Entity framework 有关系的种子项目

考虑以下几点：

public class Foo
{
    [Key]
    public int Id { get; set; }

    public string Name { get; set; }

    public string Something { get; set; }

    public virtual ICollection<Bar> Bars { get; set; }
}

public class Bar
{
    [Key]
    public int Id { get; set; }

    public string Name { get; set; }

    public string Something { get; set; }

    [ForeignKey("Foo")]
    public int FooId { get; set; }
    public virtual Foo Foo { get; set; }
}

如果我运行

updatedatabase

所有操作都如您所期望的那样工作，并且我会为每个

Foo

添加相关的

条

现在，如果我将种子数据更改为：

context.Foos.AddOrUpdate(
    r => r.Name,
    new Foo
    {
        Name = "Foo 1",
        Something = "BlahBlahBlah",
        Bars = new List<Bar>
        {
            new Bar { Name = "Foo 1 Bar 1", Something = "BlahBlahBlah" },
            new Bar { Name = "Foo 1 Bar 2", Something = "BlahBlahBlah" },
            new Bar { Name = "Foo 1 Bar 3", Something = "BlahBlahBlah" },
        }
    },
    ...

但考虑到我使用标识列作为键，我无法知道要将哪个
Bar
s添加到哪个
Foo
s。有什么想法吗？
当我输入问题时，我突然想到我可以这样做：

var bar1 = new Bar { Name = "Foo 1 Bar 1", Something = "Blah" };
var bar2 = new Bar { Name = "Foo 1 Bar 2", Something = "Blah" };
var bar3 = new Bar { Name = "Foo 1 Bar 3", Something = "Blah" };
...

context.Bars.AddOrUpdate(
    r => r.Name,
    bar1,
    bar2,
    bar3,
    ...
);

context.Foos.AddOrUpdate(
    r => r.Name,
    new Foo
    {
        Name = "Foo 1",
        Something = "Blah",
        Bars = new List<Bar>
        {
            bar1,
            bar2,
            bar3
        }
    },
    ...
);

var bar1=newbar{Name=“Foo 1 Bar 1”，Something=“Blah”}；
var bar2=newbar{Name=“Foo 1 Bar 2”，Something=“Blah”}；
var bar3=newbar{Name=“Foo 1 Bar 3”，Something=“Blah”}；
...
context.bar.AddOrUpdate(
r=>r.Name，
bar1，
bar2，
bar3，
...
);
context.Foos.AddOrUpdate(
r=>r.Name，
新富
{
Name=“Foo 1”，
某物=“诸如此类”，
条=新列表
{
bar1，
bar2，
三维直方图
}
},
...
);

然而，我还没有机会对此进行测试，无论如何，我仍然愿意接受其他想法。
您的第一种方法（问题）和第二种方法（答案）都不适用于第二次执行种子方法

第一种方法

Bars = new List<Bar>
{
    new Bar { Name = "Foo 1 Bar 1", Something = "Blah" },
    // new Bar { Name = "Foo 1 Bar 4", Something = "Blah" },
}

var bar1 = new Bar { Name = "Foo 1 Bar 1", Something = "Blah" };
context.Bars.AddOrUpdate(
    r => r.Name,
    bar1
);
context.Foos.AddOrUpdate(
    r => r.Name,
    new Foo
    {
        Name = "Foo 1",
        Something = "Blah",
        Bars = new List<Bar>
        {
            bar1
        }
    }
);

这比第一个方法更糟糕，Seed方法的第二次执行将抛出异常，因为在更新任何
条
时，它将尝试更新
条
与未找到
Foo
的关系，未指定
条的FooId
，这意味着其值将为0<找不到Id等于0的code>Foo

解决方案

您需要定义临时密钥才能正确执行“添加或更新”

var bar1 = 
    new Bar { Id = 1, Name = "Foo 1 Bar 1", Something = "Blah", FooId = 1 };
var bar2 = 
    new Bar { Id = 2, Name = "Foo 1 Bar 2", Something = "Blah", FooId = 1 };
var bar3 = 
    new Bar { Id = 3, Name = "Foo 1 Bar 3", Something = "Blah", FooId = 1 };
// var bar4 =
//     new Bar { Id = 4, Name = "Foo 1 Bar 4", Something = "Blah", FooId = 1 };

db.Bars.AddOrUpdate(
    r => r.Name,
    bar1,
    bar2,
    bar3
    // bar4
);

db.Foos.AddOrUpdate(
    r => r.Name,
    new Foo
    {
        Id = 1,
        Name = "Foo 1",
        Something = "Blah"
    }
);

第二次执行Seed方法将把

foo's和bar'Something

更新为

BlahBlahBlah

。新的

条

（注释代码）将成功添加并引用现有的

Foo

更多

var bar1 = new Bar { Name = "Foo 1 Bar 1", Something = "Blah" };
context.Bars.AddOrUpdate(
    r => r.Name,
    bar1
);
context.Foos.AddOrUpdate(
    r => r.Name,
    new Foo
    {
        Name = "Foo 1",
        Something = "Blah",
        Bars = new List<Bar>
        {
            bar1
        }
    }
);

var bar1 = 
    new Bar { Id = 1, Name = "Foo 1 Bar 1", Something = "Blah", FooId = 1 };
var bar2 = 
    new Bar { Id = 2, Name = "Foo 1 Bar 2", Something = "Blah", FooId = 1 };
var bar3 = 
    new Bar { Id = 3, Name = "Foo 1 Bar 3", Something = "Blah", FooId = 1 };
// var bar4 =
//     new Bar { Id = 4, Name = "Foo 1 Bar 4", Something = "Blah", FooId = 1 };

db.Bars.AddOrUpdate(
    r => r.Name,
    bar1,
    bar2,
    bar3
    // bar4
);

db.Foos.AddOrUpdate(
    r => r.Name,
    new Foo
    {
        Id = 1,
        Name = "Foo 1",
        Something = "Blah"
    }
);