EDIT: The problem is not related to Boxy, I've run into the same issue when I've used JQuery 's load method.
EDIT 2: When I take out link.remove() from inside the ajax callback and place it before ajax load, the problem is no more. Are there restrictions for manipulating elements inside an ajax callback function.
I am using JQuery with Boxy plugin.
When the 'Flag' link on the page is clicked, a Boxy modal pops-up and loads a form via ajax. When the user submits the form, the link (<a> tag) is removed and a new one is created from the ajax response. This mechanism works for, well, 3 times! After the 3rd, the callback function just does not remove/replace/append (tested several variations of manipulation) the element.
The only hint I have is that after the 3rd call, the parent of the link becomes non-selectable. However I can't make anything of this.
Sorry if this is a very trivial issue, I have no experience in client-side programming.
The relevant html is below:
<div class="flag-link">
<img class="flag-img" style="width: 16px; visibility: hidden;" src="/static/images/flag.png" alt=""/>
<a class="unflagged" href="/i/flag/showform/9/1/?next=/users/1/ozgurisil">Flag</a>
</div>
Here is the relevant js code:
$(document).ready(function() {
$('div.flag-link a.unflagged').live('click', function(e){
doFlag(e);
return false;
});
...
});
function doFlag(e) {
var link = $(e.target);
var url = link.attr('href');
Boxy.load(url, {title:'Inappropriate Content', unloadOnHide:true, cache:false, behaviours: function(r) {
$("#flag-form").live("submit", function(){
var post_url = $("#flag-form").attr('action');
boxy = Boxy.get(this);
boxy.hideAndUnload();
$.post(post_url, $("#flag-form").serialize(), function(data){
par = link.parent();
par.append(data);
alert (par.attr('class')); //BECOMES UNDEFINED AT THE 3RD CALL!!
par.children('img.flag-img').css('visibility', 'visible');
link.remove();
});
return false;
});
}});
}
Old and late reply, but.. I found this while googling for my answer, so.. :)
I think this is a problem with the "notmodified" error being thrown, because you return the same Ajax data.
It seems that this is happening even if the "ifModified" option is set to false (which is also the default).
Returning the same Ajax data three times will cause issues for me (jQuery 1.4). Making the data unique (just adding time/random number in the response) removes the problem.
I don't know if this is a browser (Firefox), jQuery or server (Apache) issue though..
I have had the same problem, I could not run javascript after I call boxy. So I put all my javascript code in afterShow:function one of boxy attributes. I can run almost except submit my form. My be my way can give you something.
Related
I am learning ASP.NET MVC. I have to submit a to controller side after validation in client-side(in jquery). How this can be done? Should i use <form action="#" method="post"> instead of <form action="Controller/Method" method="post"> and add an event handler in click event of submit button of , to send via ajax etc? What should i do? pls help
You are on the right track, and what you suggested will work.
A better method would be to leave the original action intact, providing backwards compatibility to older browsers. You would then create the event handler as normal, and include code to prevent the default submit behavior, and use ajax instead.
$('#submitbutton').live('click', function(e){ e.preventDefault(); });
The easiest way to do this is to use the jQuery forms plugin.
This is my go-to plugin for this type of thing. Basically it will take your existing form, action url etc and convert the submission to an ajax call automatically. From the website:
The jQuery Form Plugin allows you to easily and unobtrusively upgrade
HTML forms to use AJAX. The main methods, ajaxForm and ajaxSubmit,
gather information from the form element to determine how to manage
the submit process. Both of these methods support numerous options
which allows you to have full control over how the data is submitted.
It is extremely useful for sites hosted in low cost web hosting
providers with limited features and functionality. Submitting a form
with AJAX doesn't get any easier than this!
It will also degrade gracefully if, for some reason, javascript is disabled. Take a look at the website, there are a bunch of clear examples and demos.
This is how I do:
In jQuery:
$('document').ready(function() {
$('input[name=submit]').click(function(e) {
url = 'the link';
var dataToBeSent = $("form#myForm").serialize();
$.ajax({
url : url,
data : dataToBeSent,
success : function(response) {
alert('Success');
},
error : function(request, textStatus, errorThrown) {
alert('Something bad happened');
}
});
e.preventDefault();
});
In the other page I get the variables and process them. My form is
<form name = "myForm" method = "post">//AJAX does the calling part so action is not needed.
<input type = "text" name = "fname"/>
<input type= "submit" name = "submit"/>
<FORM>
In the action page have something like this
name = Request.QueryString("fname")
UPDATE: As one of your comment in David's post, you are not sure how to send values of the form. Try the below function you will get a clear idea how this code works. serialize() method does the trick.
$('input[name=submit]').click(function(e){
var dataToBeSent = $("form#myForm").serialize();
alert(dataToBeSent);
e.preventDefault();
})
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 () {
$(this).hide();
});
});
As an aside, you no longer need runat=server.
