I have an MVC 3 view for rendering reports from SQL Server Reporting Services. There's a form at the top where I capture parameters for the report and on submission, my controller action is dutifully called and my report is rendered into a < div >.
I'm now adding an Export to Excel function. I want the same parameters from the form, but this time, I want a full Postback, not an Ajax call to the controller, so that the user is offered the opportunity to download the report. Otherwise, my report gets rendered as binary content on the existing view.
I'm thinking that I want to be able to switch the behaviour of my form between Ajax and normal Postback, depending on which 'submit' button I click.
Any ideas?
#using (Html.BeginForm("Export", "Report"))
{
... some form fields
#Html.ActionLink("Render report", "Render", "Report", null, new { id = "generateReport" })
<input type="submit" value="Export to Excel">
}
<div id="report"></div>
and then AJAXify the Render report link in a separate js file:
$(function() {
$('#generateReport').click(function() {
var form = $(this).closest('form');
$.post(this.href, form.serialize(), function(result) {
$('#report').html(result);
});
return false;
});
});
and in your ReportController you would have both Export and Render actions.
Related
I have a View with a Form that calls a controller action Post Method to "Complete" a Package. It then needs to refresh the page its on as that contains information that will be updated, both within the view itself and also within a partial. It does use two different Controllers in different MVC Areas.
The Post works correctly and the redirect is issued, but the page is not refreshed.
I have read that instead, I should use OnSuccess within the Ajax call that calls Complete, but I thought that was for in page calls, not ones that navigate to different pages.
View Form
#using (Ajax.BeginForm("Complete", "Packages", new { Area = "Core" },
new AjaxOptions
{
HttpMethod = "POST"
}))
{
Core(Area) Packages Controller
[HttpPost]
public ActionResult Complete(int ID)
{
// Update code
// Refresh the full page
return RedirectToAction("Summary", new { Area = "Control", id = packageBuilder.CurrentPackage.ID });
}
Control (Area) Packages Controller
[HttpGet]
public ActionResult Summary(int id)
{
// Get Model
return View("Summary", model);
}
Any pointers would be warmly welcomed.
Thanks,
Chris.
The reason that your page is not refreshed after you submit the form and the redirect is not issued in the browser, is that you are submitting the request over AJAX. This is a request issued by the browser behind the scenes.
If you want to submit the form and for the page to be refreshed, I'd recommend changing your code from Ajax.BeginForm(... to Html.BeginForm(... and then it will load the page and perform the redirect as expected.
I am not quite sure how your ajax calls are structured, but if you are using the MVC Ajax helper you can just call `location.reload(); in the OnComplete method, like so:
#using (Ajax.BeginForm(new AjaxOptions{OnComplete = "javascriptfunction"}))
{
//Data and submit button
}
<script>
function javascriptfunction(){
location.reload();
}
</script>
enter code here
*MainView.cshtml*: #RadioButtonFor(m=>m.registeras,Individual) #RadioButtonFor(m=>m.registeras,BusinessEntity) <div id="loadpartial"> </div>
I have a register form in which User can register as Individual or Business entity. I kept two radio buttons.
If user selects Individual Radio Button, then that partial view should be loaded. if user selects Business entity radio Button then second partial view should be loaded on the same div.
I am new to asp.net mvc4 razor. Can someone please give the sample code.
You will have to use jQuery in order to make ajax calls.
First reference jQuery at the bottom of your page.
in your controller, define and implement a method which returns a partial view.
[HttpGet]
public ActionResult GetRegisterForm(int registeras)
{
if(registeras == Individual)
{
return PartialView("IndividualPartialViewName");
}
else
{
return PartialView("BusinessPartialViewName");
}
}
Now in your view, you can call the above action using ajax.
<script type="text/javascript">
$(function(){
$("[name=registeras]").on('change', function(){
var $radio = $(this);
$.ajax({
url:'#Url.Action("GetRegisterForm", "ControllerName")',
data: { registeras = $radio.val() }
type: 'GET',
success: function(data){
$("#loadpartial").html(data);
}
});
});
</script>
I have a partial view that the user can preform a search in, and the search results are shown in a select box. In my main view I have a section that is supposed to show the search results after a select button is pressed. Right now when I click the select button is loads the correct information into the correct model for my main view, but the main view doesn't change. When I click refresh, the page updates correctly. How do I make the page update automatically when a button is clicked in the plugin view?
