ORA-01400: 无法将 null 插入 (TABLE.COLUMN) (休眠)答案

【问题标题】：ORA-01400: Cannot insert null into (TABLE.COLUMN) (Hibernate)ORA-01400: 无法将 null 插入 (TABLE.COLUMN) (休眠)
【发布时间】：2016-08-21 17:53:41
【问题描述】：

我使用的是休眠 4.3，oracle 11：

当我想插入一个员工时，（它与类别有一对多的关系（一个类别有很多员工）），首先我将一个类别插入数据库，然后我尝试将一个员工插入数据库并得到一个例外，实体的代码是hibernate生成的，所以我不知道怎么回事，问题似乎是我插入员工时Hibernate没有插入ID_CAT，我做错了什么？我也尝试过从hibernate生成表，但给了我同样的错误，感谢您的关注，希望您能帮助我！（类别和员工都与实体“企业”相关，但这部分效果很好）

例外：

    abr 27, 2016 11:11:58 AM org.hibernate.engine.jdbc.spi.SqlExceptionHelper logExceptions
WARN: SQL Error: 1400, SQLState: 23000
abr 27, 2016 11:11:58 AM org.hibernate.engine.jdbc.spi.SqlExceptionHelper logExceptions
ERROR: ORA-01400: no se puede realizar una inserción NULL en ("WSPUSER"."TM_EMPLEADOS"."ID_CAT")

abr 27, 2016 11:11:58 AM org.hibernate.engine.jdbc.batch.internal.AbstractBatchImpl release
INFO: HHH000010: On release of batch it still contained JDBC statements
Error org.hibernate.exception.ConstraintViolationException: could not execute statement
Exception in thread "main" org.hibernate.TransactionException: Transaction not successfully started
    at org.hibernate.engine.transaction.spi.AbstractTransactionImpl.commit(AbstractTransactionImpl.java:172)
at tws.hibernate.dao.GenericDAO.endTransaction(GenericDAO.java:59)
at tws.hibernate.dao.GenericDAO.insert(GenericDAO.java:88)
at TwsTestRunner.insertarEmpleado(TwsTestRunner.java:289)
at TwsTestRunner.main(TwsTestRunner.java:27)

实体“员工”：

@Entity
@Table(name = "TM_EMPLEADOS")
public class TmEmpleados implements java.io.Serializable {

    private TmEmpleadosId id;
    private TmEmpresas tmEmpresas;
    private TmCategoria tmCategoria;
    private String nombre;

    public TmEmpleados() {
    }

    public TmEmpleados(TmEmpleadosId id,TmEmpresas tmEmpresas, TmCategoria tmCategoria,String nombre) {
        this.id = id;
        this.tmEmpresas = tmEmpresas;
        this.tmCategoria = tmCategoria;
        this.nombre = nombre;
    }

    @EmbeddedId
    @AttributeOverrides({
            @AttributeOverride(name = "idEmple", column = @Column(name = "ID_EMPLE", nullable = false, length = 25)),
            @AttributeOverride(name = "mand", column = @Column(name = "MAND", nullable = false, length = 3)),
            @AttributeOverride(name = "idEmp", column = @Column(name = "ID_EMP", nullable = false, length = 6)) })
    public TmEmpleadosId getId() {
        return this.id;
    }

    public void setId(TmEmpleadosId id) {
        this.id = id;
    }

    @ManyToOne(fetch = FetchType.LAZY)
    @JoinColumns({
            @JoinColumn(name = "ID_CAT", referencedColumnName = "ID_CAT", nullable = false, insertable = false, updatable = false),
            @JoinColumn(name = "ID_EMP", referencedColumnName = "ID_EMP", nullable = false, insertable = false, updatable = false),
            @JoinColumn(name = "MAND", referencedColumnName = "MAND", nullable = false, insertable = false, updatable = false) })
    public TmCategoria getTmCategoria() {
        return this.tmCategoria;
    }

    public void setTmCategoria(TmCategoria tmCategoria) {
        this.tmCategoria = tmCategoria;
    }

    @ManyToOne(fetch = FetchType.LAZY)
    @JoinColumns({
            @JoinColumn(name = "ID_EMP", referencedColumnName = "ID_EMP", nullable = false, insertable = false, updatable = false),
            @JoinColumn(name = "MAND", referencedColumnName = "MAND", nullable = false, insertable = false, updatable = false) })
    public TmEmpresas getTmEmpresas() {
        return this.tmEmpresas;
    }

