Question

当我尝试添加到TDEPOFAZLA表时，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];嵌套异常是org.hibernate.exception.ConstraintViolationException：无法插入：[tr.gov.tcmb.pgmtems.model.DepoFazla]

这是我的 JUnit 测试功能：

    @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，我收到此错误：

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

这是我创建 TDEPOFAZLA 表格的方式：

    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);

这是我创建 TDEPOTUR 表格的方式：

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);

关于我应该做什么的任何想法？

Answer 1

您似乎在 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;

    ...

}

Answer 2

使用FK的属性定义是错误的，因为我看到你使用可更新，可插入到false，你真正想要的是FK对象没有被修改，因为它可能是一些其他实体共用的主表。

然后您可以使用的是

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

使用DETACH，您只会将值保存在第一个表DEPOTUR列中，并且不会更新DepoTur表中的对象

同时在第一张表中添加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 3

问题是我没有正确引用连接列。这解决了这个问题：

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

org.hibernate.exception.ConstraintViolationException：无法插入

3 个答案: