Question

我正在使用hibernate 4.3，oracle 11：

当我想插入一个员工时，（它与一个类别有一个关系（一个类别有很多员工）），首先我将一个类别插入到db，然后我尝试将一个员工插入到db中并获取例外，实体的代码是由hibernate生成的，所以我不知道什么是错的，问题似乎是Hibernate在我插入时没有插入ID_CAT员工，我做错了什么？我也尝试从hibernate生成表但是给了我同样的错误，感谢您的关注，我希望您能帮助我！（类别和员工都与实体相关＆＃34;企业＆＃34;，但这部分运作良好）

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)

ENTITY＆＃34;员工＆＃34;：

@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;
    }
}

ENTITY＆＃34; CATEGORY＆＃34;：

@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hibernate配置来显示sql，控制台显示了这个：

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)

Answer 1

你确定你已经填充了字段＆＃34; nombre＆＃34;对于所有员工？

您将获得ConstraintViolationException，并根据您的代码，＆＃34; nombre＆＃34;字段不应为null（nullable = false）。

Answer 2

我解决了修改数据库的问题：我把所有外键列都作为引用表的主键的一部分，子列也是相关的，所以现在hibernate正确生成sql，也许不是最干净的方式但是有效。< / p>

Answer 3

在我的情况下，解决方法是删除实体中的insertable =“ false”和updateble =“ false”。

ORA-01400：无法插入null（TABLE.COLUMN）（Hibernate）

3 个答案: