SpringJPA：嘗試創建外鍵約束時出錯

我正在嘗試使用 JPA 映射大型資料庫的某些列，以便從中獲取一些資料。該資料庫具有復合主鍵，其中一些也是外鍵。我對 JPA 映射相當陌生，所以我需要一些幫助。這是我得到的錯誤：

org.hibernate.tool.schema.spi.CommandAcceptanceException：通過 JDBC 陳述句執行 DDL“更改表 tbpedidoentrega 添加約束 FKiwjj63py270eqhfb1olp08oox 外鍵（fk_cliente）參考 tbcadastro”時出錯

并且：

錯誤：外鍵的參考列數和被參考列數不一致

JPA 似乎沒有在命令末尾指定它需要參考的列（正確的命令應該是alter table tbpedidoentrega add constraint FKiwjj63py270eqhfb1olp08oox foreign key (fk_cliente) references tbcadastro (codigo)），只指定了表。但為什么？

這是我的代碼：客戶端類

package com.agilsistemas.construtordepedidos.model;

import java.io.Serializable;

import javax.persistence.Column;
import javax.persistence.Entity;
import javax.persistence.GeneratedValue;
import javax.persistence.GenerationType;
import javax.persistence.Id;
import javax.persistence.Table;

import lombok.Getter;
import lombok.NoArgsConstructor;
import lombok.Setter;

@Entity
@Getter
@Setter
@NoArgsConstructor
@Table(name = "tbcadastro")
public class ClienteModel implements Serializable {
    
    @Id
    @GeneratedValue(strategy = GenerationType.IDENTITY)
    @Column(name = "codigo")
    int idCliente;
    
    @Column(name = "razao")
    String razaoSocial;

    @Column(name = "logradouro")
    String rua;
    
    @Column(name = "numero")
    String numero;

    @Column(name = "bairro")
    String bairro;

    @Column(name = "complemento")
    String complemento;

    @Column(name = "cidade")
    String cidade;

    @Column(name = "fixo")
    String telefoneFixo;

    @Column(name = "celular")
    String celular;

    @Column(name = "cliente")
    String cliente;

}

訂單類：

package com.agilsistemas.construtordepedidos.model;

import java.io.Serializable;
import java.time.LocalDate;
import java.util.List;

import javax.persistence.Column;
import javax.persistence.Entity;
import javax.persistence.GeneratedValue;
import javax.persistence.GenerationType;
import javax.persistence.Id;
import javax.persistence.JoinColumn;
import javax.persistence.ManyToOne;
import javax.persistence.OneToMany;
import javax.persistence.SequenceGenerator;
import javax.persistence.Table;

import lombok.Getter;
import lombok.NoArgsConstructor;
import lombok.Setter;

@Entity
@Getter
@Setter
@NoArgsConstructor
@Table(name = "tbpedidoentrega")
public class PedidoModel implements Serializable {
    
    @Id
    @GeneratedValue(strategy = GenerationType.IDENTITY)
    @Column(name = "id_pedido", nullable = false)
    @SequenceGenerator(name = "sqpedido")
    int idPedido;

    @Column(name = "data_pedido", nullable = false)
    LocalDate dataPedido;

    @Column(name = "hora_pedido", nullable = false)
    LocalDate horaPedido;

    @ManyToOne
    @JoinColumn(name = "fk_funcionario")
    FuncionarioModel fkFuncionario;

    @ManyToOne
    @JoinColumn(name = "fk_cliente")
    ClienteModel fkCliente;

    @OneToMany(mappedBy = "pedido")
    List<ItemPedidoModel> itensPedido;

}

我期待這會創建 FK 并啟動后端。我認為問題在于hibernate生成的SQL命令的末尾（應該是references tbcadastro (codigo)），但我不知道它為什么會這樣生成。

uj5u.com熱心網友回復：

經過一些干預和一些研究，我能夠解決這個問題。問題是表 Cliente 有一個復合主鍵，所以我需要實作一個特殊的類來指定復合鍵：

package com.agilsistemas.construtordepedidos.model;

import java.io.Serializable;

import javax.persistence.Column;
import javax.persistence.Embeddable;
import javax.persistence.JoinColumn;
import javax.persistence.ManyToOne;

import lombok.AllArgsConstructor;
import lombok.EqualsAndHashCode;
import lombok.Getter;
import lombok.Setter;

@Embeddable
@Getter
@Setter
@EqualsAndHashCode
@AllArgsConstructor
public class IdClienteModel implements Serializable {
    
    @Column(name = "codigo")
    private int idCliente;

    @ManyToOne
    @JoinColumn(name = "empresa")
    private EmpresaModel idEmpresa;

}

然后我像這樣實作了 ClienteModel 類：

package com.agilsistemas.construtordepedidos.model;

import java.io.Serializable;

import javax.persistence.Column;
import javax.persistence.EmbeddedId;
import javax.persistence.Entity;
import javax.persistence.Table;

import lombok.Getter;
import lombok.NoArgsConstructor;
import lombok.Setter;

@Entity
@Getter
@Setter
@NoArgsConstructor
@Table(name = "tbcadastro")
public class ClienteModel implements Serializable {
    
    @EmbeddedId
    private IdClienteModel idCliente; //using the object as the ID
    
    @Column(name = "razao")
    String razaoSocial;

    @Column(name = "logradouro")
    String rua;
    
    @Column(name = "numero")
    String numero;

    @Column(name = "bairro")
    String bairro;

    @Column(name = "complemento")
    String complemento;

    @Column(name = "cidade")
    String cidade;

    @Column(name = "fixo")
    String telefoneFixo;

    @Column(name = "celular")
    String celular;

    @Column(name = "cliente")
    String cliente;

}

現在，在 Pedido (Order) 類中，我可以得到這樣的 OneToOne 關系：

@OneToOne
    @JoinColumns({
            @JoinColumn(name = "fk_cliente", referencedColumnName = "codigo", insertable = false, updatable = false),
            @JoinColumn(name = "fk_empresa", referencedColumnName = "empresa", insertable = false, updatable = false) })
    ClienteModel fkCliente;

我對所有其他具有復合 PK 的物體也這樣做了，現在應用程式可以正常啟動。我希望這個遮陽篷可以幫助別人。

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