Is it possible to allow multiple clicks on a sigle submit button of a form? I have a form where I want to perform an action on the first submission (first click) and a different action on the second submission (second click).
I am basically using ajax to populate a div in the form during the first submission and I want to submit the form on the second click.
I have tried to put by button in the div to by updated, and after the first click, I update update the div and re-creating the button in the updated div. But if I use this method, how can I set the action method of the newly created button in my controller method for Ajax?
My controller method returns something like
return Content( mystring + <input type='button' value='continue submission'/>
if i use this approach, how do I set the action method of the buttton, or is there another way of doing this?
Use two buttons with JavaScript:
Button 1 is shown initially. On click, it hides itself, shows button 2, and performs your action 1.
Button 2 is hidden initially. It is unhidden by button 1 and on click, it performs your second action.
This looks a little weird but I can tell you how to do this. Take an input type="submit" and make it hidden. Have a variable as var flag = false; When user first clicks you input type="button" call a function and do your stuff and make sure to make the flag=true; In the function itself check if flag=true; the trigger the event of your input type="submit".
Like as follows:
<input type="button" id="btn1" onclick="perfromAction()" value="submit"/>
<input type="submit" id="btn2" value="submit" style="display:none"/>
<script type="text/javascript">
var flag=false;
function performAction()
{
if(flag){
$("#btn2").trigger("click");
}
else{
//do processing
flag=true;
}
Related
I want to submit two form into two url and want to submit from two another functions of a Controller.
<form method="post" url="{ '/url1' }">
</form>
<form method="post" url="{ '/url2' }">
</form>
<button type="submit">Submit</button>
Is it possible to do without AJAX??
Simple answer : no, without ajax you can't send two requests.
Complicated answer: unless the first one carries the data for both then in the first response, it sends the second request with the data for it. Which is so complicating things, you should just do it in one request.
You can use jQuery to submit both forms like in the example below:
First of all, add an id to your button like this
<button type="submit" id="submitBtn">Submit</button>
Add ID to your forms:
form method="post" id="form1" url="{ '/url1' }">
form method="post" id="form2" url="{ '/url2' }">
I don't know what is wrong with the stakoverflow editor, that is why i deleted the "<" sign.
3. Now handle the jQuery click event:
$(document).ready(function(){
$('#submitBtn').on('click',function(){
$('#form1').submit();
$('#form2').submit();
});
});
You could give the forms different ids but same action.
Then in the controller called by the submit button, let IF statements check for the form ids and return the respective view desired eg
$requests = $request->all();
$form_Id = $requests['form1_Id'];
if($form_Id != 'id_of_first_form') {
return view('url2');
}else{
return view('url1');
}
If you have a main page form and a modal form and you wanna submit both in order, you can do it with javascript like this
// cause double submits at once
function doubleSubmit()
{
$.post($('#recordPaymentForm').attr("action"), $('#recordPaymentForm').serialize(), function(response) {
$('#submitEditOrderButton').trigger('click');
});
return false;
}
Leave your main page form as it is and change your modal form on submit attribute like this
<form id="recordPaymentForm" method="POST" action="/orders/{{$order->id}}/payments" onsubmit="return doubleSubmit(event)">
Please not that we first submit modal form and then we trigger on click event of submit button (submitEditOrderButton) on the main page. In this example, we first submit payment form and then we trigger click event of main page form which cause a form submit on the main page.
I have an strongly typed view for my model and what I'd like is that when the user clicks on submit, a confirmation box pop up confirming that the user does indeed wish to submit the form, if they click cancel then it shouldn't fire the HttpPost Action for that View, is this possible?
Of course it is possible. I like to use an unobtrusive approach. Here is a simplified example:
jQuery(document).ready(function () {
jQuery('[data-confirm]').click(function (e) {
if (!confirm(jQuery(this).attr("data-confirm")))
{
e.preventDefault();
}
});
});
Then you only need to add a data-confirm attribute to your submit button for example
<input type="submit" data-confirm="are u sure?" />
Of course you can use this attribute on links, buttons, etc. you are not restricted to submit buttons only, and if you want to implement a fancier confirm dialog later than you will have to replace the code only in one place.
function doSubmit()
{
if(window.confirm("ARE YOU SURE TO PERFORM THIS ACTION"))
{
return true;
}
else return false;
}
call doSubmit() function on onsubmit event of the form,
Eg- onsubmit="return doSubmit()
you can add a simply jQuery call for that.
at the end of your view add:
<script type="text/javascript">
$("form").submit(function() {
return confirm('Are you sure?');
});
</script>
or, add a
onsubmit="return confirm('Are you sure?');"
as a new element property
I believe this can be done by overriding the submit button using jquery. Jquery .submit()
This way, when the person hits submit you can show a message and either submit it or cancel it.
In my view I have this button:
<input type="submit" value="Delete" disabled="disabled" id="btnDelete" name="btnDelete" />
and I have Jquery method which makes it enabled.
