html(data);
$('#test').modal();
But, some way, it doesnt work for me. Any ideas what I am doing wrong? Thanks
I use the ajaxStart and ajaxStop events.
$("body").on({
ajaxStart: function() {
$(this).addClass("loading"); // so page knows it's in loading state
// .. your modal code
},
ajaxStop: function() {
$(this).removeClass("loading"); // not it loading state anymore
// .. What you should do if loading is done. (eg. hide modal)
}
});
In this case I set the body class to 'loading'. So you can do some magic in css if you like.
I tend to use it to disable forms as well.
body.loading div.some-class {
// your special style for during loading
}

How to use AJAX as an alternative to iframe

I'm trying to put together a snappy webapp, utilizing JS, Prototype and AJAX for all my requests once the GUI has loaded. The app is simple: A set of links and a container element to display whatever the links point to, just like an iframe. Here's an approximate HTML snippet:
<a class="ajax" href="/somearticle.html">An article</a>
<a class="ajax" href="/anotherarticle.html">Another article</a>
<a class="ajax" href="/someform.html">Some form</a>
<div id="ajax-container"></div>
The JS that accompanies the above (sorry it's a bit lengthy) looks like this:
document.observe('dom:loaded', function(event) {
ajaxifyLinks(document.documentElement);
ajaxifyForms(document.documentElement);
});
function ajaxifyLinks(container) {
container.select('a.ajax').each(function(link) {
link.observe('click', function(event) {
event.stop();
new Ajax.Updater($('ajax-container'), link.href, {
method: 'get',
onSuccess: function(transport) {
// Make sure new ajax-able elements are ajaxified
ajaxifyLinks(container);
ajaxifyForms(container);
}
});
});
});
}
function ajaxifyForms(container) {
console.debug('Notice me');
container.select('form.ajax').each(function(form) {
form.observe('submit', function(event) {
event.stop();
form.request({
onSuccess: function(transport) {
$('ajax-container').update(transport.responseText);
// Make sure new ajax-able elements are ajaxified
ajaxifyLinks(container);
ajaxifyForms(container);
}
});
});
});
}
When clicking a link, the response is displayed in the container. I'm not using an iframe for the container here, because I want whatever elements are on the page to be able to communicate with each other through JS at some point. Now, there is one big problem and one curious phenomenon:
Problem: If a form is returned and displayed in the container, the JS above tries to apply the same behavior to the form, so that whatever response is received after submitting is displayed in the container. This fails, as the submit event is never caught. Why? Note that all returned form elements have the class="ajax" attribute.
Phenomenon: Notice the console.debug() statement in ajaxifyForms(). I expect it to output to the console once after page load and then every time the container is updated with a form. The truth is that the number of outputs to the console seems to double for each time you click a link pointing to a form. Why?
I found another way to achieve what I wanted. In fact, the code for doing so is smaller and is less error prone. Instead of trying to make sure each link and form element on the page is observed at any given time, I utilize event bubbling and listen only to the document itself. Examining each event that bubbles up to it, I can determine whether it is subject for an AJAX request or not. Here's the new JS:
document.observe('submit', function(event) {
if (event.target.hasClassName('ajax')) {
event.stop();
event.target.request({
onSuccess: function(transport) {
$('ajax-container').update(transport.responseText);
}
});
}
});
document.observe('click', function(event) {
if (event.target.hasClassName('ajax')) {
event.stop();
new Ajax.Updater($('ajax-container'), event.target.href, {
method: 'get'
});
}
});
Works like a charm :)

JQM (jQueryMobile) problem with AJAX content listview('refresh') not working

This is a mock of what I'm doing:
function loadPage(pn) {
$('#'+pn).live('pagecreate',function(event, ui){
$('#'+pn+'-submit').click( function() {
$.mobile.changePage({
url: 'page.php?parm=value',
type: 'post',
data: $('form#'+pn+'_form')
},'slide',false,false);
loadAjaxPages(pn);
});
});
function loadAjaxPages(page) {
// this returns the page I want, all is working
$.ajax({
url: 'page.php?parm=value',
type: 'POST',
error : function (){ document.title='error'; },
success: function (data) {
$('#display_'+page+'_page').html(data); // removed .page(), causing page to transition, but if I use .page() I can see the desired listview
}
});
}
in the ajax call return if I add the .page() (which worked in the past but I had it out side of the page function, changing the logic on how I load pages to save on loading times), make the page transition to the next page but I can see the listview is styled the way I want:
$('#display_'+page+'_page').html(data).page();
Removing .page() fixes the transition error but now the page does not style. I have tried listview('refresh') and even listview('refresh',true) but no luck.
Any thoughts on how I can get the listview to refresh?
Solution:
$.ajax({
url: 'page.php?parm=value',
type: 'POST',
error : function (){ document.title='error'; },
success: function (data) {
$('#display_'+page+'_page').html(data);
$("div#name ul").listview(); // add div wrapper w/ name attr to use the refresh
}
});
Be sure to call .listview on the ul element
If it didn't style earlier, you just call .listview(), bot the refresh function. If your firebug setup is correct, you should have seen an error message telling you that.
I didn't have time to get down to creating some code before you posted your fix, but here's a little recommendation from me:
if(data !== null){ $('#display_'+page+'_page').html(data).find("ul").listview() }
This is a bit nicer than a new global selector. Also - you don't need the div and you can provide a detailed selector if you have multiple ULs.
caution: the above code requires data !== null. If it's null - it will throw an error.
If you add items to a listview, you'll need to call the refresh() method on it to update the styles and create any nested lists that are added. For example:
$('#mylist').listview('refresh');
Note that the refresh() method only affects new nodes appended to a list. This is done for performance reasons. Any list items already enhanced will be ignored by the refresh process. This means that if you change the contents or attributes on an already enhanced list item, these won't be reflected. If you want a list item to be updated, replace it with fresh markup before calling refresh.
more info here.