(First of all English is not my native language, I'm sorry if I'll probably be mistaken).
I've created Yii Web app where is input form on the main page which appears after button click through ajax request. There is a "Cancel" button on the form that makes div with form invisible. If I click "Show form" and "Cancel" N times and then submit a form with data the request is repeating N times. Obviously, browser binds onclick event to the submit button every time form appears. Can anybody explain how to prevent it?
Thank you!
I've had the exact same problem and there was a discussion about it in the Yii Forum.
This basically happens because you are probably returning ajax results with "render()" instead or renderPartial(). This adds the javascript code every time to activate all ajax buttons. If they were already active they will now be triggered twice. So the solution is to use renderPartial(). Either use render the first time only and then renderPartial(), or use renderPartial() from the start but make sure the "processOutput" parameter is only set to TRUE the first time.
Solved!
There was two steps:
First one. I decided to add JS code to my CHtml::ajaxSubmitButton instance that unbind 'onclick' event on this very button after click. No success!
Back to work. After two hours of digging I realized than when you click 'Submit' button it raises not only 'click' event. It raises 'submit' event too. So you need to unbind any event from whole form, not only button!
Here is my code:
echo CHtml::submitButton($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
I am using an iFrame with a form that return some content with an AJAX link.
I am then moving the returned content out of the iFrame into the main page.
However, then the ajax link does not work and the error "Element is null" is created once the link is clicked.
How can I move content from the iFrame and still have the AJAX link working?
Here's the code returned by the iFrame:
<span id="top">
<a id="link8" onclick=" event.returnValue = false; return false;" href="/item_pictures/delete/7">
<img src="/img/delete.bmp"/>
</a>
<script type="text/javascript">
parent.Event.observe('link8', 'click', function(event) {
new Ajax.Updater('top','/item_pictures/delete/3', {
asynchronous:true, evalScripts:true,
onCreate:function(request, xhr) {
document.getElementById("top").innerHTML = "<img src=\"/img/spinner_small.gif\">";
},
requestHeaders:['X-Update', 'top']
})
}, false);
</script>
</span>
I see two problems with your code.
First the solution (I think :-))
When your iframe loads, the javascript in it runs. Javascript attaches an observer to parent document's link8.
Inside the observer you define another function (onCreate). This function will run in iframe context, making document object refer to iframe and not to main document. When you remove link8 from iframe to move it to main document, document.getElementById("top") will become null - hence error.
Perhaps change it to:
parent.document.getElementById("top").innerHTML = "<img src=\"/img/spinner_small.gif\">";
Second problem (that is not really a problem in this particular case) is, if you move whole span (including the script) to main document, the javascript will run again in main document's context. In your case, you should see an error or warning after you move the content, stating that parent is null (or similar).
To remove the second problem, return your iframe data in two divs or similar. Then copy only div with html to main document.
What I did was move the AJAX call out to an external js file and called the function once the link was clicked. It works now.
I am using JQuery with Boxy plugin.
When a user clicks on a link on one page, I call Boxy.load to load a form onto the pop-up. The form loads and is displayed inside the pop-up without problems.
However, I can't bind the form to a submit event, since I can't select the form element.
This is the event handler:
$('#flag-link a.unflagged').click (function(e) {
url = $(e.target).attr('href');
Boxy.load(url, {behaviours: function(r) {
alert ($("#flag-form").attr('id'));
}
});
});
The alert reads "undefined" when it is displayed.
And this is the form:
<form id="flag-form" method="POST" action="somepage">
<table>
<tr><td><input type="text" name = "name"></td></tr>
<tr><td><input type="submit" value="OK"></td></tr>
</table>
</form>
What am I doing wrong?
First (a minor point, but a potential source of trouble), it should be id="flag-form" not id = "flag-form" (no spaces).
Second, you shouldn't need r.find(). Just do $("#flag-form").attr("id")
As far as I understand, live() method must be used to bind an element to an event in this case:
$("#flag-form").live("submit", function(){ ... }
Presently, live method is documented to be not supporting the submit event. However, I could work it out with Chrome and FF. On the other hand, I couldn't get it working in IE. A better way for cross-browser compatibility seems to be binding the submit button of the form to the click event.
$("#flag-form-submit").live("click", function(){
I learnt that declaring methods in behaviours: function (e) {} works, in addition to using live() methods.
E.g.:
$('#flag-link a.unflagged').click (function() {
Boxy.load(this.href, {
behaviours: function(r) {
r.find('#flag-form').bind('submit', function() {
// do on submit e.g. ajax calls etc.
});
}
});
return false;
});
Boxy opens the URL (url = $(e.target).attr('href');) in an iframe. So you cannot find the form from the opening page(parent page). Your code to bind the form should be in the child page (ie, the Boxy iframe). You can check the iframe URL using your code, url = $(e.target).attr('href');