    @Column(name = "NOMBRE", nullable = false, length = 50)
    public String getNombre() {
        return this.nombre;
    }

    public void setNombre(String nombre) {
        this.nombre = nombre;
    }
}

实体“类别”：

@Entity
@Table(name = "TM_CATEGORIA")
public class TmCategoria implements java.io.Serializable {

    private TmCategoriaId id;
    private String descripcion;
    private Set<TmEmpleados> tmEmpleadoses = new HashSet<TmEmpleados>(0);

    public TmCategoria() {
    }

    public TmCategoria(TmCategoriaId id, String descripcion) {
        this.id = id;
        this.descripcion = descripcion;
    }

    public TmCategoria(TmCategoriaId id, String descripcion,Set<TmEmpleados> tmEmpleadoses) {
        this.id = id;
        this.descripcion = descripcion;
        this.tmEmpleadoses = tmEmpleadoses;
    }

    @EmbeddedId
    @AttributeOverrides({
            @AttributeOverride(name = "idCat", column = @Column(name = "ID_CAT", nullable = false, length = 3)),
            @AttributeOverride(name = "idEmp", column = @Column(name = "ID_EMP", nullable = false, length = 6)),
            @AttributeOverride(name = "mand", column = @Column(name = "MAND", nullable = false, length = 3)) })
    public TmCategoriaId getId() {
        return this.id;
    }

    public void setId(TmCategoriaId id) {
        this.id = id;
    }

    @Column(name = "DESCRIPCION", nullable = false, length = 50)
    public String getDescripcion() {
        return this.descripcion;
    }

    public void setDescripcion(String descripcion) {
        this.descripcion = descripcion;
    }

    @OneToMany(fetch = FetchType.LAZY, targetEntity = TmEmpleados.class, mappedBy = "tmCategoria")
    public Set<TmEmpleados> getTmEmpleadoses() {
        return this.tmEmpleadoses;
    }

    public void setTmEmpleadoses(Set<TmEmpleados> tmEmpleadoses) {
        this.tmEmpleadoses = tmEmpleadoses;
    }

}

我编辑了休眠配置以显示 sql 和控制台显示：

abr 28, 2016 10:55:39 AM org.hibernate.validator.internal.util.Version <clinit>
INFO: HV000001: Hibernate Validator 5.1.3.Final
        Hibernate: select tmcategori_.ID_CAT, tmcategori_.ID_EMP, tmcategori_.MAND, tmcategori_.DESCRIPCION as DESCRIPCION4_6_ from WSPUSER.TM_CATEGORIA tmcategori_ where tmcategori_.ID_CAT=? and tmcategori_.ID_EMP=? and tmcategori_.MAND=? 
    Hibernate: select tmempresas_.ID_EMP, tmempresas_.MAND, tmempresas_.DESCRIPCION as DESCRIPCION5_25_ from WSPUSER.TM_EMPRESAS tmempresas_ where tmempresas_.ID_EMP=? and tmempresas_.MAND=?
    Hibernate: insert into WSPUSER.TM_EMPLEADOS ( NOMBRE, ID_EMP, ID_EMPLE, MAND) values (?, ?, ?, ?)
abr 28, 2016 10:55:40 AM org.hibernate.engine.jdbc.spi.SqlExceptionHelper logExceptions
WARN: SQL Error: 1400, SQLState: 23000
abr 28, 2016 10:55:40 AM org.hibernate.engine.jdbc.spi.SqlExceptionHelper logExceptions
ERROR: ORA-01400: no se puede realizar una inserción NULL en ("WSPUSER"."TM_EMPLEADOS"."ID_CAT")

abr 28, 2016 10:55:40 AM org.hibernate.engine.jdbc.batch.internal.AbstractBatchImpl release
INFO: HHH000010: On release of batch it still contained JDBC statements
Exception in thread "main" org.hibernate.TransactionException: Transaction not successfully started
    at org.hibernate.engine.transaction.spi.AbstractTransactionImpl.commit(AbstractTransactionImpl.java:172)
    at tws.hibernate.dao.GenericDAO.endTransaction(GenericDAO.java:59)
    at tws.hibernate.dao.GenericDAO.insert(GenericDAO.java:88)

【问题讨论】：

当您插入TmEmpleados 实例时，您的tmCategoria 属性是否被填充？
是的，我在使用构造函数创建实体 tmcategory 时将其添加为属性
但是您的TmCategoria 实例是否填充了id 字段？根据您收到的错误，这似乎是问题所在。您可能会将 new TmCategoria() 传递给您的 TmEmpleados 实例，这将导致后者的所有属性为空。
是的，我调试了应用程序并且 TmCategoriaId 已正确填写...

标签： hibernate jpa jakarta-ee oracle11g persistence

【解决方案1】：

就我而言，解决方案是删除实体中的 insertable="false", updateble="false"。

【讨论】：

【解决方案2】：

您确定已为所有员工填写了“名词”字段吗？

您收到 ConstraintViolationException 异常，根据您的代码，“nombre”字段不应为 null（可为 null = false）。

【讨论】：

是的，我使用了hibernate生成的构造函数并正确填充了它

【解决方案3】：

我解决了修改数据库的问题：我将所有外键列作为引用表的主键的一部分和相关的子列，所以现在hibernate正确生成了sql，可能不是最干净的方式但有效。

【讨论】：