This is in the view.
How to check if the button is enabled or disabled in the controller?
I have this method:
public void MyMethod(FormCollection form)
{...}
but form doesn't contain anything about the button.
also, form["btnDelete"] is null.
Yes because input button can not hold any value and returning it is useless.
If you want to have that attribute value in your controller make a hidden input which hold the button value on page load and will be changed on btnDelete click event.
I am having multiple dynamic buttons on my asp.net mvc 3 page. what is the best way to handle button click in asp.net mvc 3? there is no event handling in asp.net, so what is the best practice to hadle.?
You could handle the buttons clicks using javascript by subscribing to their click event. For example with jQuery you could give those buttons a class and then:
$(function() {
$('.someClass').click(function() {
// a button was clicked, this will point to the actual button
});
});
or if those are submit buttons of a form you could give them the same name and different values and then on the server test the value of the name parameter. It's value will equal to the button that was clicked.
Let's suppose for example that you have the following form with multiple submit buttons:
#using (Html.BeginForm())
{
... some input fields
<button type="submit" name="Button" value="delete">Delete data</button>
<button type="submit" name="Button" value="save">Save data</button>
}
Now inside the controller action you are posting to you could determine which button was clicked:
[HttpPost]
public ActionResult Index(MyViewModel model)
{
var button = Request["button"];
if (button == "save")
{
// the save button was clicked
}
else if (button == "delete")
{
// the delete button was clicked
}
...
}
If the buttons do not require the same form data, then you can create two forms with different action methods. This is the easiest solution.
If you need to use the same form data, then there are a number of methods, inclduing Darin and tvanfosson's approaches. There is also an approach based on attributes that will select the correct action method based on which button is clicked.
http://www.dotnetcurry.com/ShowArticle.aspx?ID=724
Depends on what the buttons are doing. If they are logically separate actions, then you could have each postback to a separate action on the server side. This often also works they are variants of the same action, Save vs. Cancel, for instance where Save posts back the form and Cancel redirects to you the previous url (say, going back to details from edit). If the buttons represent different data that would get posted back to the same action, you can give them different values. If the buttons are named, the values will get posted back along with the rest of the form, assuming they are included in the form. If posting back from AJAX, you might need to explicitly serialize the button value along with the form.
Example of Save/Cancel
#using (Html.BeginForm())
{
//...
<button type="submit" class="submit-button button">Save</button>
#Html.ActionLink( "Cancel", "details", new { ID = Model.ID }, new { #class = "cancel-button button" } )
}
Then use CSS, perhaps in conjunction with jQuery UI to style the buttons.
<script type="text/javascript">
$(function() {
$('.button').button();
...
});
</script>
I have an controller which has check like that
if (form["submit"].ToString() == "Continue")
{
}
and i have button which is doing submit
<button name="submit" value="Continue">Continue</button>
It was all working well until i decided to disable Continue button on submit to prevent double click using this function:
$('form').submit(function () {
if ($(this).valid()) {
$(':submit', this).attr('disabled', 'disabled');
}
});
So now i don't get value form["submit"] posted on controller.
Any thoughts how may i fix that?
I want still prevent second click but be able to get form["submit"] value posted on controller.
Can you control the submit value in a hidden field in the form? I can't tell what other logic you might need, but when the form renders, you could set the hidden field's value to the submit button's value and change it when necessary using the first script below. As long as it has a name attribute and is enabled (which you'd rarely disable a hidden field) then it will post when the form is submitted.
$(function() {
// this assumes your button has id="myButton" attribute
$(':hidden[name="submit"]').val($('#myButton').val());
});
And of course in your form, you would need a hidden field with name="submit"
<input type="hidden" name="submit" value="Continue" />
Then, whenever the state of your form changes, modify the disabled state of the button and the value of the hidden field to reflect the value (if it changed at all).
There are also frameworks you may find useful for UI features like this. KnockoutJS comes to mind. It can be used to "value" bind input elements. It's probably overkill for this small example, but it could be useful if your UI expands. I've added markup, script and comments below if you're interested.
$(function () {
var viewModel = {
submitValue: ko.observable("Continue")
};
ko.applyBindings(viewModel);
$('form').submit(function() {
if($(this).valid()) {
// the following line will change the both the hidden field's value
// as well as the button's value attribute
viewModel.submitValue("some other value");
// I couldn't follow your selector here, but please note I changed
// the name of the submit button in the markup below.
$(':submit, this).attr('disabled', 'disabled');
}
});
});
KnockoutJS requires you use the data-bind attribute to setup your elements. In your case, you'd bind one property to multiple elements like this:
<button name="submitButton" data-bind="value: submitValue"/>Continue</button>
<!-- and bind the same value similarly in the hidden field-->
<input type="hidden" name="submit" data-bind="value: submitValue"/>