I have called scripts on _Layout.cshtml page and my Index.cshtml page has partial view into it. So on page load, SignalR scripts working perfect on partial view, on page end I make another ajax request and load the partial view with another data filled in that and embed under already displayed data, and then the SignalR does not work on the newly embedded record.
This is my index page code:
<div class="col-md-6">
<div class="profile-body">
<div class="row infinite-scroll">
#Html.Partial("_AlbumRow", Model)
</div>
</div>
</div>
This is my partial View Code:
#model IEnumerable<SmartKids.Lib.Core.ViewModels.FileMediaAlbumsVM>
#foreach (var item in Model)
{
<div class="widget">
<div class="block rounded">
<img src="#Url.Content(item.ImageUrl)" alt="#item.Title">
<input type="button" data-image-id="#item.imageId" class="btn btn-sm btn-default">Like</input>
</div>
</div>
}
Kindly help me how to resolve this issue that after making an ajax request I am not able to get those SignalR working. Here is more to say when I put the SignalR scripts on PartialView that works but it also sucks that on each ajax request there is again SignalR loaded on the page and when I click on LIke button it makes many calls to the function behind it.
Kindly help me to resolve this issue, I am stuck at this point since 1 week.
Here is signalR Code:
$(".btn.btn-sm.btn-default").on("click", function () {
var imageId = $(this).attr("data-image-id");
albumClient.server.like(imageId);
});
Problem: You are binding event to elements directly, So when you remove this element and replace it with a different one the events are also removed along with that element, This is something like strongly coupled.
Solution: Use Jquery event delegation. This will make sure the events will be triggered on the current elements and also all the elements that can come in future.
syntax is as below.
$(document).on("click", ".btn.btn-sm.btn-default",function () {
var imageId = $(this).attr("data-image-id");
albumClient.server.like(iamgeId);
});
NOTE: This was never a singlaR issue, it was Jquery issue.
Efficient Way: The problem in using $(document).on("click"... is that when ever there is a click happening on the entire page the Jquery framework will bubble the events from the clicked element upwards(its parent, and its parent and so on..) unless the element specified in the selector arrives, So its kind of performance hit as we don't want this check's to run if we are clicking outside the required area ( button .btn.btn-sm.btn-default in this example).
So best practice is to bind this event delegation to the closest parent possible which will not be removed, <div class="row infinite-scroll"> in this question. So that only when the click happens within this element the event bubbling will happen and also will be stopped once it reaches the parent element,it acts kind of a boundary for event bubbling.
$('.row.infinite-scroll').on("click", ".btn.btn-sm.btn-default",function () {
var imageId = $(this).attr("data-image-id");
albumClient.server.like(iamgeId);
});
I have a view with a button and a DIV
I am trying to have this kind of functionality:
if the button is clicked - a controller method is executed ( i have the method, db.insert, etc.)
- if test (inside the controller method) is passed the button dissapears and the div appears ( I thought at using ajax - not to refresh the hole page)
whenever the page is refreshed the test has to be made again for the button to be visible or not
thanks
Something like this?
{{=DIV(A('click me',callback=URL('mycallback'),target="me"),_id="me")}}
def mycallback():
# do whatever you need to do
return DIV("I will appear in place of he link when you click")
I looked in more of your examples and I think my problem was simpler ( if there isn't any other solution)
So what I did was I used eval:
button in view :
<input id="b_normal" type="button" value="normal" onClick="ajax('{{=URL('db_test')}}',[],':eval')" />
and the controller method:
def db_test()
#tests and updates
return "jQuery('#b_normal').fadeOut();jQuery('#commDiv').show();"
for further refresh i used jquery, in view:
jQuery(document).ready(function(){
var flag = '{{=flag_normal}}';
if(flag == 'da')
jQuery('#b_normal').hide();
else jQuery('#commDiv').hide();
});
where *flag_normal* is sent by the main controller
I hope this is not too inefficient and if so, useful
jQuery.validate stops my form from being submitted. I would like it to just show the user what is wrong but allow them to submit anyway.
I am using the jquery.validate.unobtrusive library that comes with ASP MVC.
I use jquery.tmpl to dynamically create the form and then I use jquery.datalink to link the input fields to a json object on the page. So my document ready call looks something like this.
jQuery(function ($) {
// this allows be to rebind validation after the dynamic form has been created
$("form").removeData("validator");
$("form").removeData("unobtrusiveValidation");
$.validator.unobtrusive.parse($("form"));
// submit the answers
$("form").submit(function(e) {
$("input[name=jsonResponse]").val(JSON.stringify(answerArray));
return true;
});
}
I note that there is an option
$("form").validate({ onsubmit: false });
but that seems to kill all validation.
So just to recap when my form is rendered I want to show all errors immediately but I don't want to prevent the submit from working.
So after some research (reading the source code) I found I needed to do 2 things
add the class cancel to my submit button
<input id="submitButton" type="submit" class="cancel" value="OK" />
This stops the validation running on submit.
To validate the form on load I just had to add this to my document ready function
$("form").valid();
Hope this helps someone else
(First of all English is not my native language, I'm sorry if I'll probably be mistaken).
I've created Yii Web app where is input form on the main page which appears after button click through ajax request. There is a "Cancel" button on the form that makes div with form invisible. If I click "Show form" and "Cancel" N times and then submit a form with data the request is repeating N times. Obviously, browser binds onclick event to the submit button every time form appears. Can anybody explain how to prevent it?
Thank you!
