I want to create a page that has a next button and previous button that switches the image displayed.
For that purpose I created an Ajax.BeginForm and inserted into it, an image and two submit buttons.
Can I (should I) have multiple submit buttons inside an Ajax.BeginForm?
How would the controller handle each submit separately?
Try this,
View
@model TwoModelInSinglePageModel.RegisterModel
@using (Ajax.BeginForm("DYmanicControllerPage", "Test", FormMethod.Post,null, new { id = "frmSignUp" }))
{
<div>
<input type="hidden" id="" name="hidden2" id="hdasd" />
@Html.HiddenFor(m => m.hidden1)
@Html.LabelFor(m => m.Name)
@Html.TextBoxFor(m => m.Name)
@Html.ValidationMessageFor(m => m.Name)
</div>
<br />
<div>
@Html.LabelFor(m => m.Address)
@Html.TextBoxFor(m => m.Address)
@Html.ValidationMessageFor(m => m.Address)
</div>
<br />
<div>
@Html.LabelFor(m => m.PhoneNo)
@Html.TextBoxFor(m => m.PhoneNo)
@Html.ValidationMessageFor(m => m.PhoneNo)
</div>
<input type="submit" value="Save" id="btnSave" name="ButtonType"/>
<input type="submit" value="Next" id="btnNext" name="ButtonType" />
}
Controller
[HttpPost]
public ActionResult DYmanicControllerPage(RegisterModel model, string ButtonType)
{
if(ButtonType == "Next")
{
// Do Next Here
}
if (ButtonType == "Save")
{
//Do save here
}
return JavaScript("REturn anything()");
}
egisterModel model using this in your view page @model TwoModelInSinglePageModel.RegisterModel. i just update my post.
I would recommend that you have two buttons and then depending on what button was clicked you could set the action on the form:
Razor
$(function (){
$("#btn-prev").click(function() {
$("#form").attr
(
"action",
"@Url.Action("Action", "Controller", new {area="Area" })",
).submit();
});
$("#btn-next").click(function() {
$("#form").attr
(
"action",
"@Url.Action("Action", "Controller", new {area="Area" })",
).submit();
});
});
I am using jQuery here to do this, but I think you can get the idea.
Ajax.BeginForm. It will output the form tag on the client which you can change via JavaScript.Ajax.BeginForm? Do I need a special signature for the action in the control? I called @using (Ajax.BeginForm(new AjaxOptions { UpdateTargetId = "result" })).I had the same requirement/issue and tried both solutions here and they both work for me. I LIKE the idea of setting the action via jquery when clicking so I can keep my actions separate so they can be used by other views.
HOWEVER, I've found that when I do this while I debug, it posts TWICE and BOTH the OnSuccess and OnFailure are triggered. It only happens when debugging though. Keep this in mind when picking.
