I have a Razor/ASP/MVC3 web application with a form and a Submit button, which results in some action on the server and then posts back to the form. There is often some delay, and it's important that users know they should wait for it to complete and confirm before closing the page or doing other things on the site, because it seems users are doing that and sometimes their work has not been processed when they assume it has.
So, I added a "Saving, Please Wait..." spinner in a hidden Div that becomes visible when they press the Submit button, which works very nicely, but I haven't been able to find a way to get the Div re-hidden when the action is complete.
My spinner Div is:
<div id="hahuloading" runat="server">
<div id="hahuloadingcontent">
<p id="hahuloadingspinner">
Saving, Please Wait...<br />
<img src="../../Content/Images/progSpin.gif" />
</p>
</div>
</div>
Its CSS is:
#hahuloading
{
display:none;
background:rgba(255,255,255,0.8);
z-index:1000;
}
I get the "please wait" spinner to appear in a JS method for the visible button, which calls the actual submit button like this:
$(document).ready(function () {
$("#submitVisibleButton").click(function () {
$(this).prop('disabled', true);
$("#myUserMessage").html("Saving...");
$("#myUserMessage").show();
$("#hahuloading").show();
document.getElementById("submitHiddenButton").click();
});
});
And my view model code gets called, does things, and returns a string which sets the usermessage content which shows up fine, but when I tried doing some code in examples I saw such as:
// Re-hide the spinner:
Response.write (hahuloading.Attributes.Add("style", "visibility:hiddden"));
It tells me "hahuloading does not exist in the current context".
Is there some way I am supposed to define a variable in the view model which will correspond to the Div in a way that I can set its visibility back from the server's action handler?
Or, can I make the div display conditional on some value, in a way that will work when the page returns from the action?
Or, in any way, could anyone help me figure out how to get my div re-hidden after the server action completes?
Thanks!
Is this done with ajax? I would assume so because the page is not being redirected. Try this:
$(document).ready(function () {
$("#submitVisibleButton").click(function () {
$(this).prop('disabled', true);
$("#myUserMessage").html("Saving...");
$("#myUserMessage").show();
$("#hahuloading").show();
document.getElementById("submitHiddenButton").click();
});
$("#hahuloading").ajaxStop(function () {
$(this).hide();
});
});
As an aside, you no longer need runat=server.
I think there is a way to do this in ajax. But if you have a better way please let me know.
I have this code:
$interactionBox= '<input type="button" value="Pending Friend Requests('.$num_rows.')" onclick="return false" onmousedown="javascript:toggleInteractContainers(\'friend_requests\');"/>';
This code opens up a togle box where user can accept friends if there is a pending request.
So if button looks like this:
Pending Friend Request(1)
And user accepts that request user has to refresh the page for this to show as:
Pending Friend Request(0)
Is there a way to do this without refreshing the page using ajax or any other way?
Here is HTML for above code:
<div class="interactContainers" id="add_friend">
<div align="right">cancel </div>
Add <?php echo $username; ?> as a friend?
Yes
You can just get the value between brackets with JavaScript and then reduce by 1,
You can get that value (if it isn't a variable get via regexpresion:
var button = document.getElementById('buttonid'); // fill in the id
button.value = "Pending Friend Requests("+ (parseInt(button.value.match(/\d{1,10}/)) - 1) + ')';
just excute that after each accept (or not-accept)
note you should just use the function in onclick, it's not a link so it will only excute the javacript bit: so no onmousedown just onclick="javascript:toggleInteractContainers(\'friend_requests\');"
look ma: no jQuery!
jQuery.validate stops my form from being submitted. I would like it to just show the user what is wrong but allow them to submit anyway.
I am using the jquery.validate.unobtrusive library that comes with ASP MVC.
I use jquery.tmpl to dynamically create the form and then I use jquery.datalink to link the input fields to a json object on the page. So my document ready call looks something like this.
jQuery(function ($) {
// 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.