org.hibernate.AssertionFailure：找不到表medici.person

时间：2016-06-21 18:11:55

标签： hibernate jpa assertion

我正在创建一个名为Person的抽象类，并且从Person创建两个称为Employee和Customer的Concrete类，我使用JPA 2.1和Hibernate 5.3，但是当我运行Wildfly 10时，我得到以下内容错误。我尝试了很多东西，但没有成功，无法在网上找到任何帮助，mabe在这里我将能够找到。我也会发布这三个课程。

这是Person Class 

package com.medici.general.entity;

import java.util.Date;
import java.util.List;

import javax.persistence.Column;
import javax.persistence.DiscriminatorColumn;
import javax.persistence.DiscriminatorType;
import javax.persistence.Entity;
import javax.persistence.FetchType;
import javax.persistence.GeneratedValue;
import javax.persistence.GenerationType;
import javax.persistence.Id;
import javax.persistence.Inheritance;
import javax.persistence.InheritanceType;
import javax.persistence.JoinColumn;
import javax.persistence.OneToMany;
import javax.persistence.OneToOne;
import javax.persistence.Table;
import javax.persistence.Temporal;
import javax.persistence.TemporalType;
import javax.persistence.Version;

@SuppressWarnings("serial")
@Entity
@Table(name="person", schema="medici")
@Inheritance(strategy=InheritanceType.JOINED)
@DiscriminatorColumn(name="per_type", discriminatorType=DiscriminatorType.INTEGER)
public class Person extends Base {

    @Id
    @GeneratedValue(strategy=GenerationType.IDENTITY)
    @Column(name="per_id", unique=true, nullable=false)
    private Integer id;

    @OneToOne(fetch=FetchType.EAGER)
    @JoinColumn(name="per_situation", nullable=false)
    private Register situation;

    @OneToOne(fetch=FetchType.EAGER)
    @JoinColumn(name="per_type", nullable=false, updatable=false, insertable=false)
    private Register type;

    @OneToMany(mappedBy="person", fetch=FetchType.EAGER)
    @Column(name="per_address", nullable=false)
    private List<Address> addresses;

    @Temporal(value=TemporalType.DATE)
    @Column(name="per_creation_date", nullable=false)
    private Date creationDate;

    @Version
    @Temporal(TemporalType.TIMESTAMP)
    @Column(name="per_last_update", nullable=false)
    private Date lastUpdate;

    @Column(name="per_user", nullable=false)
    private String User;

    public Integer getId() {
        return id;
    }

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

    public Register getSituation() {
        return situation;
    }

    public void setSituation(Register situation) {
        this.situation = situation;
    }

    public Register getType() {
        return type;
    }

    public void setTipo(Register type) {
        this.type = type;
    }

    public List<Address> getAddresses() {
        return addresses;
    }

    public void setAddresses(List<Address> addresses) {
        this.addresses = addresses;
    }

    public Date getCreationDate() {
        return dataCriacao;
    }

    public void setCreationDate(Date creationDate) {
        this.creationDate = creationDate;
    }

    public Date getLastUpdate() {
        return lastUpdate;
    }

    public void setLastUpdate(Date lastUpdate) {
        this.lastUpdate = lastUpdate;
    }

    public String getUser() {
        return user;
    }

    public void setUser(String user) {
        this.user = user;
    }
}

这是员工类

package com.medici.general.entity;

import java.beans.Transient;
import java.util.Date;

import javax.persistence.Column;
import javax.persistence.DiscriminatorValue;
import javax.persistence.Entity;
import javax.persistence.FetchType;
import javax.persistence.ForeignKey;
import javax.persistence.JoinColumn;
import javax.persistence.OneToOne;
import javax.persistence.PrimaryKeyJoinColumn;

@SuppressWarnings("serial")
@Entity
@DiscriminatorValue("1")
@PrimaryKeyJoinColumn(name = "emp_id")
public class Employee extends Person {

    @Column(name="emp_sin", unique=true, nullable=false)
    private Integer sin;

    @Column(name="emp_name", nullable=false)
    private String name;

    @Column(name="emp_lastname", nullable=false)
    private String lastname;

    @Column(name="emp_date_birth", nullable=false)
    private Date dateBirth;

    @Column(name="emp_telephone", nullable=true)
    private Integer dateBirth;

    @Column(name="emp_mobile", nullable=true)
    private Integer mobile;

    @Column(name="emp_email", nullable=true)
    private String email;

    @OneToOne(fetch=FetchType.EAGER)
    @JoinColumn(name="emp_sex", nullable=false, foreignKey=@ForeignKey(name="fk_sex"))
    private Register sex;

    @OneToOne(fetch=FetchType.EAGER)
    @JoinColumn(name="emp_martial_status", nullable=false, foreignKey=@ForeignKey(name="fk_martial_status"))
    private Register martialStatus;

    public Integer getSin() {
        return sin;
    }

    public void setSin(Integer sin) {
        this.sin = sin;
    }

    public String getName() {
        return name;
    }

    public void setName(String name) {
        this.name = name;
    }

    public String getLastName() {
        return lastName;
    }

    public void setLastName(String lastName) {
        this.lastName = lastName;
    }

    public Date getDateBirth() {
        return dateBirth;
    }

    public void setDateBirth(Date dateBirth) {
        this.dateBirth = dateBirth;
    }

    public Integer getTelephone() {
        return telephone;
    }

    public void setTelephone(Integer telephone) {
        this.telephone = telephone;
    }

    public Integer getMobile() {
        return mobile;
    }

    public void setMobile(Integer mobile) {
        this.mobile = mobile;
    }

    public String getEmail() {
        return email;
    }

    public void setEmail(String email) {
        this.email = email;
    }

    public Register getSex() {
        return sex;
    }

    public void setSex(Register sex) {
        this.sex = sex;
    }

    public Register getMartialStatus() {
        return martialStatus;
    }

    public void setMartialStatus(Register martialStatus) {
        this.martialStatus = martialStatus;
    }

    @Transient
    @Override
    public String toString() {

        return this.getSin() + " - " + this.getName() + " " + this.getLastName();
    }
}

以下是我收到的控制台错误。

12:29:06,369 WARN  [org.hibernate.cfg.AnnotationBinder] (ServerService Thread Pool -- 58) HHH000457: Joined inheritance hierarchy [com.medici.general.entity.Person] defined explicit @DiscriminatorColumn.  Legacy Hibernate behavior was to ignore the @DiscriminatorColumn.  However, as part of issue HHH-6911 we now apply the explicit @DiscriminatorColumn.  If you would prefer the legacy behavior, enable the `hibernate.discriminator.ignore_explicit_for_joined` setting (hibernate.discriminator.ignore_explicit_for_joined=true)
12:29:06,778 ERROR [org.hibernate.AssertionFailure] (ServerService Thread Pool -- 58) HHH000099: an assertion failure occured (this may indicate a bug in Hibernate, but is more likely due to unsafe use of the session): org.hibernate.AssertionFailure: Table medici.person not found
12:29:06,779 ERROR [org.jboss.msc.service.fail] (ServerService Thread Pool -- 58) MSC000001: Failed to start service jboss.persistenceunit."Medici.war#mediciPU": org.jboss.msc.service.StartException in service jboss.persistenceunit."Medici.war#mediciPU": javax.persistence.PersistenceException: [PersistenceUnit: mediciPU] Unable to build Hibernate SessionFactory
      at org.jboss.as.jpa.service.PersistenceUnitServiceImpl$1$1.run(PersistenceUnitServiceImpl.java:172)
      at org.jboss.as.jpa.service.PersistenceUnitServiceImpl$1$1.run(PersistenceUnitServiceImpl.java:117)
      at org.wildfly.security.manager.WildFlySecurityManager.doChecked(WildFlySecurityManager.java:667)
      at org.jboss.as.jpa.service.PersistenceUnitServiceImpl$1.run(PersistenceUnitServiceImpl.java:182)
      at java.util.concurrent.ThreadPoolExecutor.runWorker(ThreadPoolExecutor.java:1142)
      at java.util.concurrent.ThreadPoolExecutor$Worker.run(ThreadPoolExecutor.java:617)
      at java.lang.Thread.run(Thread.java:745)
      at org.jboss.threads.JBossThread.run(JBossThread.java:320)
Caused by: javax.persistence.PersistenceException: [PersistenceUnit: mediciPU] Unable to build Hibernate SessionFactory
      at org.hibernate.jpa.boot.internal.EntityManagerFactoryBuilderImpl.persistenceException(EntityManagerFactoryBuilderImpl.java:954)
      at org.hibernate.jpa.boot.internal.EntityManagerFactoryBuilderImpl.build(EntityManagerFactoryBuilderImpl.java:882)
      at org.jboss.as.jpa.hibernate5.TwoPhaseBootstrapImpl.build(TwoPhaseBootstrapImpl.java:44)
      at org.jboss.as.jpa.service.PersistenceUnitServiceImpl$1$1.run(PersistenceUnitServiceImpl.java:154)
      ... 7 more
Caused by: org.hibernate.MappingException: Could not instantiate persister org.hibernate.persister.entity.JoinedSubclassEntityPersister
      at org.hibernate.persister.internal.PersisterFactoryImpl.createEntityPersister(PersisterFactoryImpl.java:112)
      at org.hibernate.persister.internal.PersisterFactoryImpl.createEntityPersister(PersisterFactoryImpl.java:77)
      at org.hibernate.internal.SessionFactoryImpl.<init>(SessionFactoryImpl.java:346)
      at org.hibernate.boot.internal.SessionFactoryBuilderImpl.build(SessionFactoryBuilderImpl.java:444)
      at org.hibernate.jpa.boot.internal.EntityManagerFactoryBuilderImpl.build(EntityManagerFactoryBuilderImpl.java:879)
      ... 9 more
Caused by: org.hibernate.AssertionFailure: Table medici.person not found
      at org.hibernate.persister.entity.AbstractEntityPersister.getTableId(AbstractEntityPersister.java:5107)
      at org.hibernate.persister.entity.JoinedSubclassEntityPersister.<init>(JoinedSubclassEntityPersister.java:433)
      at sun.reflect.NativeConstructorAccessorImpl.newInstance0(Native Method)
      at sun.reflect.NativeConstructorAccessorImpl.newInstance(NativeConstructorAccessorImpl.java:62)
      at sun.reflect.DelegatingConstructorAccessorImpl.newInstance(DelegatingConstructorAccessorImpl.java:45)
      at java.lang.reflect.Constructor.newInstance(Constructor.java:423)
      at org.hibernate.persister.internal.PersisterFactoryImpl.createEntityPersister(PersisterFactoryImpl.java:96)
      ... 13 more

