将项目移动到Entity Framework迁移中的项目列表

Question

我正在使用带有Entity Framework 6的Asp.Net MVC 5.我有一个这样的表:

public class Recipe
{
    [Required]
    public virtual Food Food { get; set; }

    //...rest
}

现在我想把Food一份食物列入清单.像这样:

public class Recipe
{
    public virtual IList<Food> Foods { get; set; }

    //... rest
}

如果我尝试这个并添加迁移,我将丢失与FoodIds 相关的数据.并且不知道哪种食谱适合哪种食物.

我试着保留Food并添加如下列表:

public class Recipe
{
    [Required]
    public virtual Food Food { get; set; }

    public virtual IList<Food> Foods { get; set; }

    //... rest
}

但是迁移添加了第二个外键并且无法更新.这是失败的迁移代码:

    public override void Up()
    {
        DropForeignKey("dbo.Recipes", "Food_ID", "dbo.Foods");
        DropIndex("dbo.Recipes", new[] { "Food_ID" });
        AddColumn("dbo.Foods", "Recipe_Id", c => c.Int());
        AddColumn("dbo.Foods", "Recipe_Id1", c => c.Int());
        AlterColumn("dbo.Recipes", "Food_ID", c => c.Int());
        CreateIndex("dbo.Foods", "Recipe_Id");
        CreateIndex("dbo.Foods", "Recipe_Id1");
        CreateIndex("dbo.Recipes", "Food_ID");
        AddForeignKey("dbo.Foods", "Recipe_Id", "dbo.Recipes", "Id");
        AddForeignKey("dbo.Foods", "Recipe_Id1", "dbo.Recipes", "Id");
        AddForeignKey("dbo.Recipes", "Food_ID", "dbo.Foods", "ID");
    }

如何将单个Food列表移动到Foods 列表而不丢失表中的当前数据？

Answer 1

创建迁移后，您需要对其进行自定义。

1. 修改您的实体类

   public class Recipe
   {
       public virtual IList<Food> Foods { get; set; }
   
       //... rest
   }
   

2. 用于Add-Migration为迁移搭建脚手架

3. 修改生成的迁移以使用数据填充新列。您需要使用先前相关行的 ID填充新Recipe_Id列Foods。

   public override void Up()
   {
       DropForeignKey("dbo.Recipes", "Food_ID", "dbo.Foods");
       DropIndex("dbo.Recipes", new[] { "Food_ID" });
       AddColumn("dbo.Foods", "Recipe_Id", c => c.Int());
   
       // Update values from existing data, not sure if the syntax is perfect
       Sql(@"UPDATE dbo.Foods SET Recipe_Id = r.Id
             FROM (SELECT Id, Food_ID FROM dbo.Recipes) AS r
             WHERE Foods.ID = r.Food_ID");
   
       DropColumn("dbo.Recipes", "Food_ID");
       CreateIndex("dbo.Foods", "Recipe_Id");
       AddForeignKey("dbo.Foods", "Recipe_Id", "dbo.Recipes", "Id");
   }