I'm using a so as to edit an object, with an <input type="submit"> for validation
I've also another input, type="button" with an onclick event for the cancel button (with a redirect). However, this use a JS call, which I'd like to avoid.
I'd rather prefer to process the validation or cancel choice within the controller, so as to be NoScript compliant.
So is it possible to retrieve, within the controller post method, the id of the <input> that was clicked in the <form>?
Thanks
You can set the name of your
<input type="submit" name="submitButton" />
Then in your controller:
if(Request["submitButton"] != null) {
// ...
}
I'm not sure if I understand you, but if your cancel button does a redirect, it should end in a GET, and your submit button does a POST, that's how you should difference both request.
Check the [HttpPost] attribute in controller action methods.
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.
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;
}
i wish to change the inner html of a view on button click but maintain the view. I know how to change the html content of a div in javascript, but how can I have the action of the button not return a different view?
My buton looks like
<input type="submit" value="submit" onchange="myfunc()"/>
where myfunc() is the function in Javascript changing the div content.
Assuming you want a link to render content using ajax (and hopefully using razor) you can do something like the following:
First, setup the action to render the content partially. this can be done a few ways, but I'll keep with the logic in the action (and make it callable directly or by ajax):
[HttpPost]
public ActionResult Save(SomeModel model)
{
/* build view */
return Request.IsAjaxRequest() ? PartialView(model) : Wiew(model);
}
Next, setup a container in your page where the content will be populated along with the form you're looking to submit. If you want the form to disappear on a save, wrap it in the container. Otherwise, keep the container separated. In the below example, the from will submit and on success it'll come back, otherwise the new content will appear in its place:
<div id="ajaxContentPlaceholder">
#using (Ajax.BeginForm("Save", new AjaxOptions { UpdateTargetId = "ajaxContentPlaceholder" })) {
<!-- form elements -->
<input type="submit" value="save" />
}
</div>
I have a scenario in which a button value is not being posted while submitting form in ASP.NET MVC controller.
This is happening only for Google Chrome browser. For all other browsers Firefox, IE, I am getting Submit_0 value in Controller.
//Server side
public ActionResult Answers(string id, SurveyViewModel model, string Submit, string button)
{
string[] buttonParts = button.Split(new char[]{'_'});
..
...
}
//Client Side
#using (Html.BeginForm("Answers", "Survey"))
{
<button value="Submit_0" name="button" onclick="document.forms[0].submit();"><span><span>Submit</span></span></button>
}
Kindly suggest.
I wonder if it is because the way you wired the button. Why don't you use something like this instead?
<button type="submit" name="button" value="submit_0">Submit</button>
Notice that the button has a type of submit (which eliminate the need to call document.forms[0].submit() manually)
I need a little help. I'm trying to make a little project in MVC 3 with Razor. A page with 2 buttons: Button 1 and Button 2. When I click on Button 1 I want to go at Page 1. The same with Button 2 ( to Page 2). It's not difficult, BUT I want the redirection to be made in Controller, not in View (cshtml). I know that I need to use ActionName and RedirectToAction, but I don't know how. Please help me!
What you'll need to do is check which button was pressed in the HttpPost part of the controllers action then redirect accordingly.
As a very basic example you could add two
<input type="submit" name="submit" value="<val>">
controls into your forms view each having the same name and a different value (instead of ) then add a string parameter called submit to the HttpPost action. Assuming the buttons have values "button1" and "button2" Then in your action's code you could use:
if(submit == "button1") {
RedirectToAction("Page1");
} else {
RedirectToAction("Page2");
}
to redirect based on which button was pressed
This is a simplified example, but I think you will get my meaning. You simply need to name your buttons and check the formcollection to see which exists in the collection thus indicating which what clicked. see code below:
#using (Html.BeginForm("Test", "Home", FormMethod.Post))
{
<input type="submit" value="Go 1" name="go-1" />
<input type="submit" value="Go 2" name="go-2" />
}
and now the Action implementation.
[HttpPost]
public ActionResult Test(FormCollection collection)
{
if (collection.AllKeys.Contains("go-1")) return View("Page1");
if (collection.AllKeys.Contains("go-2")) return View("Page2");
return View("Index");
}
and thats it.
In your controller action for page 1, you can use RedirectToAction:
public ActionResult Process()
{
// do processing
// redirect to page 2
return this.RedirectToAction("Index", "Page2");
}
You can invoke the Process action from the Page 1 button using either a GET or POST request, depending on if the Process action is idempotent. E.g your page 1 view:
#Html.BeginForm("Process", "Page1", FormMethod.Post)
{
<input type="submit" name="button" value="Submit" />
}
Alternatively, you could use an ActionLink:
#Html.ActionLink("Redirect to Page 2", "Process", "Page1")