Question

我遇到一个问题，当我有一个复杂的模型时，如果我提交表单它不会给我所有模型属性的所有值，在下面的例子中，我没有得到gridModel属性：< / p>

模型

public class InventoryModel {
    public GridModel GridModel { get; set; }
    public Int32 UserKey { get; set; }
}

public class GridModel {
    public String GridId { get; set; }
    public String GridName { get; set; }
    public List<String> columns { get; set; }
}

控制器

public ActionResult Index(){
    InventoryModel model = new InventoryModel();

    model.UserKey= 20014;
    model.GridModel = new GridModel();
    model.GridModel.GridId = "jqgInventory";
    model.GridModel.GridName = "Inventory Grid";
    return View(model);
}

 [HttpPost]
 public ActionResult Index(InventoryModel model){
    Int32 userId = model.UserKey; // This has a value
    String gridId = model.GridModel.GridId;  // This doesn't have a value
    String gridName= model.GridModel.GridName; // This doesn't have a value
}

查看

@model InventoryModel
@using (Html.BeginForm()) {
    @Html.TextBoxFor(m => m.UserKey, new { @class = "w200" })
    @Html.TextBoxFor(m => m.GridModel.GridId , new { @class = "w200" })
    @Html.TextBoxFor(m => m.GridModel.GridName, new { @class = "w200" })

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

任何建议都将不胜感激。

谢谢，阿拉

Answer 1

您可以使用ViewModel而不是实际的Model。这将是一个更平坦的类，它反映了View专门的数据。

public class InventoryViewModel{
    Int32 UserKey {get; set; }
    public String GridId { get; set; }
    public String GridName { get; set; }        
}

如有必要，您的控制器可以将您的模型映射到ViewModel

当表单发布时，MVC在模型中提交多个对象

1 个答案: