多对多关系基本示例(MVC3)

本文关键字：MVC3 关系 | 更新日期: 2023-09-27 17:54:38

我有一个mvc c#项目，我有一个FoodItem和FoodItemCategory的模型。两个模型如下图所示:

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_CreateOrEdit。cshtml视图，它最初是由脚手架生成的，我修改了它，以包括所有的类别，并勾选食品项目所属的框。一种食物可以有许多或所有的类别。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)

正如您所看到的，我有一个嵌套循环，它为每个类别创建一个复选框，当它创建每个类别时，我循环并检查模型的Categories属性中的特定类别。如果存在，则设置复选框的checked属性。如果您选中一个框并单击save，那么在控制器上的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);
    }

我可以保存一次分类。如果我返回到视图中，并单击save，就会收到以下错误:

不能在唯一索引中插入重复的值。[表名=>FoodItemCategoryFoodItems，约束名= PK_FoodItemCategoryFoodItems_00000000000000A8]描述:在执行当前web请求期间发生了未处理的异常。请查看堆栈跟踪，了解有关错误及其在代码中的起源的更多信息。

Exception Details: System.Data.SqlServerCe.SqlCeException:一个重复的值不能插入到一个唯一索引中。[Table name = FoodItemCategoryFoodItems,Constraint name =>PK_FoodItemCategoryFoodItems_00000000000000A8]

源错误:

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

不确定是否重要，但我正在使用SQLServer精简版4。我这样做对吗?对于这样的代码，通常的编码实践是什么?我知道同样的情况每天都会发生，因为同样的关系模型在博客等许多情况下都使用。

多对多关系基本示例(MVC3)

试试这样做(未测试):

[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，但因为你处理的是多对多关系，关系可以在视图中添加或删除，这是唯一的方法。原因如下:

  • 你必须说EF哪些关系被添加，哪些被删除
  • 如果您只是添加所有相关类别，则需要再次插入关系
  • 你不能说EF哪些关系被删除了，因为你不传递未选中类别的信息

更多关于分离对象图和处理关系的描述在这里。这是关于ObjectContext API，但DbContext API只是包装，所以同样的限制仍然存在。

除了Ladislav的答案之外，您还可以通过使用http://haacked.com/archive/2008/10/23/model-binding-to-a-list.aspx来删除Request[].split()部分，这使得:

[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);
}
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