ORA-01400: Cannot insert null into (TABLE.COLUMN) (Hibernate)

Question

I'm using hibernate 4.3, oracle 11:

When I want to insert an employee, (it have a relationship one to many with category (one category have many employees)), first I Insert a category to db, then I try to insert an employee to the db and get an exception, the code of the entities was generated by hibernate, so I don't know what's wrong, The problem seems to be that Hibernate is not inserting ID_CAT when I'm inserting an employee, what im doing wrong? I also tried to generate tables from hibernate but gives me the same error, thanks for your attention and I hope you can help me! (category and employees are both related with entity "enterprise", but this part works well)

Exception:

    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 "EMPLOYEES":

@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 "CATEGORY":

@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;
    }

}

I edited hibernate config to show sql and console showed this:

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

In my case, the solucion was removing insertable="false", updateble="false" in the entity.

Answer 2

I solved the issue modifying the database: I made all foreign key columns part of the primary key of the referenced table and child columns related too, so now hibernate properly generate the sql, maybe is not the cleanest way but works.

Answer 3

Are you sure that you have populated field "nombre" for all the employees?

You are getting a ConstraintViolationException and according to your code, "nombre" field should not be null (nullable = false).