Develop Reference

How to prevent repeatative submit while using ajax or javascript?

I have a controller in c# and inside the controller there is a save method. The save method saves/updates data that is submitted by submit button click and javascript. The problem is, if you click on the button multiple time, it should only process the very first click and rest of them should be identified as duplicate submit and should be discarded by controller. How to do this in c# mvc web application?

Disable the button after it's clicked. So it can just be clicked once.

Simple way
when button clicked disabled it then actived again after you got response result from ajax! u can also add loader that make ur web look so cool!
<button id="btnSend" onClick="send()">submit</button>
<script>
btnSend=document.getElementById("btnSend");
function send(){
btnSend.disabled=true;
//set disabled button here
$.ajax({ type: "GET",
url: "http://www.google.de",
async: false,
success : function(text)
{
btnSend.disabled=false;
//set active to button
// add your code here
},
fail : function(text)
{
btnSend.disabled=false;
//set active to button
// add your code here
}
});
}
</script>

I would also disable the button on the client side. But you could also check if the submitted data is different from the stored data. If no changes were made you could just return without further saving logic.
Should it be possible to just save the data once? Maybe a redirect to a different view after saving could be a possible solution in special cases.

When I disable submit button (after form submit) the page does not load

I have an HTML form which submits to another page via POST. Nothing special about it, except that after the form validates I try to hide and/or disable the submit button so that it cannot be double-submit, while also telling the user the next page might take a while to load.
The relevant code is:
jQuery(document).ready(function () {
jQuery("form#form").submit(function() {
var result = validate();
jQuery(this).find('input[type=submit]').prop('disabled', true);
jQuery("#submit-button-wrapper").html(jQuery("#submit-button-wrapper").html()+
"<br/><br/><span style='margin: 25px; padding: 5px; background: yellow; "+
"width: 100%; font-weight: bold;'>Loading... this may take a few minutes! "+
"<i class='fa fa-spinner fa-spin' style='color: blue;'></i></span>");
return result;
});
});
function validate() {
return true; // Does stuff, then returns a simple true or false
}
By request, here is the (very simple) button wrapper HTML:
<div class="col-sm-12" id="submit-button-wrapper">
<input type="submit" value="One More Step" />
</div>
When the I remove the which changes the button wrapper's HTML, the form submits just as you'd expect. When I have that line in, however, it still calls the next page and executes that code, without the displayed page ever changing.
I have tested in both Chrome and Firefox, so I know it's not a browser issue, but this is really weird behavior. What am I doing wrong?
My goal: (1) validate the user's input, (2) give the user a clue that the page is going to take a while to load and (3) display the output from the action="complete.php" page once the PHP on it has run.

Maybe you can achieve this with $ajax and show results on the same page.
Send POST data to /some.php
After sending data, give feedback to user changing button behavior
When the task is complete, receive data and verify success or error and act accordingly. If OK, change button text to "complete!" or something else, and append response data to some div. If NOT OK, give feedback as well.
In code:
$("form#form").submit(function(e) {
e.preventDefault();
$.ajax({
method: "POST",
url: "some.php",
data: $(this).serialize(),
dataType : 'json',
timeout: 2000,
cache: false,
afterSend: function() {
/*change button behavior here*/
},
success: function(result) {
if (result === "ok"){
/*maybe append data to div and update button text to complete*/
} else {
/*if result not ok, send feedback*/
}
}
});
});
BTW: ajax documentation http://api.jquery.com/jquery.ajax/

When you return true (if validated) then the form is submitted. However, you're changing the DOM / submit button wrapper and essentially removing the submit button, right? Likely that is what is causing your problem. Leave the submit button wrapper alone. If you want to display a message display it as an overlay or hide the submit button wrapper and show the message wrapper in its place, don't remove the submit button altogether.

I know that you're just showing a slight jQuery and HTML portion of the script, but that isn't quite enough to figure out your problem. Since not all of it seems to be there because you mention action="complete.php" but I don't see that within your jQuery or HTML sample of code. So I have a few questions of my own.
Is the form small or large. If it's a small form then why aren't you displaying the output on the submit page? You could do that with what you currently have but a single PHP or ASP page could save you on amount of pages to make and what not. As a side note, depending on size of the form, you can do the validation on same page or continue to use action="" for it if large.
Do you have need for a database file or are you saving to one? If you do/are, you could write to the DB file, have the submit open the next page and view what was saved in the database on that new page. Again, you can probably use a single PHP or ASP page.
This last part sounds more valid for your purpose. You could use location.href="http://www.domain.com/home.html";
or use window.location("http://www.domain.com/home.html"); to redirect to the new page.
On another matter about some of the comments others made.
You don't exactly need the + unless you're dividing each of those out into their own separate lines when you could just use one line to do that. That's probably what confused Rajesh about the '. In fact I'm not sure why you yourself mentioned the + when referring to Rajesh comment about "concat string" and "append" ,because those two have nothing to do with the +. In fact to take a guess, he might have been referring to your jQuery("#submit-button-wrapper").html(jQuery("#submit-button-wrapper").html() which kinda looks like a concat string.
Speaking about append, not really needed unless for example you're doing something like giving the user the option to add rows to a form question.

How to stay on same position after submitting an html form?

