JPA 无法创建（持久化）EJB

Question

现在两天试图解决这个问题。问题似乎来自 DAO class.

Caused by: projet.helpdesk.dao.DAOException: Exception [EclipseLink-4002] (Eclipse Persistence Services - 2.6.1.v20150605-31e8258): org.eclipse.persistence.exceptions.DatabaseException
Internal Exception: java.sql.SQLSyntaxErrorException: ORA-02289: sequence does not exist

Error Code: 2289
Call: SELECT SEQ_GEN_IDENTITY.NEXTVAL FROM DUAL
Query: ValueReadQuery(sql="SELECT SEQ_GEN_IDENTITY.NEXTVAL FROM DUAL")
    at projet.helpdesk.dao.UserDao.creer(UserDao.java:25)

这是实体：

    package projet.helpdesk.beans;

import java.sql.Timestamp;
import javax.persistence.*;
@Entity
@Table(name = "Users")
public class Utilisateur {
    @Column(name = "nom")
    private String nom;
    @Column(name = "prenom")
    private String prenom;
    @Column(name = "email")
    private String email;
    @Column(name = "departement")
    private String dept;
    @Column(name = "poste")
    private String poste;
    @Column(name = "agence")
    private String agence;
    @Column(name = "mdp")
    private String mdp;
    @Column(name = "type")
    private String type;
    @Column(name = "date_inscr")
    private Timestamp date_inscr;

    @Id
    @GeneratedValue( strategy = GenerationType.IDENTITY )
    @Column(name = "id_emp")
    private int idemp;
    public String getNom() {
        return nom;
    }
    public void setNom(String nom) {
        this.nom = nom;
    }
    public String getPrenom() {
        return prenom;
    }
    public void setPrenom(String prenom) {
        this.prenom = prenom;
    }
    public String getEmail() {
        return email;
    }
    public void setEmail(String email) {
        this.email = email;
    }
    public String getDept() {
        return dept;
    }
    public void setDept(String dept) {
        this.dept = dept;
    }
    public String getPoste() {
        return poste;
    }
    public void setPoste(String poste) {
        this.poste = poste;
    }
    public String getAgence() {
        return agence;
    }
    public void setAgence(String agence) {
        this.agence = agence;
    }
    public int getIdemp() {
        return idemp;
    }
    public void setIdemp(int id) {
        this.idemp = id;
    }
    public String getMdp() {
        return mdp;
    }
    public void setMdp(String mdp) {
        this.mdp = mdp;
    }
    public String getType() {
        return type;
    }
    public void setType(String type) {
        this.type = type;
    }
    public Timestamp getDate_inscr() {
        return date_inscr;
    }
    public void setDate_inscr(Timestamp date_inscr) {
        this.date_inscr = date_inscr;
    }

}

编辑： 执行查询时发生错误。这是堆栈跟踪：

Caused by: java.lang.IllegalArgumentException: An exception occurred while creating a query in EntityManager: 
Exception Description: Problem compiling [SELECT u FROM Users u WHERE u.email=:email]. 
[14, 19] The abstract schema type 'Users' is unknown.
[28, 35] The state field path 'u.email' cannot be resolved to a valid type.
    at org.eclipse.persistence.internal.jpa.EntityManagerImpl.createQuery(EntityManagerImpl.java:1616)
    at com.sun.enterprise.container.common.impl.EntityManagerWrapper.createQuery(EntityManagerWrapper.java:456)
    at projet.helpdesk.dao.UserDao.trouver(UserDao.java:32)

错误来自方法"trouver"

@Stateless
public class UserDao {
    private static final String JPQL_SELECT_PAR_EMAIL = "SELECT u FROM Users u WHERE u.email=:email";
    private static final String PARAM_EMAIL           = "email";

这是方法"trouver"

 public Utilisateur trouver( String email ) throws DAOException {
        Utilisateur utilisateur = null;
        Query requete = em.createQuery( JPQL_SELECT_PAR_EMAIL );
        requete.setParameter( PARAM_EMAIL, email );
        try {
            utilisateur = (Utilisateur) requete.getSingleResult();
        } catch ( NoResultException e ) {
            return null;
        } catch ( Exception e ) {
            throw new DAOException( e );
        }
        return utilisateur;
    }

知道 table 用户已声明。这是bean Utilisateur。

@Entity
@Table(name = "Users")
public class Utilisateur {...

Answer 1

消息清楚地说明了问题所在：

Internal Exception: java.sql.SQLSyntaxErrorException: ORA-02289: sequence does not exist

sql="SELECT SEQ_GEN_IDENTITY.NEXTVAL FROM DUAL")

代码正在尝试从名为 "SEQ_GEN_IDENTITY" 的数据库序列中读取，但这个序列不存在。

我不确定为什么会这样，你的代码中有这个：

@GeneratedValue( strategy = GenerationType.IDENTITY )

这应该告诉 JPA 它应该使用数据库标识列来获取它想要保留的对象的 ID。

如果您没有特定理由使用 GenerationType.IDENTITY，您应该将其更改为 GenerationType.SEQUENCE。

为此，您必须将 class 更改为如下所示：

@Id
@GeneratedValue( strategy = GenerationType.SEQUENCE )
@Column(name = "id_emp")
private int idemp;

如果您使用 EclipseLink（默认），您必须创建一个名为 "seq_gen_sequence" 的数据库序列。如果您使用的是 Hibernate，则必须创建一个名为 "hibernate_sequence".

的数据库序列

另请参阅：

what is the use of annotations @Id and @GeneratedValue(strategy = GenerationType.IDENTITY)? Why the generationtype is identity?
@GeneratedValue(strategy=“IDENTITY”) 与@GeneratedValue(strategy=“SEQUENCE”)
How to choose the id generation strategy when using JPA and Hibernate

JPA 无法创建（持久化）EJB

JPA is'nt able to create ( persist ) an EJB

ejb

jpa

glassfish

jakarta-ee