In my view, I have a form I want to dynamically render. This form is wrapped around a larger form:
<form>
//......stuff...
@using (Ajax.BeginForm("FindWorkOrder", new AjaxOptions {
UpdateTargetId = "workOrders" }))
{
<input type="text" name="workOrder" />
<input type="submit" value="Find" />
}
<div id="workOrders">
@{ Html.RenderPartial("DisplayWorkOrder"); }
</div>
</form>
In my controller:
public ActionResult FindWorkOrder()
{
// do query, return a model
return View();
}
I have a partial view named DisplayWorkOrder.cshtml
.
Several questions:
FindWorkOrder
controller? My intended functionality is for the ajax form to submit (without the whole form submitting) and populate <div id="workOrders">
with data that I queried.
Thanks.
To render the partial view , you can do this
@Html.RenderPartial("DisplayWorkOrder")
If you want to pass a model to the Partial view,you can do this
@Html.RenderPartial("DisplayWorkOrder",Model)
If you want to pass a proprty of the model to the Partial view,you can do this
@Html.RenderPartial("DisplayWorkOrder",Model.MyProperty)
Assiming that you have a model binded to the initial (parent) view from where you want to call the partial view. You should be returning the Model /View Model in your Action called "FindWorkOrder" .Something like this
public ActionResult FindWorkOrder()
{
CustomerViewModel objCustVM=CustomerService.GetCustomerViewModel(); // just to get the customer model.
return View(objCustVM);
}
and in your Main View
@model MyProject.ViewModel.CustomerViewModel
<h2>This will show the content from Partial View</h2>
@{ @Html.RenderPartial("DisplayWorkOrder",Model)}
To Avoid submitting the enitire form, you could do a jquery ajax call from your script with the data you want to send. I would keep the (only one) form tag at the outer level and change the submit button to normal button control.
$("#submitWorkOrder").click(function(){
//Do validation
var id=233; //get customer id from wherever you have it
ajaxUrl="Customer/SaveWorkOrder/"+id+"?workOrderId=$("#workOrder").val();
$.ajax({
url: ajaxUrl,
success: function(data){
//do whatever with the result data.
}
});
});
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