My section in the main view (Index.vbhtml) in my main app:
#Section CUInfo
Credit Union Name: #Model.CUInfo.CUName
end section
Here is my controller method in my Plugin:
Function ChangeCUInfo(strCUName As String) As ActionResult
m_hostApp.CUInfo.CUName = strCUName
m_hostApp.blnPluginRefreshButtonPressed = True
Return View("Index", m_hostApp)
End Function
I've tried to set a boolean value in the hostApp object and then in my main razor view call this function if it is true:
#code
If Model.blnPluginRefreshButtonPressed = True Then
#<script type="text/javascript">
$(function () {
window.location.reload();
});
</script>
End If
Model.blnPluginRefreshButtonPressed = False
End Code
EDIT:
JS function called when the select button is clicked:
function loadCU(CUInfo) {
strCU = CUInfo.split('|');
strCUName = strCU[0];
$.ajax({
type: "POST",
url: "/CUContractNumberPlugin/ChangeCUInfo",
data: { "strCUName": strCUName }
});
}
Form that is used in the plugin view:
#Using (Html.BeginForm("ChangeCUInfo", "CUContractNumberPlugin"))
#<div id="LogoSigSearch" style="height:300px;width:500px;position:relative;">
<span style="display:inline-block;height:20px;width:166px;position:absolute;top:35px;left:5px;">Credit Union Name</span>
<br />
#Html.TextBox("strCUName")
<input type="submit" name="LogoSigSearch$ctl02" value="Search" id="LogoSigSearch_ctl02" tabindex="3" style="width:60px;position:absolute;top:5px;left:352px;" />
<input name="LogoSigSearch$ctl05" type="button" onclick="javascript:clearSearch()" value="Clear" style="position:absolute;top:35px;left:352px;width:60px;" />
<select size="4" name="LogoSigSearch$ctl06" id="LogoSigSearch_ctl06" tabindex="5" style="height:230px;width:342px;position:absolute;top:65px;left:5px;"></select>
<input type="button" name="SelectCU" value="Select" onclick="javascript:loadCU(LogoSigSearch_ctl06.options[LogoSigSearch_ctl06.selectedIndex].value)" tabindex="4" style="width:60px;position:absolute;top:65px;left:352px;" />
</div>
End Using
Are both buttons part of a form? A button won't invoke an action without you attaching it to script or making it part of a form with an associated action.
Use a partial view to render the results of the query, even on the main page load. This simplifies your development.
Add a jQuery event handler (jQuery.on()) to watch for the button click on your main page, or if the button is returned in the partial view, just use an on ready handler in your partial and attach a button.click() event, again using jQuery.
The jQuery event handler can take care of submitting the values of the query, posting to your controller, and displaying the results. I have a number of older articles here but they are still relevant to your question and demonstrate submitting data and fetching partials.
Your client-side code will end up looking something like this:
$("#your-button").click(function () {
var fetchUrl = '#Url.Action("ActionName", "Controller")';
$.post(fetchUrl, { searchParams: $("#your-search-box").val() })
.success(function (data) {
// replace the contents of the DIV with the results. 'data'
// here has whatever you sent back from your partial view
})
.error(function (data) {
// handle the error, use a DIV with some kind of alert message etc
});
});
Hope this helps some.
I have a form which loads within a fancybox so that if the user clicks on a link, a form loads up in fancybox, its an #Ajax.BeginForm(). Like so:
#using (Ajax.BeginForm("AddToBasket", new { controller = "Orders" }, new AjaxOptions()
{
InsertionMode = InsertionMode.Replace,
UpdateTargetId = "SuccessBasket",
OnSuccess = "goToCheckout"
}, new { #class = "product-order-form" }))
{
#* Form elements for Model *#
<div id="SuccessBasket"></div>
}
This gets loaded up in a fancybox window. When I submit the form, my SuccessBasket div does not get updated with the new content passed from the controller, is there some way to enable Ajax calls to be updated within Fancybox? This works fine if I don't use fancybox but I wish to use it.
EDIT:
The form is loaded like so:
#Ajax.ActionLink("Order", "OrderProduct", new { controller = "Orders", id = ViewData["ID"], search = ViewData["search"] }, new AjaxOptions()
{
UpdateTargetId = "ModalWindow",
InsertionMode = InsertionMode.Replace,
OnFailure = "NotAuthorised"
}, new { #class = "lightbox order-link" })
I have an empty div called ModalWindow which is wrapped by a div set to display: none as per the instructions:
<div style="display: none">
<div id="ModalWindow">
</div>
</div>
OrderProduct Action in my controller returns a PartialView:
return PartialView("_OrderProduct", basket);
Where basket is my model BasketModel basket = new BasketModel();
My _OrderProduct PartialView is the view which contains the Ajax form at the start of my post. Which has the SuccessBasket div.
Upto this point, it works perfectly. The form loads up in fancybox.
The AddToBasket Action returns a PartialView:
return PartialView("_BasketSuccess");
This view simply tells the user that their item was added to the basket:
<p>This item has been added to your basket. Search again or goto #Html.ActionLink("checkout", "Order", new { controller = "Orders" }, new { #class = "checkout-link" }) to continue.</p>
The problem is, the SuccessBasket div does not update with the text above but through debugging, I can see that it does load the view _BasketSuccess, it just doesn't update in the modal window.
The way you are loading the fancybox is strange. You are sending 2 AJAX requests: one for the Ajax.ActionLink and one by the fancybox. All that is not necessary. Also you don't need a hidden div in your main view, the fancybox does all this automatically.
So to recap, in your main view all you need is a simple HTML link to the controller actin which will return a partial containing the form:
#Html.ActionLink("Order", "OrderProduct", "Orders", null, new { #class = "lightbox" })
and in a separate javascript file you will attach the fancybox to this anchor so that when it is clicked it will automatically send an AJAX request (the fancybox, not you), fetch the partial form and show it:
$(function () {
$('.lightbox').fancybox();
});
Alright, now you have a partial form shown in a fancybox. This partial form is actually an Ajax.BeginForm. So when you submit this form it will send an AJAX request to the AddToBasket action and upon success it will update the <div id="SuccessBasket"></div> which is inside this form with the result returned by this action.
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>