I have a simple html form:
<form action="addToCart" method="post"><br><br>
<input name="productId"
value="${product.id}" type="hidden"> <input class="submit" onclick="addedCart()" value="<fmt:message key='AddToCart'/>"
type="submit">
</form>
Every time I click on "submit", it brings me straight back to top which creates a poor user experience because the user would have to scroll back down where he was browsing the products... Can I do this without using a script?

-Every time I click on "submit", it brings me straight back to top.
Yes that is what default functionality when submitting forms, it always repaints the dom so it causes a jump and page's top position is rendered.
-which creates a poor user experience because the user would have to scroll back down where he was browsing the products
To make a good user experience you can use ajax for this functionality, as you tagged jQuery in your question then you can try with jquery ajax:
$('form').submit(function(e){
e.preventDefault(); // stops the form submission
$.ajax({
url:$(this).attr('action'), // action attribute of form to send the values
type:$(this).attr('method'), // method used in the form
data:$(this).serialize(), // data to be sent to php
dataType:"text",
success:function(data){
alert(data); // you can alert the success message.
},
error:function(err){
console.log(err);
}
});
});
I have used a dataType:"text", just assuming if you are going to echo "Added in the cart."; kind of message from the php.

Ajax is your best bet if you want to achieve what you want
$('.submit').on('click',function(event){
event.preventDefault(); //this is important else page will get submitted
$.ajax({
url:'where you want to process data',
dataType:'html',
data: your form data as json or whatever type
success: function(result){
//here you can update any thing on the frontside
}
});
});

Since you're using onclick attribute on your button, you have to change type attribute from submit to button. That way you can get your addedCart() method to fire. Then handle form submission inside that method (if your submit handler isn't already there).
<input class="submit" onclick="addedCart()" value="<fmt:message key='AddToCart'/>" type="button">
Script:
function addedCart(){
// ... your method on click
// $.ajax({...});
};
If you're not using jQuery, you can handle form submission with XMLHttpRequest :
How to make an AJAX call without jQuery?

On Click: Open a Pop-up Div on a Different Page

On page1.php I have a click event that causes the user to be redirected to page2.php. It goes something like this:
$("#someButton").click(function() {
window.location = "page2.php";
});
And that works great. But what I really want is to open a hidden, UI-blocking <div> on page2. The user can already open this <div> manually by clicking another button on page2, that goes something like this:
$('#someOtherButton').click(function() {
$("#pageContainer").block({message: $("#theDivIWant2See")});
});
Can I make a click event from the JavaScript on one page call the JavaScript on another? Or will I need to add in some HTML-parsing to pass information between pages? (I'm not looking for a JavaScript hand-out here, just a strategy to help me move forward.)

When you redirect from the first page, add a querystring value in your url. and in the second page, using your server side page language, set in in a hidden field and in the document ready event check the value of that hidden field. If the value is expected, call a javascript function to show the popup.
Some thing like this
$("#someButton").click(function() {
window.location = "page2.php?showpopup=yes";
});
and in page2.php set it (forgive for errors, i am not a php guy)
<input type='<?php $_GET["showpopup"] ?>' id='hdnShow' />
and in the script
$(function(){
if($("#hdnShow").val()=="yes")
{
//Call here the method to show pop up
}
});

You need to do your stuff when DOM for page2 is ready. You can use jQuery's ready function for that.
$(document).ready(function() {
// put code for showing your div here
});
Hope that helps.

Could you pass a query string argument or assign a cookie that the other page could then check when the document loads? If the value exists then present a modal dialog (e.g. jQuery UI Modal Popup)
http://jqueryui.com/demos/dialog/

how can you have the same form handle by javascript multiple times on the same page?

I have a thumb gallery where I am using ajax/javascript to submit a form per image to report the image as broken seamlessly along with php. The form and script is templated so the script is in the header and then the form is printed multiple times on the same page with a hidden field with a different id for the value per thumb. So basically this is what i have.
javascript in header
just a quick idea of the forms i have. Just a quick idea not what I actually have.
image1 followed by the form
image2 followed by the form
So when you hit the button it basically submits all of the forms at the same time. I am sure it can be fixed with a (this) or something like that so it only submits a single form at a time. Let me know please.
$(function() {
$(".submit").click(function() {
var imgId = $("#imgId").val();
var dataString = 'imgId='+ imgId;
if(imgId==''){
$('.success').fadeOut(200).hide();
$('.error').fadeIn(200).show();
$('.error').fadeOut(200).hide();
}else{
$.ajax({
type: "POST",
url: "inc/brokenImgReport.php",
data: dataString,
success: function(){
});
$('.error').fadeOut(200).hide();
$('.success').fadeIn(200).show();
setTimeout(function() {
$('.success').fadeOut(200); }, 2000);
}
return false;
});
});

in a page with multiple forms. You can refer individual form as an array element forms[0],forms[1] and likewise, on click of submit button. you can do
forms[0].submit
forms[1].submit
and likewise

$(document).ready( function(){
$('form.your_class_if_needed').submit( function(){
do_your_stuff_here()
})
})
Every form with class your_class_if_needed will have the function you define attached to its submit method. Within that function, $(this) will refer to each specific form -- not all of them. You can then easily select, process, and assemble your form fields and values within the context of each form, and submit them with $.ajax() or $.post().