12:29:06,784 ERROR [org.jboss.as.controller.management-operation] (Controller Boot Thread) WFLYCTL0013: Operation ("deploy") failed - address: ([("deployment" => "Medici.war")]) - failure description: {"WFLYCTL0080: Failed services" => {"jboss.persistenceunit.\"Medici.war#mediciPU\"" => "org.jboss.msc.service.StartException in service jboss.persistenceunit.\"Medici.war#mediciPU\": javax.persistence.PersistenceException: [PersistenceUnit: mediciPU] Unable to build Hibernate SessionFactory
    Caused by: javax.persistence.PersistenceException: [PersistenceUnit: mediciPU] Unable to build Hibernate SessionFactory
    Caused by: org.hibernate.MappingException: Could not instantiate persister org.hibernate.persister.entity.JoinedSubclassEntityPersister
    Caused by: org.hibernate.AssertionFailure: Table medici.person not found"}}
12:29:06,871 INFO  [org.jboss.as.server] (ServerService Thread Pool -- 34) WFLYSRV0010: Deployed "Medici.war" (runtime-name : "Medici.war")
12:29:06,874 INFO  [org.jboss.as.controller] (Controller Boot Thread) WFLYCTL0183: Service status report
WFLYCTL0186:   Services which failed to start:      service jboss.persistenceunit."Medici.war#mediciPU": org.jboss.msc.service.StartException in service jboss.persistenceunit."Medici.war#mediciPU": javax.persistence.PersistenceException: [PersistenceUnit: mediciPU] Unable to build Hibernate SessionFactory

1 个答案:

答案 0 :(得分：0)

如果从Person类中删除schema =“medici”，则代码可以正常工作（减去很多拼写错误）。

在这里放置架构并不是最好的做法，所以我建议删除它。如果您的项目使用多个模式，则更好的方法是在persistence.xml中设置两个不同的持久性单元，然后根据您使用的模式调用另一个实体管理器。

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