I've had the exact same problem and there was a discussion about it in the Yii Forum.
This basically happens because you are probably returning ajax results with "render()" instead or renderPartial(). This adds the javascript code every time to activate all ajax buttons. If they were already active they will now be triggered twice. So the solution is to use renderPartial(). Either use render the first time only and then renderPartial(), or use renderPartial() from the start but make sure the "processOutput" parameter is only set to TRUE the first time.
Solved!
There was two steps:
First one. I decided to add JS code to my CHtml::ajaxSubmitButton instance that unbind 'onclick' event on this very button after click. No success!
Back to work. After two hours of digging I realized than when you click 'Submit' button it raises not only 'click' event. It raises 'submit' event too. So you need to unbind any event from whole form, not only button!
Here is my code:
echo CHtml::submitButton($diary->isNewRecord ? 'Создать' : 'Сохранить', array('id' => 'newRecSubmit'));
Yii::app()->clientScript->registerScript('btnNewRec', "
var clickNewRec = function()
{
jQuery.ajax({
'success': function(data) {
$('#ui-tabs-1').empty();
$('#ui-tabs-1').append(data);
},
'type': 'POST',
'url': '".$this->createUrl('/diary/newRecord')."',
'cache': false,
'data': jQuery(this).parents('form').serialize()
});
$('#new-rec-form').unbind();
return false;
}
$('#newRecSubmit').unbind('click').click(clickNewRec);
");
Hope it'll help somebody.
I just run into the same problem, the fix is in the line that starts with 'beforeSend'. jQuery undelegate() function removes a handler from the event for all elements which match the current selector.
<?php echo CHtml::ajaxSubmitButton(
$model->isNewRecord ? 'Add week(s)' : 'Save',
array('buckets/create/'.$other['id'].'/'.$other['type']),
array(
'update'=>'#addWeek',
'type'=>'POST',
'dataType'=>'json',
'beforeSend'=>'function(){$("body").undelegate("#addWeeksAjax","click");}',
'success'=>'js:function(data) {
var a=[];
}',
),
array('id'=>'addWeeksAjax')
); ?>
In my example I've added the tag id with value 'addWeeksAjax' to the button generated by Yii so I can target it with jQuery undelegate() function.
I solved this problem in my project this way, it may not be a good way, but works fine for me: i just added unique 'id' to ajax properties (in my case smth like
<?=CHtml::ajaxLink('<i class="icon-trash"></i>',
$this->createUrl('afisha/DeletePlaceAjax',
array('id'=>$value['id'])),
array('update'=>'.data',
'beforeSend' => 'function(){$(".table").addClass("loading");}',
'complete' => 'function(){$(".table").removeClass("loading");}'),
array('confirm'=>"Уверены?",'id'=>md5($value['id']).time()))
?>
).
Of course, you should call renderPartial with property 'processOutput'=true. After that, everything works well, because every new element has got only one binded js-action.
text below copied from here http://www.yiiframework.com/forum/index.php/topic/14562-ajaxsubmitbutton-submit-multiple-times-problem/
common issue...
yii ajax stuff not working properly if you have more than one, and if
you not set unique ID
you should make sure that everything have unique ID every time...
and you should know that if you load form via ajax - yii not working
well with it, cause it has some bugs in the javascript code, with die
and live
In my opinion you should use jQuery.on function. This will fire up event on dynamically changed content. For example: you're downloading some list of images, and populate them on site with new control buttons (view, edit, remove). Example structure could looks like that:
<div id="img_35" class='img-layer'>
<img src='path.jpg'>
<button class='view' ... />
<button class='edit' ... />
<button class='delete' ... />
</div>
Then, proper JS could look like this ( only for delete, others are similiar ):
<script type="text/javascript">
$(document).on('click', '.img-layer .delete', function() {
var imgId = String($(this).parent().attr('id)).split('_')[1]; //obtain img ID
$.ajax({
url: 'http://www.some.ajax/url',
type : 'POST',
data: {
id: imgId
}
}).done({
alert('Success!');
}).fail({
alert('fail :(');
});
}
</script>
After that you don't have to bind and unbind each element when it has to be appear on your page. Also, this solutiion os simple and it's code-clean. This is also easy to locate and modify in code.
I hope, this could be usefull for someone.
Regards
. Simon
Is there a JQuery plugin that allows me to 'unhide' a form by after clicking a link? Like I have an invite link that can take me to a one text field form for an email address but I want this form to just drop down (pushing the rest of the content down also) and shows the form to submit the email. If you guys can think of a JQuery plugin that lets me do this, please let me know
Edit:
So I did this
<div class='add-link'>
<div id='invite_link'><a href=''>Invite User</a></div>
<div id='invitation_form'>
<form>
<input type='text'/>
</form>
</div>
</div>
and my jquery looks like
<script type="text/javascript">
$(function()
{
$("table").tablesorter({sortList:[[0,0],[2,1]], widgets: ['zebra']});
$('#invitation_form').hide();
}
);
$('#invite_link').click(function() {
$('#invitation_form').slideDown();
});
Do you guys see any error that causes the form not to slide down. It hides the form when the page loads but when I click the link it is not sliding down.
$('a.mylink').click(function() {
$('#MyForm').slideDown();
});
I don't think you need a jQuery plugin for this. The base jQuery library should be sufficient.
$('#showFormLink').click(function () {
$('#form').slideDown();
});
If you're looking for animation, that's possible as well by passing in a duration argument to slideDown.
Take a look at the jQuery show documentation.