【发布时间】:2018-12-13 06:22:51
【问题描述】:
我有一个“ViewModel”的工作示例,我认为该示例适用于下拉列表。目前我正在使用它来做一个 Foreach 并且它可以正常工作,但我需要在我的表单上的下拉列表中获得相同的结果。我尝试了不同的想法,我认为这些想法可能会奏效。似乎没有人能做到这一点。我没有示例,因为我通常只是阅读一堆帖子并开始向其抛出代码以查看哪些内容。我仍在学习,所以希望有人可以为我指明正确的方向,或者向我展示一些可以填充我需要的代码。我相信我的 ViewModel 中有我需要的一切,我想远离 Viewbag。下面是我的“视图模型”。
除非有更好的方法,否则我想用这条线来填充下拉菜单。
result.Addresses = db.Addresses.Where(a => a.CompanyId == result.CompanyId);
编辑 - 这是此“视图模型”的当前工作下拉列表
@Html.DropDownList("AddressId", new SelectList(Model.Addresses, "AddressId", "LocationName"), "Select Address user belongs to", new { @class = "form-control" })
这是整个“ViewModel”
public class RegisterUserViewModel
{
public static RegisterUserViewModel GetCompanyId(string userId, CustomerEntities db)
{
var qCust = from ad in db.Addresses
join anu in db.Users on ad.AddressId equals anu.AddressId
join cus in db.CompanyNames on ad.CompanyId equals cus.CompanyId
where (anu.Id == userId)
select new RegisterUserViewModel()
{
UserId = userId,
CompanyId = cus.CompanyId,
AddressId = ad.AddressId
};
var result = qCust.SingleOrDefault();
if (result != null)
{
result.Addresses = db.Addresses.Where(a => a.CompanyId == result.CompanyId);
};
return result;
}
[Required]
[Display(Name = "User Name")]
public string UserName { get; set; }
[Required]
[EmailAddress]
[Display(Name = "Email")]
public string Email { get; set; }
[Required]
[StringLength(100, ErrorMessage = "The {0} must be at least {2} characters long.", MinimumLength = 6)]
[DataType(DataType.Password)]
[Display(Name = "Password")]
public string Password { get; set; }
[DataType(DataType.Password)]
[Display(Name = "Confirm password")]
[Compare("Password", ErrorMessage = "The password and confirmation password do not match.")]
public string ConfirmPassword { get; set; }
[Required]
[Display(Name = "User Address")]
public int AddressId { get; set; }
[Required]
[Display(Name = "User Roles")]
public string UserRoles { get; set; }
public string UserId { get; set; }
public int CompanyId { get; set; }
public virtual IEnumerable<Addresses> Addresses { get; set; }
}
我希望将 AddressId 作为值,将 LocationName 作为文本。
【问题讨论】:
标签: c# asp.net-mvc asp.net-mvc-viewmodel