jqGrid trigger "Loading..." overlay

Does anyone know how to trigger the stock jqGrid "Loading..." overlay that gets displayed when the grid is loading? I know that I can use a jquery plugin without much effort but I'd like to be able to keep the look-n-feel of my application consistent with that of what is already used in jqGrid.
The closes thing I've found is this:
jqGrid display default "loading" message when updating a table / on custom update
n8
If you are searching for something like DisplayLoadingMessage() function. It does not exist in jqGrid. You can only set the loadui option of jqGrid to enable (default), disable or block. I personally prefer block. (see http://www.trirand.com/jqgridwiki/doku.php?id=wiki:options). But I think it is not what you wanted.
The only thing which you can do, if you like the "Loading..." message from jqGrid, is to make the same one. I'll explain here what jqGrid does to display this message: Two hidden divs will be created. If you have a grid with id=list, this divs will look like following:
<div style="display: none" id="lui_list"
class="ui-widget-overlay jqgrid-overlay"></div>
<div style="display: none" id="load_list"
class="loading ui-state-default ui-state-active">Loading...</div>
where the text "Loading..." or "Lädt..." (in German) comes from $.jgrid.defaults.loadtext. The ids of divs will be constructed from the "lui_" or "load_" prefix and grid id ("list"). Before sending ajax request jqGrid makes one or two of this divs visible. It calls jQuery.show() function for the second div (id="load_list") if loadui option is enable. If loadui option is block, however, then both divs (id="lui_list" and id="load_list") will be shown with respect of .show() function. After the end of ajax request .hide() jQuery function will be called for one or two divs. It's all.
You will find the definition of all css classes in ui.jqgrid.css or jquery-ui-1.8.custom.css.
Now you have enough information to reproduce jqGrid "Loading..." message, but if I were you I would think one more time whether you really want to do this or whether the jQuery blockUI plugin is better for your goals.
I use
$('.loading').show();
$('.loading').hide();
It works fine without creating any new divs
Simple, to show it:
$("#myGrid").closest(".ui-jqgrid").find('.loading').show();
Then to hide it again
$("#myGrid").closest(".ui-jqgrid").find('.loading').hide();
I just placed below line in onSelectRow event of JQ grid it worked.
$('.loading').show();
The style to override is [.ui-jqgrid .loading].
You can call $("#load_").show() and .hide() where is the id of your grid.
its is worling with $('div.loading').show();
This is also useful even other components
$('#editDiv').dialog({
modal : true,
width : 'auto',
height : 'auto',
buttons : {
Ok : function() {
//Call Action to read wo and
**$('div.loading').show();**
var status = call(...)
if(status){
$.ajax({
type : "POST",
url : "./test",
data : {
...
},
async : false,
success : function(data) {
retVal = true;
},
error : function(xhr, status) {
retVal = false;
}
});
}
if (retVal == true) {
retVal = true;
$(this).dialog('close');
}
**$('div.loading').hide();**
},
Cancel : function() {
retVal = false;
$(this).dialog('close');
}
}
});
As mentioned by #Oleg the jQuery Block UI have lots of good features during developing an ajax base applications. With it you can block whole UI or a specific element called element Block
For the jqGrid you can put your grid in a div (sampleGrid) and then block the grid as:
$.extend($.jgrid.defaults, {
ajaxGridOptions : {
beforeSend: function(xhr) {
$("#sampleGrid").block();
},
complete: function(xhr) {
$("#sampleGrid").unblock();
},
error: function(jqXHR, textStatus, errorThrown) {
$("#sampleGrid").unblock();
}
}
});
If you want to not block and not make use of the builtin ajax call to get the data
datatype="local"
you can extend the jqgrid functions like so:
$.jgrid.extend({
// Loading function
loading: function (show) {
if (show === undefined) {
show = true;
}
// All elements of the jQuery object
this.each(function () {
if (!this.grid) return;
// Find the main parent container at level 4
// and display the loading element
$(this).parents().eq(3).find(".loading").toggle(show);
});
return show;
}
});
and then simple call
$("#myGrid").loading();
or
$("#myGrid").loading(true);
to show loading on all your grids (of course changing the grid id per grid) or
$("#myGrid").loading(false);
to hide the loading element, targeting specific grid in case you have multiple grids on the same page
In my issues I used
$('.jsgrid-load-panel').hide()
Then
$('.jsgrid-load-panel').show()

Resources