I have an form with following url:
CreateEntity?officeCodeId=5
When I send form to validate and if validation is fail it returns just CreateEntity url. No officeCodeId=5.
if user click enter on URL or F5 - my site fail - it require missing officecodeId param. I can save it to the session or in the other storage. But I want to have it in the URL
My view:
[HttpGet]
public virtual ActionResult CreateEntity(int? officeCodeId)
{
var model = new CreateViewModel();
FillViewModel(model, officeCodeId);
return View("Create", model);
}
[HttpPost]
protected virtual ActionResult CreateEntity(TEditViewModel model)
{
if (ModelState.IsValid)
{
//Do some model stuff if
}
return View("Create", model);
}
EDIT.
My View:
using (Html.BeginForm("CreateEntity", "Employee", FormMethod.Post, new { enctype = "multipart/form-data" }))
{
@Html.HiddenFor(x => x.OfficeCodeId)
<div>
@Html.LabelFor(model => model.FirstName, CommonRes.FirstNameCol)
@Html.TextBoxFor(model => model.FirstName, Model.FirstName)
@Html.ValidationMessageFor(model => model.FirstName)
</div>
<div>
@Html.LabelFor(model => model.LastName, CommonRes.LastNameCol)
@Html.TextBoxFor(model => model.LastName, Model.LastName)
@Html.ValidationMessageFor(model => model.LastName)
</div>
<div> <div class="input-file-area"></div>
<input id="Agreements" type="file" name="Agreements"/>
</div>
}
Edit 2.
Adding:
@using (Html.BeginForm("CreateEntity", "Employee", FormMethod.Post, new { officeCodeId = Model.OfficeCodeId, enctype = "multipart/form-data" }))
Haven`t help.
It produce the following form:
<form action="/PhoneEmployee/CreateEntity" enctype="multipart/form-data" method="post" officecodeid="5">
Solution Is
<form action="@Url.Action("CreateEntity", "Employee")[email protected]" enctype="multipart/form-data" method="post">
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