我有一个HomeController用于登陆页面(主页,关于,联系我们),在家庭视图中我注册了两种类型(2个独立的视图模型)和登录(另一个视图模型)。我想知道是否有一种从索引视图(HomeController)将表单发布到另一个控制器的方法。我试图用标签帮助器来做这件事,但似乎它不起作用。这是我的代码,我试图这样做
<form asp-controller="Employee" asp-action="Register" method="post" role="form" class="ui large form">
....
</form>
这是我在员工控制器中的注册操作
[HttpPost]
[AllowAnonymous]
[ValidateAntiForgeryToken]
public async Task<IActionResult> Register(EmployeeRegisterViewModel employeeRegisterModel)
{
if (ModelState.IsValid)
{
IdentityResult result = await _service.CreateEmployeeAccount(employeeRegisterModel);
if (result.Succeeded)
{
return RedirectToAction(nameof(EmployeeController.Index), "Employee");
}
AddErrors(result);
}
return View(employeeRegisterModel);
}
感谢您的帮助
答案 0 :(得分:3)
使用剃刀语法Stream<MyType> stream = // get the stream from somewhere
List<MyType> list = stream.collect(Collectors.toList()); // materialize the stream contents
list.stream().doSomething // create a new stream from the list
list.stream().doSomethingElse // create one more stream from the list
Html.BeginForm
或者在纯HTML中执行此操作
@using (Html.BeginForm("ActionName", "ControllerName", FormMethod.Post, new { @class = "ui large form"}))
{
}
答案 1 :(得分:2)
@using (Html.BeginForm("Register", "YourController", FormMethod.Post, null))
{
<input type="submit" value="Html PsBk Click" />
}