Question

我有两张桌子：

DocumentSet:
setId (int)
name (varchar(50))

DocIdentifier:
docIdId (int)
docSetId (int) <-- (setId from DocumentSet)
customId (char(9))

在SQL中执行的两个表的典型查询如下所示：

SELECT ds.name
FROM DocumentSet ds
INNER JOIN DocIdentifiers di ON ds.setId = di.docSetId
WHERE di.customId == 'someId'

请注意命名差异：setId和docSetId是用于外键的列。当我尝试在Entity Framework中复制它时，问题出现了。我有明显的模型：

class DocumentSet {
    public int Id {get;set;}
    public string Name {get;set;}
    public virtual ICollection<DocumentIdentifier> Identifiers {get;set;}
}

class DocumentIdentifier {
    public int Id {get;set;}
    public int SetId {get;set;}
    public string CustomId {get;set;}
    public DocumentSet DocSet {get;set;}
}

同样，为了清晰起见，代码中存在一些命名差异。所以我为两者创建了EntityTypeConfiguration：

internal class DocumentSetTypeConfiguration : EntityTypeConfiguration<DocumentSet> {
    public DocumentSetTypeConfiguration () {
        HasKey(f => f.Id);
        Property(f => f.Id).HasColumnName("setId");
        HasMany(f => f.DocumentIdentifiers);
    }
}

internal class DocumentIdentifierTypeConfiguration : EntityTypeConfiguration<DocumentIdentifier> {
    public DocumentIdentifierTypeConfiguration () {
        ToTable("DocIdentifier");
        HasKey(f => f.Id);
        Property(f => f.Id).HasColumnName("docIdId");
        Property(f => f.SetId).HasColumnName("docSetId");
    }
}

问题在于：当我发送这样的LINQ查询时：

_context.Set<DocumentSet>().Where(d => d.DocumentIdentifiers.Any(pt=>pt.CustomId == customId)).Single();

生成的内容有点拙劣：

SELECT TOP (2) 
    [Extent1].[SetId] AS [SetId], 
    [Extent1].[Name] AS [Name]
    FROM [dbo].[DocumentSet] AS [Extent1]
    WHERE  EXISTS (SELECT 
        1 AS [C1]
        FROM [dbo].[DocIdentifier] AS [Extent2]
        WHERE ([Extent1].[SetId] = [Extent2].[DocumentSet_Id]) AND (([Extent2].CustomId = @p__linq__0) OR (([Extent2].[CustomId] IS NULL) AND (@p__linq__0 IS NULL)))
    )

我的猜测问题在于setId和docSetId是不同的名称，但我似乎无法弄清楚如何让EF看到[Extent2].[DocumentSet_Id]应该是[Extent2].[docSet_Id]。在DocumentIdentifierTypeConfiguration中我是否遗漏了哪些内容可以帮助EF理解这一点？我认为Property(f => f.SetId).HasColumnName("docSetId");行会这样做，但显然不行。

简而言之，我怎么说

`DocumentSet has many DocumentIdentifiers with a foreign key relationship of DocumentSet.setId = DocIdentifier.docSetId`?

Answer 1

尝试添加WithRequired和HasForeignKey。

HasMany(f => f.Identifiers).WithRequired().HasForeignKey(x => x.SetId);

完整代码

public class AppContext : DbContext
{
    public DbSet<DocumentSet> DocumentSets { get; set; }
    protected override void OnModelCreating(DbModelBuilder modelBuilder)
    {
        modelBuilder.Configurations.Add(new DocumentSetTypeConfiguration());
        modelBuilder.Configurations.Add(new DocumentIdentifierTypeConfiguration());
    }
}
public class DocumentSetTypeConfiguration : EntityTypeConfiguration<DocumentSet>
{
    public DocumentSetTypeConfiguration()
    {
        HasKey(f => f.Id);
        Property(f => f.Id).HasColumnName("setId");
        HasMany(f => f.DocumentIdentifiers).WithRequired().HasForeignKey(x => x.SetId);
    }
}
public class DocumentIdentifierTypeConfiguration : EntityTypeConfiguration<DocumentIdentifier>
{
    public DocumentIdentifierTypeConfiguration()
    {
        ToTable("DocIdentifier");
        HasKey(f => f.Id);
        Property(f => f.Id).HasColumnName("docIdId");
        Property(f => f.SetId).HasColumnName("docSetId");
    }
}
public class DocumentSet
{
    public int Id { get; set; }
    public string Name { get; set; }
    public virtual ICollection<DocumentIdentifier> DocumentIdentifiers { get; set; }
}
public class DocumentIdentifier
{
    public int Id { get; set; }
    public int SetId { get; set; }
    public string CustomId { get; set; }
}

数据库结果

Database Result

查询结果

SELECT TOP (2) 
    [Extent1].[setId] AS [setId], 
    [Extent1].[Name] AS [Name]
FROM [dbo].[DocumentSets] AS [Extent1]
WHERE  EXISTS (SELECT 
    1 AS [C1]
    FROM [dbo].[DocIdentifier] AS [Extent2]
    WHERE ([Extent1].[setId] = [Extent2].[docSetId])
    AND   (([Extent2].[CustomId] = @p__linq__0) 
          OR (([Extent2].[CustomId] IS NULL) AND (@p__linq__0 IS NULL)))
)

EF Code First：与不同密钥名称的关系

1 个答案: