org.hibernate.exception.ConstraintViolationException：无法插入

Question

当我尝试添加到TDEPOFAZLA 表时，我收到以下错误：

org.springframework.dao.DataIntegrityViolationException：无法插入：[tr.gov.tcmb.pgmtems.model.DepoFazla]； SQL [插入 PGMTEMS.TDEPOFAZLA (ID, FAZLABULUNDURMAORANI, GRUP) 值 (默认, ?, ?)];约束 [null];嵌套异常是 org.hibernate.exception.ConstraintViolationException：无法插入：[tr.gov.tcmb.pgmtems.model.DepoFazla]

这是我的 JUnit 测试函数：

    @Test
public void testSaveDepoFazla() {
    DepoTur depoTur = new DepoTur("my tür", 5);
    depoTurService.saveDepoTur(depoTur);

    List<DepoTur> list = depoTurService.getDepoTurList();
    assertNotNull(list.get(0));

    BigDecimal fazlaBulundurmaOrani = new BigDecimal(6000);
    DepoFazla depoFazla = new DepoFazla(1, list.get(0), fazlaBulundurmaOrani);

    depoFazlaService.saveDepoFazla(depoFazla);
}

这是我的 DepoFazla.java：

@Entity
@Table(schema = "PGMTEMS", name = "TDEPOFAZLA")
public class DepoFazla implements Serializable {
    private static final long serialVersionUID = -2800365387332643658L;

    @Id
    @GeneratedValue
    @Column(name = "ID", nullable = false, updatable = false)
    private Long id;

    @Column(name = "GRUP", nullable = false, columnDefinition = "INTEGER")
    private Integer grup;

    @ManyToOne(fetch = FetchType.LAZY, targetEntity = DepoTur.class)
    @JoinColumn(name = "ID", insertable = false, updatable = false)
    @NotNull
    private DepoTur depoTur;

    @Column(name = "FAZLABULUNDURMAORANI", nullable = false, columnDefinition = "DECIMAL(6, 2)")
    private BigDecimal fazlaBulundurmaOrani;

    public DepoFazla() {
        super();
    }

    public DepoFazla(Integer grup, DepoTur depoTur, BigDecimal fazlaBulundurmaOrani) {
        super();
        this.grup = grup;
        this.depoTur = depoTur;
        this.fazlaBulundurmaOrani = fazlaBulundurmaOrani;
    }
//GETTER AND SETTER METHODS
}

这里是 DepoTur.java：

@Entity
@Table(schema = "PGMTEMS", name = "TDEPOTUR")
public class DepoTur implements Serializable {
    private static final long serialVersionUID = 6203672609079710060L;

    @Id
    @GeneratedValue
    @Column(name = "ID", nullable = false, updatable = false)
    @Index(name = "XUTDEPOTURP", columnNames = { "id" })
    private Long id;

    @Column(name = "ACIKLAMA", nullable = false)
    private String aciklama;

    @Column(name = "BLOKESIRASI", nullable = false)
    private Integer blokeSirasi; //

    @Column(name = "DEPOCINSI")
    private String depoCinsi;

    public DepoTur() {
        super();
    }

    public DepoTur(String aciklama, Integer blokeSirasi, String depoCinsi) {
        super();
        this.aciklama = aciklama;
        this.depoCinsi = depoCinsi;
        this.blokeSirasi = blokeSirasi;
    }

public DepoTur(String aciklama, Integer blokeSirasi) {
    super();
    this.aciklama = aciklama;
    this.blokeSirasi = blokeSirasi;
}
//GETTER AND SETTER METHODS

当我调试 JUnit 测试时，我得到这个错误：

错误：DB2 SQL 错误：SQLCODE=-407，SQLSTATE=23502，SQLERRMC=TBSPACEID=2，TABLEID=75，COLNO=2，DRIVER=3.50.152 SQLState：23502 错误代码：-407

当我搜索错误时，我发现我尝试插入 NULL 但我无法弄清楚我在哪里添加了 null 值。

这就是我创建 TDEPOFAZLA 表的方式：

    CREATE TABLE TDEPOFAZLA
(
   ID decimal(20,0) PRIMARY KEY NOT NULL,
   GRUP int NOT NULL,
   DEPOTUR decimal(20,0) NOT NULL,
   FAZLABULUNDURMAORANI decimal(6,2) NOT NULL
);
CREATE UNIQUE INDEX XUTDEPOFAZLAP ON TDEPOFAZLA(ID);

这就是我创建 TDEPOTUR 表的方式：

CREATE TABLE TDEPOTUR
(
   ID decimal(20,0) PRIMARY KEY NOT NULL,
   ACIKLAMA varchar(100) NOT NULL,
   DEPOCINSI char(1),
   BLOKESIRASI int NOT NULL
);
CREATE UNIQUE INDEX XUTDEPOTURP ON TDEPOTUR(ID);

对我应该做什么有什么想法吗？

Answer 1

您使用 FK 的属性定义是错误的，因为我看到您使用可更新、可插入为 false，您真正想要的是不修改 FK 对象，因为它可能是某些其他实体通用的主表。

那你可以用的是这个

@ManyToOne(cascade= {CascadeType.DETACH})
@JoinColumn(name = "DEPOTUR")
@NotNull
private DepoTur depoTur;

使用 DETACH，您将仅将值保存在第一个表 DEPOTUR 列中，并且不会更新 DepoTur 表中的对象

在你的第一个表中添加 FK

CREATE TABLE TDEPOFAZLA
(
  ID decimal(20,0) PRIMARY KEY NOT NULL,
  GRUP int NOT NULL,
  DEPOTUR decimal(20,0) NOT NULL,
  FAZLABULUNDURMAORANI decimal(6,2) NOT NULL
);
CREATE UNIQUE INDEX XUTDEPOFAZLAP ON TDEPOFAZLA(ID);
CONSTRAINT fk_column FOREIGN KEY (DEPOTUR) REFERENCES TDEPOTUR(ID);

Answer 2

问题是我没有正确引用连接列。这解决了问题：

@ManyToOne(fetch = FetchType.LAZY, optional = false)
@JoinColumn(name = "DEPOTUR", referencedColumnName = "ID", nullable = false, columnDefinition = "DECIMAL(20,0)")
@NotNull
private DepoTur depoTur;

Answer 3

您在 DepoFazla 模型中为连接列 ID 指定了两次相同的名称，您必须更改其名称，例如：

@Entity
@Table(schema = "PGMTEMS", name = "TDEPOFAZLA")
public class DepoFazla implements Serializable {
    private static final long serialVersionUID = -2800365387332643658L;

    ...

    @ManyToOne(fetch = FetchType.LAZY, targetEntity = DepoTur.class)
    @JoinColumn(name = "ID_depoTur", insertable = false, updatable = false)
    @NotNull
    private DepoTur depoTur;

    ...

}