1

我有一个 MVC3 C# 项目,我有一个 FoodItem 和 FoodItemCategory 模型。两种模型如下图所示:

public class FoodItem
{
    public int ID { get; set; }
    [Required]
    public string Name { get; set; }
    public string Description { get; set; }
    public virtual ICollection<FoodItemCategory> Categories { get; set; }
    public DateTime CreateDate { get; set; }
}

public class FoodItemCategory {
    public int ID { get; set; }
    public string Name { get; set; }
    public string Description { get; set; }
    public virtual ICollection<FoodItem> FoodItems { get; set; }
    public DateTime CreateDate { get; set; }
} 

我有一个最初从 Scaffolder 生成的 _CreateOrEdit.cshtml 视图,我对其进行了修改以包含所有类别并选中食品所属的框。一个食品项目可以有许多或所有类别。视图如下所示:

@model StackOverFlowIssue.Models.FoodItem
<div class="editor-label">
    @Html.LabelFor(model => model.Name)
</div>
<div class="editor-field">
    @Html.EditorFor(model => model.Name)
    @Html.ValidationMessageFor(model => model.Name)
</div>
<div class="editor-label">
    @Html.LabelFor(model => model.Description)
</div>
<div class="editor-field">
    @Html.EditorFor(model => model.Description)
    @Html.ValidationMessageFor(model => model.Description)
</div>
<div class="editor-label">
    @Html.LabelFor(model => model.Categories)
</div>
<div class="editor-field">
    @foreach (var FoodItemCategory in (IEnumerable<StackOverFlowIssue.Models.FoodItemCategory>)ViewBag.Categories){
        <input type="checkbox" name="FoodItemCategoryId" value="@FoodItemCategory.ID" 
        @foreach(var c in Model.Categories){
            if(c.ID == FoodItemCategory.ID){ 
                @String.Format("checked=\"checked\"")
            } 
        } 
        />
        @FoodItemCategory.Name 
        <br />
    } 
</div>
@Html.Hidden("CreateDate", @DateTime.Now)

如您所见,我有一个嵌套循环,它为每个类别创建一个复选框,并且在创建每个类别时,我循环并在模型的 Categories 属性中检查该特定类别。如果存在,我设置复选框的选中属性。如果您选中一个框并单击保存,则在控制器上的 HttpPost 操作上,我将执行以下操作:

    [HttpPost]
    public ActionResult Edit(FoodItem foodItem)
    {
        if (ModelState.IsValid)
        {
            var cList = Request["CategoryId"].Split(',');
            List<FoodItemCategory> categories = new List<FoodItemCategory>();

            foreach (var c in cList) {
                var ci = Convert.ToInt32(c);
                FoodItemCategory category = context.FoodItemCategories.Single(x => x.ID == ci);
                categories.Add(category);
            }

            context.Entry(foodItem).State = EntityState.Modified;
            restaurant.Categories = categories;
            context.SaveChanges();
            return RedirectToAction("Index");
        }
        return View(foodItem);
    }

我可以一次保存类别。如果我返回视图,然后单击保存,我会收到以下错误:

不能将重复值插入唯一索引。[表名=>FoodItemCategoryFoodItems,约束名= PK_FoodItemCategoryFoodItems_00000000000000A8 ] 描述:当前Web请求执行过程中出现未处理的异常。请>查看堆栈跟踪以获取有关错误及其源自代码的位置的更多信息。

异常详细信息:System.Data.SqlServerCe.SqlCeException:无法将重复值插入 > 唯一索引。[ 表名 = FoodItemCategoryFoodItems,约束名称 = >PK_ FoodItemCategoryFoodItems _00000000000000A8 ]

源错误:

第 97 行: context.Entry(foodItem).State = EntityState.Modified; 第 98 行:foodItem.Categories = 类别;第 99 行:context.SaveChanges(); 第 100 行:返回 RedirectToAction("Index"); 第 101 行:}

不确定这是否重要,但我使用的是 SQLServer Compact Edition 4。我这样做的方式是否正确?像这样的事情的正常编码实践是什么?我知道每天都会发生同样的情况,因为在博客等许多情况下都使用了同样的关系模型。

4

2 回答 2

4

尝试这样的事情(未经测试):

[HttpPost]
public ActionResult Edit(FoodItem foodItem)
{
    if (ModelState.IsValid)
    {
        int id = foodItem.Id;
        // Load food item with related categories first
        var item = context.FoodItems
                          .Include(f => f.Categories)
                          .Single(f => f.Id == id);

        // Process changed scalar values
        context.Entry(item).CurrentValues.SetValues(foodItem);

        // Brute force processing of relations
        // This can be optimized - instead of deleting all and adding all again
        // you can manually compare which relations already exists, add new and
        // remove non existing but let's make that as a homework
        item.Categories.Clear();

        var cList = Request["CategoryId"].Split(',');

        foreach (var c in cList) 
        {
            var ci = Convert.ToInt32(c);
            // Use find - if category was already loaded in the first query, it will
            // be reused without additional query to DB
            var category = context.Categories.Find(ci);
            // Now add category to attached food item to create new relation
            item.Categories.Add(category);
        }

        context.SaveChanges();
        return RedirectToAction("Index");
    }

    return View(foodItem);
}

这看起来效率很低,但是因为您正在处理可以在视图中添加或删除关系的多对多关系,所以这是唯一的方法。原因是:

  • 你必须说 EF 添加了哪些关系,删除了哪些关系
  • 如果您只是添加所有相关类别,您将再次插入关系
  • 您不能说 EF 删除了哪些关系,因为您不传输有关未检查类别的信息

有关分离对象图和处理关系的更多信息在此处进行了描述。它与 ObjectContext API 有关,但 DbContext API 只是围绕它进行包装,因此仍然存在相同的限制。

于 2011-05-01T19:47:02.773 回答
0

除了 Ladislav 的回答之外,您还可以使用http://haacked.com/archive/2008/10/23/model-binding-to-a-list.aspx摆脱 Request[].split()部分使:

[HttpPost]
public ActionResult Edit(FoodItem foodItem, ICollection<int> CategoryId)
{
    if (ModelState.IsValid)
    {
        int id = foodItem.Id;
        // Load food item with related categories first
        var item = context.FoodItems
                          .Include(f => f.Categories)
                          .Single(f => f.Id == id);

        // Process changed scalar values
        context.Entry(item).CurrentValues.SetValues(foodItem);

        // Brute force processing of relations
        // This can be optimized - instead of deleting all and adding all again
        // you can manually compare which relations already exists, add new and
        // remove non existing but let's make that as a homework
        item.Categories.Clear();

        foreach (var id in CategoryID) 
        {
            // Use find - if category was already loaded in the first query, it will
            // be reused without additional query to DB
            var category = context.Categories.Find(id);
            // Now add category to attached food item to create new relation
            item.Categories.Add(category);
        }

        context.SaveChanges();
        return RedirectToAction("Index");
    }

    return View(foodItem);
}
于 2011-06-28T09:49:01.830 回答