Question

这是来自我的控制器的动作

[HttpPost]
[AjaxAction]
public ActionResult Registration(RegisterUserModel registerUser)
{
    var data;

    if (ModelState.IsValid)
    {
        if (!IsUserExist(registerUser.Email))
        {    

            var crypto = new SimpleCrypto.PBKDF2();

            var encrpPass = crypto.Compute(registerUser.Password);

            var newUser = _db.Users.Create();

            newUser.Name = registerUser.Name;
            newUser.Email = registerUser.Email;
            newUser.Type = UserType.User.ToString();

            newUser.Password = encrpPass;
            newUser.PasswordSalt = crypto.Salt;

            _db.Users.Add(newUser);
            _db.SaveChanges();

            data = new { status = "OK", message = "Success" };

        }
        else
        {

            data = new { status = "ERROR", message = "User already exists" };
        }
    }
    else
    {

        data = new { status = "ERROR", message = "Data is incorrect" };
    }
    return Json(data, JsonRequestBehavior.AllowGet);
}

但我不知道如何以正确的方式初始化data变量，因为我需要在不同的情况下设置不同的值。这样做的正确方法是什么？

Answer 1

我通常使用多个return语句来避免声明像

这样的对象

if(something){
return Json(new{status = "status 1", message = "message1"})
}
else{
return Json(new{status = "status 2", message = "message2"})
}

Answer 2

以下是其中一个选项

[HttpPost]
[AjaxAction]
public ActionResult Registration(RegisterUserModel registerUser)
{
    JsonResult data;

    if (ModelState.IsValid)
    {
        if (!IsUserExist(registerUser.Email))
        {    

            var crypto = new SimpleCrypto.PBKDF2();

            var encrpPass = crypto.Compute(registerUser.Password);

            var newUser = _db.Users.Create();

            newUser.Name = registerUser.Name;
            newUser.Email = registerUser.Email;
            newUser.Type = UserType.User.ToString();

            newUser.Password = encrpPass;
            newUser.PasswordSalt = crypto.Salt;

            _db.Users.Add(newUser);
            _db.SaveChanges();

            data = Json(new { status = "OK", message = "Success" }, JsonRequestBehavior.AllowGet);

        }
        else
        {

            data = Json(new { status = "ERROR", message = "User already exists"}, JsonRequestBehavior.AllowGet);
        }
    }
    else
    {

        data = Json(new { status = "ERROR", message = "Data is incorrect" }, JsonRequestBehavior.AllowGet);
    }
    return data;
}

Answer 3

你可以试试这个：

var data = new object();

Answer 4

您可以使用dynamic关键字。

dynamic data;

声明变量以返回Json结果（ASP.NET MVC4）

4 个答案: