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[英]How to insert dan update value from HTML <select> multiple Attributes in ASP.NET MVC to database?
[英]select multiple entities from database in asp.net mvc
我嘗試從數據庫中獲取性別實體(男性或女性),並希望在下拉框中顯示它:
控制器:
List<Gender> list2 = db.Gender.ToList();
ViewBag.GenderList = new SelectList(list2, "GenderID", "GenderType");
視圖:
@Html.DropDownListFor(model => model.GenderID, ViewBag.GenderList as SelectList,"--select--",new { @class="form-control"})
我的錯在哪里? 我需要幫助
性別階層
public class GenderViewModel
{
public int GenderID { get; set; }
public string GenderN { get; set; }
}
個人班
public class PersonalViewModel
{
public int PersonalID { get; set; }
public string PName { get; set; }
public string PLastname { get; set; }
public System.DateTime DOB { get; set; }
public string Email { get; set; }
public string Phone { get; set; }
public string PPassword { get; set; }
public int GenderID { get; set; }
public int PAuthorisation { get; set; }
public bool PStatus { get; set; }
public int ShopId { get; set; }
public System.DateTime JobStartDate { get; set; }
public string GenderType { get; set; }
public string ShopName { get; set; }
}
您的Gender類具有GenderN(我假設是GenderName?),但是綁定到GenderType,因此僅不匹配將導致異常。 ViewBag是動態的,因此您無需在視圖中進行強制轉換(因為它是在運行時評估的)...
更改GenderClass:
public class GenderViewModel
{
public int GenderID { get; set; }
public string GenderName { get; set; }
}
並更改控制器:
ist<Gender> list2 = db.Gender.ToList();
ViewBag.GenderList = new SelectList(list2, "GenderID", "GenderName");
發布特定錯誤會有所幫助。
不知道您看到的確切錯誤,但是我確定您可以像這樣使用ViewBag,因為它是一個匿名對象。 將ViewBag分配給一個單獨的變量:
@{
var genderlist = (SelectList)ViewBag.GenderList;
}
@Html.DropDownList("genderID", genderlist, "Gender", new { @class = "classnameifneeded" })
就個人而言,我會這樣做:
類/ C#/控制器端:
List<Gender> list2 = db.Gender.ToList();
ViewBag.GenderList = list2;
側面:
@{
var genderlist = ViewBag.GenderList;
}
Html.DropDownList("genderID", new SelectList(genderlist, "GenderID", "GenderType", 0), "Gender", new { @class="classnameifneeded" })
在html上嘗試:@ Html.DropDownList(“ GenderList”,null,“ --select--”,htmlAttributes:new {@class =“ form-control”})
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