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我有一个下拉视图,它与视图模型相关联......

视图模型的属性之一:

public int selectDTypeID { get; set; }

视图有一个下拉菜单:

   Html.DropDownListFor(
       model => model.selectDTypeID , 
       new SelectList(
              new List<Object>{ 
                   new { value = 1 , text = "Test1"  },
                   new { value = 2 , text = "Test2" },
                   new { value = 3 , text = "Test3"}
                },
              "value",
              "text"
       )
    )

如何在不使用额外参数的情况下获得回发下拉列表的值...我只想能够做到这一点...

[HttpPost]
public ActionResult ContinueDonation(AddDonationViewModel model)
{
 var id = model.selectDTypeID;

}

我可以做这样的事情吗?

4

1 回答 1

2

如果您的视图是这样的强类型视图,

@model AddDonationViewModel
@using (Html.BeginForm())
{
   @Html.DropDownListFor(
       model => model.selectDTypeID , 
       new SelectList(
              new List<Object>{ 
                   new { value = 1 , text = "Test1"  },
                   new { value = 2 , text = "Test2" },
                   new { value = 3 , text = "Test3"}
                },
              "value",  "text"))    

    <input type="submit" value="go" />
}

当您单击提交时,所选值将在您的 HttpPost Action 方法的 SelectdTYpeID 属性中可用

[HttpPost]
public ActionResult ContinueDonation(AddDonationViewModel model)
{
  var id = model.selectDTypeID;  // You have your selected ID here
}
于 2012-05-05T21:03:02.013 回答