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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
我尝试从数据库中获取性别实体(男性或女性),并希望在下拉框中显示它:
控制器:
List<Gender> list2 = db.Gender.ToList();
ViewBag.GenderList = new SelectList(list2, "GenderID", "GenderType");
视图:
@Html.DropDownListFor(model => model.GenderID, ViewBag.GenderList as SelectList,"--select--",new { @class="form-control"})
我的错在哪里? 我需要帮助
性别阶层
public class GenderViewModel
{
public int GenderID { get; set; }
public string GenderN { get; set; }
}
个人班
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类具有GenderN(我假设是GenderName?),但是绑定到GenderType,因此仅不匹配将导致异常。 ViewBag是动态的,因此您无需在视图中进行强制转换(因为它是在运行时评估的)...
更改GenderClass:
public class GenderViewModel
{
public int GenderID { get; set; }
public string GenderName { get; set; }
}
并更改控制器:
ist<Gender> list2 = db.Gender.ToList();
ViewBag.GenderList = new SelectList(list2, "GenderID", "GenderName");
发布特定错误会有所帮助。
不知道您看到的确切错误,但是我确定您可以像这样使用ViewBag,因为它是一个匿名对象。 将ViewBag分配给一个单独的变量:
@{
var genderlist = (SelectList)ViewBag.GenderList;
}
@Html.DropDownList("genderID", genderlist, "Gender", new { @class = "classnameifneeded" })
就个人而言,我会这样做:
类/ C#/控制器端:
List<Gender> list2 = db.Gender.ToList();
ViewBag.GenderList = list2;
侧面:
@{
var genderlist = ViewBag.GenderList;
}
Html.DropDownList("genderID", new SelectList(genderlist, "GenderID", "GenderType", 0), "Gender", new { @class="classnameifneeded" })
在html上尝试:@ Html.DropDownList(“ GenderList”,null,“ --select--”,htmlAttributes:new {@class =“ form-control”})
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