Java org.hibernate.exception.ConstraintViolationException:无法插入

当我尝试添加到

TDEPOFAZLA

表中时，出现以下错误：

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

这里是DepoTur.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
}

@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

调试JUnit测试时，出现以下错误：

错误：db2sql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，我发现我试图插入NULL，但我无法确定在哪里添加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);

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);

这就是我创建TDEPOTUR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);

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);

---

有什么想法吗？

在您的

DepoFazla

模型中，您似乎两次为联接列Id指定了相同的名称。您必须更改其名称使用，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;

    ...

}

---

您对FK的属性定义是错误的，正如我看到您使用可更新、可插入到false一样，您真正想要的是FK对象不被修改，因为它可能是一些其他实体的公共主表

那么你能用的就是这个

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

使用DETACH，您将仅在第一个表DEPOTUR列中保存值，并且不会更新DEPOTUR表中的对象

还可以在第一个表中添加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);

---

问题是我没有正确引用联接列。这就解决了问题：

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

---

我尝试过在数据库和代码中更改ID列名。但是，我仍然得到相同的错误。@supaplexy您能在准备对象并持久化它的地方添加一段Java代码吗？您是否为所有属性定义了getter/setter？

TDEPOFAZLA

有四个不可为空的列，而insert语句提供的值较少。表定义中没有默认的

ID

，并且

DEPOTUR

完全缺失。为什么要用MySQL>@cralfaro标记这个问题？我已经为所有属性定义了getter/setter，并相应地更新了问题。@supalexy你能试试我在回答中添加的内容吗？更改映射并创建FK，如果仍然失败，可能我需要您发送一段代码，在其中准备对象并做出承诺感谢您的响应。我已将我的答案添加到这个问题上。