Question

当我运行程序并在控制器方法中设置断点时，我可以单步执行并查看recipe.Name和recipe.ID已正确填充，但recipe.Ingredients为空。

我在俯瞰什么？这是我的相关代码：

型号：

public class Recipe
{
    public int ID { get; set; }
    public string Name { get; set; }
    public virtual ICollection<Ingredient> Ingredients { get; set; }
}

查看：

@model Cookbook.Models.Recipe
// removed irrelevant code for this question
<dt>@Html.DisplayNameFor(model => model.Name)</dt>
<dd>@Html.DisplayFor(model => model.Name)</dd>
<dd>
   <table class="table">
        <tr><th>Ingredient</th></tr>
        @foreach (var item in Model.Ingredients)
        {
            <tr><td>@Html.DisplayFor(model => item.Name)</td></tr>
        }
    </table>
</dd>
<p> @Html.ActionLink("Export Data", "ExportData", Model)</p>

控制器：

public ActionResult ExportData (Recipe recipe)
{
    //I am dynamically building an XML file by constructing it line by line
    string xml = recipe.Name + "\r\n\r\n"; //Here, Name is populated
    //Here, recipe.Ingredients is empty even though it appears in the view
    foreach(Ingredient ing in recipe.Ingredients)
    {
        xml = xml + ing.Ingredient.Name + "\r\n";
    }
}

Answer 1

您不能使用@Html.ActionLink()将包含属性集合（或复杂对象）的模型传递给get方法。在内部，该方法通过在模型中的每个属性上调用.ToString()方法来生成查询字符串值。在你的情况下它正在生成

...?ID=someValue&Name=someValue&Ingredients=System.Collections.Generic.ICollection<Ingredient>

为了绑定你的模型，它需要

...?....&Ingredients[0].Name=someValue&Ingredients[1].Name=someValue&...

相反。将模型的ID传递给方法，然后再次获取模型（就像在生成此视图的GET方法中所做的那样）来构建xml文件

@Html.ActionLink("Export Data", "ExportData", new { id = Model.ID })

public ActionResult ExportData (int ID)
{
  var recipe = db.Recipies.Where(r => r.ID = ID).FirstOrDefault();
  ....
}

将模型从视图传递到控制器时，不会填充虚拟属性

1 个答案: