java.lang.NoSuchMethodError：javax.persistence.JoinColumn.foreignKey [英] java.lang.NoSuchMethodError: javax.persistence.JoinColumn.foreignKey

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问题描述

请帮助我解决这个错误。按照我的代码如下：



这是我的班级/实体Person： 

 包实体; 
 
 import java.util.Calendar; 
 import java.util.List; 
 
 import javax.persistence.CascadeType; 
 import javax.persistence.Column; 
 import javax.persistence.Entity; 
 import javax.persistence.FetchType; 
 import javax.persistence.GeneratedValue; 
 import javax.persistence.Id; 
 import javax.persistence.OneToMany; 
 import javax.persistence.Temporal; 
 import javax.persistence.TemporalType; 
 
 @Entity 
公共类人员{
 
 @Id 
 @GeneratedValue 
私人长ID; 
 
 @Column 
私人字符串名称; 
 
 @Temporal（TemporalType.DATE）
私人日历dateBirth; 
 
 @Column 
私人字符串电子邮件; 
 
 // MappedBy：实例的名称，表示一个人属于谁，即一种体验。 
 //应该说哪个人是关联的
 // TargetEntity：通知我们正在关联的实体
 // FetchType.Lazy：被选为表现
 // Cascade：ALL以改变所有关系。 
 @OneToMany（mappedBy =person，targetEntity = Experience.class，fetch = FetchType.LAZY，cascade = CascadeType.ALL）
 private List< Experience>经验; 
 
 // Getters和Setters 
 public long getId（）{
 return id; 
} 
 
 public void setId（long id）{
 this.id = id; 
} 
 
 public String getName（）{
 return name; 
} 
 
 public void setName（String name）{
 this.name = name; 
} 
 
 public Calendar getDateBirth（）{
 return dateBirth; 
} 
 
 public void setDateBirth（Calendar dateBirth）{
 this.dateBirth = dateBirth; 
} 
 
 public String getEmail（）{
 return email; 
} 
 
 public void setEmail（String email）{
 this.email = email; 
} 
 
公开列表<体验> getExperiences（）{
返回体验; 
} 
 
 public void setExperiences（List< Experience> experience）{
 this.experiences = experience; 
} 
 
}

这是我的班级/实体体验：

 包实体; 
 
 import javax.persistence.Column; 
 import javax.persistence.Entity; 
 import javax.persistence.GeneratedValue; 
 import javax.persistence.Id; 
 import javax.persistence.JoinColumn; 
 import javax.persistence.ManyToOne; 
 
 @实体
公共类体验{
 
 @Id 
 @GeneratedValue 
私人长ID; 
 
 @ManyToOne 
 @JoinColumn（name =idPerson，referencedColumnName =id）
 private Person person; 
 
 @Column 
私人字符串公司; 
 
 @Column 
私人字符串活动; 
 
 // Getters和setters 
 
 public long getId（）{
 return id; 
} 
 
 public void setId（long id）{
 this.id = id; 
} 
 
 public Person getPerson（）{
 return person; 
} 
 
 public void setPerson（Person person）{
 this.person = person; 
} 
 
 public String getCompany（）{
 return company; 
} 
 
 public void setCompany（String company）{
 this.company = company; 
} 
 
 public String getActivities（）{
 return activities; 
} 
 
 public void setActivities（String activities）{
 this.activities = activities; 
} 
 
}

我的persistence.xml：

 <？xml version =1.0encoding =UTF-8？> 
 
< persistence version =1.0xmlns =http://java.sun.com/xml/ns/persistence> 
 
< persistence-unit name =professionalmanagementDB> 
 
 <！ -   -  JPA实现 - > 
< provider> org.hibernate.ejb.HibernatePersistence< / provider> 
 
<！ - 实体 - > 
< class> entities.Person< / class> 
< class> entities.Experience< / class> 
 
<属性> 
< property name =hibernate.dialectvalue =org.hibernate.dialect.MySQLDialect/> 
< property name =hibernate.hbm2ddl.autovalue =update/> 
< property name =hibernate.show_sqlvalue =true/> 
< property name =hibernate.format_sqlvalue =true/> 
< property name =javax.persistence.jdbc.drivervalue =com.mysql.jdbc.Driver/> 
< property name =javax.persistence.jdbc.urlvalue =jdbc：mysql：// localhost / professionalmanagementDB/> 
< property name =javax.persistence.jdbc.uservalue =root/> 
< property name =javax.persistence.jdbc.passwordvalue =/> 
< / properties> 
 
< / persistence-unit> 
 
< /持久性>

所以，我试图用这段代码来进行查询：

  EntityManagerFactory factory = Persistence.createEntityManagerFactory（professionalmanagementDB）; 
 EntityManager manager = factory.createEntityManager（）; 
 
 String jpql =select p; 
 jpql + =from Person; 
 
查询查询= manager.createQuery（jpql）; 
 
列表< Person> list = query.getResultList（）;

但总会发生这样的错误：

  java.lang.NoSuchMethodError：javax.persistence.JoinColumn.foreignKey（）Ljavax / persistence / ForeignKey; 
 org.hibernate.cfg.AnnotationBinder.bindManyToOne（AnnotationBinder.java:2881）
 org.hibernate.cfg.AnnotationBinder.processElementAnnotations（AnnotationBinder.java:1795）
 org.hibernate.cfg。 AnnotationBinder.processIdPropertiesIfNotAlready（AnnotationBinder.java:963）
 org.hibernate.cfg.AnnotationBinder.bindClass（AnnotationBinder.java:796）
 $ org.hibernate.cfg.Configuration MetadataSourceQueue.processAnnotatedClassesQueue（Configuration.java： 3790）
 $ org.hibernate.cfg.Configuration MetadataSourceQueue.processMetadata（Configuration.java:3744）
 org.hibernate.cfg.Configuration.secondPassCompile（Configuration.java:1410）
组织。 hibernate.cfg.Configuration.buildSessionFactory（Configuration.java:1844）
 org.hibernate.jpa.boot.internal.EntityManagerFactoryBuilderImpl $ 4.perform（EntityManagerFactoryBuilderImpl.java:850）
 org.hibernate.jpa.boot .internal.EntityManagerFactoryBuilderImpl $ 4.perform（EntityManagerFactoryBuilderImpl.java:843）
 org.hibernate.boot.registry.cl assloading.internal.ClassLoaderServiceImpl.withTccl（ClassLoaderServiceImpl.java:397）
 org.hibernate.jpa.boot.internal.EntityManagerFactoryBuilderImpl.build（EntityManagerFactoryBuilderImpl.java:842）
 org.hibernate.jpa.HibernatePersistenceProvider。的createEntityManagerFactory（HibernatePersistenceProvider.java:73）
 org.hibernate.ejb.HibernatePersistence.createEntityManagerFactory（HibernatePersistence.java:54）
 javax.persistence.Persistence.createEntityManagerFactory（Persistence.java:63）
 javax.persistence.Persistence.createEntityManagerFactory（Persistence.java:47）
 dao.PersonDAO.readAll（PersonDAO.java:65）
 manager.PersonManager.getPersons（PersonManager.java:17）
 org.apache.jsp.HomeScreen_jsp._jspService（HomeScreen_jsp.java:87）
 org.apache.jasper.runtime.HttpJspBase.service（HttpJspBase.java:70）
 javax.servlet.http.HttpServlet。服务（HttpServlet.java:727）
 org.apache.jasper.servlet.JspServletWrapper.service（JspServletWrapper.java:432）
 org.apac he.jasper.servlet.JspServlet.serviceJspFile（JspServlet.java:390）
 org.apache.jasper.servlet.JspServlet.service（JspServlet.java:334）
 javax.servlet.http.HttpServlet。服务（HttpServlet.java:727）
 org.apache.tomcat.websocket.server.WsFilter.doFilter（WsFilter.java:52）

我的maven依赖关系如下：

pre code $ javaee-api 7.0
hibernate-jpa-2.0-api：1.0.1.Final
hibernate-entitymanager：4.3.4.Final
mysql-connector-java：5.1.29


解决方案

我不知道为什么Hibernate想要 foreignKey 上的 @JoinColumn 注释属性，但它已添加到JPA 2.1中。使用JPA 2.1 API依赖项而不是2.0。


Please help me with this error. Follow my code below:

This my class/entity Person:

package entities;

import java.util.Calendar;
import java.util.List;

import javax.persistence.CascadeType;
import javax.persistence.Column;
import javax.persistence.Entity;
import javax.persistence.FetchType;
import javax.persistence.GeneratedValue;
import javax.persistence.Id;
import javax.persistence.OneToMany;
import javax.persistence.Temporal;
import javax.persistence.TemporalType;

@Entity
public class Person {

@Id
@GeneratedValue
private long id;

@Column
private String name;

@Temporal(TemporalType.DATE)
private Calendar dateBirth;

@Column
private String email;

// MappedBy: Name of the instance that will indicate to whom One belongs, that is, an experience.
//           It should say which person is associated
// TargetEntity: Informs the entity we are associating
// FetchType.Lazy: Was chosen for performance
// Cascade: ALL to allow changes in all relationships.
 @OneToMany(mappedBy = "person", targetEntity = Experience.class, fetch = FetchType.LAZY, cascade = CascadeType.ALL)
 private List<Experience> experiences;

// Getters and setters
public long getId() {
    return id;
}

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

public String getName() {
    return name;
}

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

public Calendar getDateBirth() {
    return dateBirth;
}

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

public String getEmail() {
    return email;
}

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

public List<Experience> getExperiences() {
    return experiences;
}

public void setExperiences(List<Experience> experiences) {
    this.experiences = experiences;
}

}

This is my class/entity Experience:

package entities;

import javax.persistence.Column;
import javax.persistence.Entity;
import javax.persistence.GeneratedValue;
import javax.persistence.Id;
import javax.persistence.JoinColumn;
import javax.persistence.ManyToOne;

@Entity
public class Experience {

@Id
@GeneratedValue
private long id;

@ManyToOne
@JoinColumn(name="idPerson", referencedColumnName="id")
private Person person;

@Column
private String company;

@Column
private String activities;

// Getters and setters

public long getId() {
    return id;
}

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

public Person getPerson() {
    return person;
}

public void setPerson(Person person) {
    this.person = person;
}

public String getCompany() {
    return company;
}

public void setCompany(String company) {
    this.company = company;
}

public String getActivities() {
    return activities;
}

public void setActivities(String activities) {
    this.activities = activities;
}

}

My persistence.xml:

<?xml version="1.0" encoding="UTF-8"?>

<persistence version="1.0" xmlns="http://java.sun.com/xml/ns/persistence"> 

<persistence-unit name="professionalmanagementDB">

    <!-- JPA implementation -->
    <provider>org.hibernate.ejb.HibernatePersistence</provider>

    <!-- Entities --> 
    <class>entities.Person</class>
    <class>entities.Experience</class>

    <properties> 
        <property name="hibernate.dialect"               value="org.hibernate.dialect.MySQLDialect" /> 
        <property name="hibernate.hbm2ddl.auto"          value="update" /> 
        <property name="hibernate.show_sql"              value="true" /> 
        <property name="hibernate.format_sql"            value="true" />  
        <property name="javax.persistence.jdbc.driver"   value="com.mysql.jdbc.Driver" /> 
        <property name="javax.persistence.jdbc.url"      value="jdbc:mysql://localhost/professionalmanagementDB" /> 
        <property name="javax.persistence.jdbc.user"     value="root" /> 
        <property name="javax.persistence.jdbc.password" value="" /> 
    </properties>

