org.hibernate.AssertionFailure：テーブルのmedici.personが見つかりません

Question

私はPersonという抽象クラスを作成しています.Personalから2つのConcreteクラスを作成します。従業員と顧客と呼ばれ、JPA 2.1とHibernate 5.3を使用しています。私はWildfly 10を実行します。次のエラーが表示されます。私は多くのことを試みましたが、成功しなかったので、オンラインで助けを見つけることができませんでした、ここで私は見つけることができます。私は3つのクラスも投稿します。ここで

はここで人クラス

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; 
    } 
} 

は、従業員クラスは、ここで

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(); 
    } 
} 

である私が取得していますコンソールのエラーです。

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 

Answer 1

Personクラスからschema = "medici"を削除した場合、コードは機能します（マイナスの多くのタイプミス）。

ここにスキーマを置くことは、実際にはベストプラクティスではありませんので、削除することをお勧めします。プロジェクトで複数のスキーマを使用する場合は、persistence.xmlに2つの異なる永続性単位を設定し、使用しているスキーマに応じて異なるエンティティ・マネージャを呼び出す方がよいでしょう。