C# 带有复合键的实体框架一对多关系API

下面是我正在使用的类。我想创建一个从分层到耐心医生的单向参考。如果实体框架需要的话，我也可以建立一对多的关系，但这将始终是1-1，即使这是一个链接器表

我必须为这个ClientId，PatientId，PhysicianId使用一个复合键。我经常犯的错误是

One or more validation errors were detected during model generation:

SPM.Data.Stratification_Physician: : Multiplicity conflicts with the referential constraint in Role 'Stratification_Physician_Target' in relationship 'Stratification_Physician'. Because all of the properties in the Dependent Role are non-nullable, multiplicity of the Principal Role must be '1'.
Stratification_PrimaryPhysician_Target_Stratification_PrimaryPhysician_Source: : The number of properties in the Dependent and Principal Roles in a relationship constraint must be identical.
PatientClientPhysician_Stratifications_Target_PatientClientPhysician_Stratifications_Source: : The number of properties in the Dependent and Principal Roles in a relationship constraint must be identical.

什么是正确的fluent api调用

如果可能的话，我真的很想通过PatientClientMedical课程直接到内科医师

谢谢

public class PatientClientPhysician : BaseEntity
{
    public Guid ClientId { get; set; }
    public Guid PatientId { get; set; }
    public Guid PhysicianId { get; set; }

    public virtual Client Client { get; set; }
    public virtual Patient Patient { get; set; }
    public virtual Physician Physician { get; set; }

    public PatientClientPhysician()
        : base()
    {

    }
}


public class PatientClientPhysicianConfiguration : 
    EntityConfigurationBase<PatientClientPhysician>
{
    public PatientClientPhysicianConfiguration()
    {
        ToTable("patientclientphysician");

        //Property(m => m.Id).HasColumnName("id");
        Property(m => m.ClientId).HasColumnName("clientid");
        Property(m => m.PatientId).HasColumnName("patientid");
        Property(m => m.PhysicianId).HasColumnName("physicianid");

        HasRequired(m => m.Patient).WithMany().HasForeignKey(p => p.PatientId);
        HasRequired(m => m.Client).WithMany().HasForeignKey(c => c.ClientId);
        HasRequired(m => m.Physician).WithMany().HasForeignKey(p => p.PhysicianId);
    }
}

public class Stratification : BaseEntity
{
    public Stratification()
    {
        LabOrders = new List<StratLabOrder>();
        Observations = new List<StratObs>();
        Contacts = new List<StratContact>();
        StratIcd9s = new List<StratIcd9>();
    }

    public Guid ClientLocationId { get; set; }
    public Guid PatientId { get; set; }
    public Guid StratInputId { get; set; }
    public Guid ClientId { get; set; }
    public Guid PhysicianId { get; set; }

    public virtual ClientLocation ClientLocation { get; set; }
    public virtual Patient Patient { get; set; }
    public virtual StratInput StratInput { get; set; }
    public string EpisodeNumber { get; set; }
    public virtual Physician Physician { get; set; }
    public PrintStatus LetterPrintStatus { get; set; }

    public virtual ICollection<StratLabOrder> LabOrders { get; set; }
    public virtual ICollection<StratObs> Observations { get; set; }
    public virtual ICollection<StratContact> Contacts { get; set; }
    public virtual ICollection<StratIcd9> StratIcd9s { get; set; }
    public virtual PatientClientPhysician PrimaryPhysician { get; set; }
}

//this is in EntityConfigurationBase<Stratification> class
HasRequired(m => m.PrimaryPhysician).WithRequired().HasForeignKey(p => new { p.ClientId, p.PatientId, p.PhysicianId });

转到PatientClientPhysicianConfiguration类并获取此错误

One or more validation errors were detected during model generation:

SPM.Data.Stratification_Physician: : Multiplicity conflicts with the referential constraint in Role 'Stratification_Physician_Target' in relationship 'Stratification_Physician'. Because all of the properties in the Dependent Role are non-nullable, multiplicity of the Principal Role must be '1'.

这就是我现在设定的关系

HasRequired(m => m.PatientClientPhysicians).WithMany().HasForeignKey(p => new { p.ClientId, p.PatientId, p.PhysicianId });

试试这个：

public class PatientClientPhysicianConfiguration : 
    EntityConfigurationBase<PatientClientPhysician>
{
    public PatientClientPhysicianConfiguration()
    {
        ToTable("patientclientphysician");

        // Composite key:
        HasKey(t => new { t.ClientId, t.PatientId, t.PhysicianId });

        Property(m => m.ClientId).HasColumnName("clientid");
        Property(m => m.PatientId).HasColumnName("patientid");
        Property(m => m.PhysicianId).HasColumnName("physicianid");

        HasRequired(m => m.Patient).WithMany().HasForeignKey(p => p.PatientId);
        HasRequired(m => m.Client).WithMany().HasForeignKey(c => c.ClientId);
        HasRequired(m => m.Physician).WithMany().HasForeignKey(p => p.PhysicianId);
    }
}

公共类PatientClientPhysician配置：
实体配置库
{
公共患者客户端物理配置（）
{
ToTable（“患者-医生”）；
//组合键：
HasKey（t=>new{t.ClientId，t.PatientId，t.PhysicianId}）；
属性（m=>m.ClientId）.HasColumnName（“ClientId”）；
属性（m=>m.PatientId）.HasColumnName（“PatientId”）；
属性（m=>m.PhysicianId）.HasColumnName（“PhysicianId”）；
HasRequired（m=>m.Patient）.WithMany（）.HasForeignKey（p=>p.PatientId）；
HasRequired（m=>m.Client）.WithMany（）.HasForeignKey（c=>c.ClientId）；
HasRequired（m=>m.medical）.WithMany（）.HasForeignKey（p=>p.PhysicianId）；
}
}

我添加了HasKey，这就是我现在得到的错误。在模型生成过程中检测到一个或多个验证错误：SPM.Data.stratization\u medicine:：多重性与关系“stratization\u medicine”中角色“stratization\u medicine\u Target”中的引用约束冲突。由于从属角色中的所有属性都不可为Null，因此主体角色的多重性必须为“1”。这与另一个属性可为Null是不同的问题。这很有效。谢谢

public class PatientClientPhysicianConfiguration : 
    EntityConfigurationBase<PatientClientPhysician>
{
    public PatientClientPhysicianConfiguration()
    {
        ToTable("patientclientphysician");

        // Composite key:
        HasKey(t => new { t.ClientId, t.PatientId, t.PhysicianId });

        Property(m => m.ClientId).HasColumnName("clientid");
        Property(m => m.PatientId).HasColumnName("patientid");
        Property(m => m.PhysicianId).HasColumnName("physicianid");

        HasRequired(m => m.Patient).WithMany().HasForeignKey(p => p.PatientId);
        HasRequired(m => m.Client).WithMany().HasForeignKey(c => c.ClientId);
        HasRequired(m => m.Physician).WithMany().HasForeignKey(p => p.PhysicianId);
    }
}