</persistence-unit>

</persistence>

So, I'm trying to use this code to make a query:

EntityManagerFactory factory = Persistence.createEntityManagerFactory("professionalmanagementDB");
EntityManager manager = factory.createEntityManager();

String jpql = "select p ";
jpql += "from Person";

Query query = manager.createQuery(jpql);

List<Person> list = query.getResultList();

But always happen this error:

java.lang.NoSuchMethodError: javax.persistence.JoinColumn.foreignKey()Ljavax/persistence/ForeignKey;
org.hibernate.cfg.AnnotationBinder.bindManyToOne(AnnotationBinder.java:2881)
org.hibernate.cfg.AnnotationBinder.processElementAnnotations(AnnotationBinder.java:1795)
org.hibernate.cfg.AnnotationBinder.processIdPropertiesIfNotAlready(AnnotationBinder.java:963)
org.hibernate.cfg.AnnotationBinder.bindClass(AnnotationBinder.java:796)
org.hibernate.cfg.Configuration$MetadataSourceQueue.processAnnotatedClassesQueue(Configuration.java:3790)
org.hibernate.cfg.Configuration$MetadataSourceQueue.processMetadata(Configuration.java:3744)
org.hibernate.cfg.Configuration.secondPassCompile(Configuration.java:1410)
org.hibernate.cfg.Configuration.buildSessionFactory(Configuration.java:1844)
org.hibernate.jpa.boot.internal.EntityManagerFactoryBuilderImpl$4.perform(EntityManagerFactoryBuilderImpl.java:850)
org.hibernate.jpa.boot.internal.EntityManagerFactoryBuilderImpl$4.perform(EntityManagerFactoryBuilderImpl.java:843)
org.hibernate.boot.registry.classloading.internal.ClassLoaderServiceImpl.withTccl(ClassLoaderServiceImpl.java:397)
org.hibernate.jpa.boot.internal.EntityManagerFactoryBuilderImpl.build(EntityManagerFactoryBuilderImpl.java:842)
org.hibernate.jpa.HibernatePersistenceProvider.createEntityManagerFactory(HibernatePersistenceProvider.java:73)
org.hibernate.ejb.HibernatePersistence.createEntityManagerFactory(HibernatePersistence.java:54)
javax.persistence.Persistence.createEntityManagerFactory(Persistence.java:63)
javax.persistence.Persistence.createEntityManagerFactory(Persistence.java:47)
dao.PersonDAO.readAll(PersonDAO.java:65)
manager.PersonManager.getPersons(PersonManager.java:17)
org.apache.jsp.HomeScreen_jsp._jspService(HomeScreen_jsp.java:87)
org.apache.jasper.runtime.HttpJspBase.service(HttpJspBase.java:70)
javax.servlet.http.HttpServlet.service(HttpServlet.java:727)
org.apache.jasper.servlet.JspServletWrapper.service(JspServletWrapper.java:432)
org.apache.jasper.servlet.JspServlet.serviceJspFile(JspServlet.java:390)
org.apache.jasper.servlet.JspServlet.service(JspServlet.java:334)
javax.servlet.http.HttpServlet.service(HttpServlet.java:727)
org.apache.tomcat.websocket.server.WsFilter.doFilter(WsFilter.java:52)

My maven dependencies is as follows:

javaee-api:7.0
hibernate-jpa-2.0-api:1.0.1.Final
hibernate-entitymanager:4.3.4.Final
mysql-connector-java:5.1.29

解决方案

I don't know why Hibernate is wanting that foreignKey attribute on the @JoinColumn annotation, but it was added in JPA 2.1. Use the JPA 2.1 API dependency instead of the 2.0.

这篇关于java.lang.NoSuchMethodError：javax.persistence.JoinColumn.foreignKey的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！

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