JPA:如何使用@NamedQuery. Schema.Table指定来自不同DB2模式的表不能是FROM子句的第一个声明

pexxcrt2  于 2022-11-07  发布在  DB2
关注(0)|答案(1)|浏览(139)

我在从Postman发送GET请求来测试我的端点时遇到了一个问题。相同的NamedQueries以前只使用一个Schema就可以在Derby中工作,所以不需要区分。我现在已经改为在本地Docker示例中运行DB2,并且正在使用Maven运行Open-Liberty Framework来创建端点。我收到了下面的错误。我是否需要创建一个带有实体Map的orm.xml文件,或者做一些其他的事情来解决这个问题?如果可能的话,我更愿意在没有更多XML文件的情况下这样做。
Postman :
错误500:Java.lang.空指针异常:无法调用〉“javax.persistence.EntityManager.createNamedQuery(字符串,java.lang.Class)”,因为“this.em”为空
玛文:
[信息] [错误] CWWJP 0015 E:org.eclipse.persistence.jpa.PersistenceProvider持久性提供程序在尝试为jpa单元持久性单元创建容器实体管理器工厂时出错。出现以下错误:异常错误[Eclipse链接-28019](Eclipse持久性服务- 2.7.9.v20210604- 2c 549 e2208):实体管理器设置异常
[INFO]异常描述:部署持久性单元[jpa-unit]失败。请关闭此持久性单元的所有工厂。
[INFO]内部异常:异常错误[Eclipse链接-0](Eclipse持久性服务- 2.7.9.v20210604- 2c 549 e2208):org.eclipse.persistence.exceptions.JPQLException
[INFO]异常描述:编译[SELECT u FROM Sankofa.用户u]时出现问题。
[INFO] [14,29]“Sankofa.Users u”不能是FROM子句的第一个声明。
用户道

package dao;

import java.util.List;
import java.util.concurrent.TimeoutException;

import javax.persistence.EntityManager;
import javax.persistence.PersistenceContext;

import org.hyperledger.fabric.gateway.ContractException;

import javax.enterprise.context.RequestScoped;
import models.*;

@RequestScoped
public class UserDao {

//DB2 Methods
    @PersistenceContext(name = "jpa-unit")
    private EntityManager em;

    public void createUser(Users user){
        em.persist(user);
    }

    public Users readUser(int userID){
        return em.find(Users.class, userID);
    }
    //NEED TO DO set return limit to 20
    public List<Users> readAllUsers(){
        return em.createNamedQuery("Users.findAll", Users.class).getResultList();
    }

    public void updateUser(Users user){
        em.merge(user);
    }

    public void deleteUser(Users userID){
        em.remove(userID);
    }

    public List<Users> findUser(String email){
        return em.createNamedQuery("Users.findUser", Users.class)
            .setParameter("email", email)
            .getResultList(); 
    }

    public void createHistory(History hist){
        em.persist(hist);
    }
    //wait this doesnt do anything? 
    public Users readHistory(int userID){
        return em.find(Users.class, userID);
    }

    public List<History> readAllHistory(){
        return em.createNamedQuery("History.findAll", History.class).getResultList();
    }
}

使用者

package models;

import java.util.HashSet;
import java.util.Objects;
import java.util.Set;

import javax.json.Json;
import javax.json.JsonArrayBuilder;
import javax.json.JsonObject;
import javax.json.JsonObjectBuilder;
import javax.persistence.*;
import java.time.LocalDate;

@Entity
@Table(name = "Users")
@NamedQueries({
    @NamedQuery(name = "Users.findAll", query = "SELECT u FROM Users u"),
    @NamedQuery(name = "Users.findUser", query = "SELECT usr FROM Users usr WHERE usr.email = :email")
})
public class Users {

    private static JsonObjectBuilder builder = Json.createObjectBuilder();

    @GeneratedValue(strategy = GenerationType.AUTO)
    @Id
    @Column(name = "userId")
    private int id;

    @Column(name = "firstName")
    private String firstName;
    @Column(name = "lastName")
    private String lastName;
    @Column(name = "gender")
    private String gender;
    @Column(name = "address")
    private String address;
    @Column(name = "email")
    private String email;
    @Column(name = "password")
    private String password;
    @Column(name = "dateOfBirth")
    private LocalDate dateOfBirth;

Persistence.xml

<?xml version="1.0" encoding="UTF-8"?>
<persistence version="2.2"
    xmlns="http://xmlns.jcp.org/xml/ns/persistence"
    xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
    xsi:schemaLocation="http://xmlns.jcp.org/xml/ns/persistence 
                        http://xmlns.jcp.org/xml/ns/persistence/persistence_2_2.xsd">
    <persistence-unit name="jpa-unit" transaction-type="JTA">
        <mapping-file>orm.xml</mapping-file>

        <properties>
            <!-- Connection Specific -->
            <property name="hibernate.dialect" value="org.hibernate.dialect.DB2Dialect"/>

            <property name="javax.persistence.schema-generation.database.action" value="create" />
            <property name="javax.persistence.schema-generation.create-database-schemas" value="true" />
            <property name="javax.persistence.schema-generation.scripts.action" value="create" />
            <property name="javax.persistence.schema-generation.scripts.create-target" value="create.ddl"/>

        </properties>
    </persistence-unit>
</persistence>

Server.xml

<server description="Obdoblock REST Server">
    <featureManager>
        <feature>jaxrs-2.1</feature>
        <feature>openapi-3.1</feature>
        <feature>jpa-2.2</feature>
        <feature>cdi-2.0</feature>
    </featureManager>

    <httpEndpoint 
        httpPort="${default.http.port}" 
        httpsPort="${default.https.port}"
        id="defaultHttpEndpoint" 
        host="*" 
    />

    <webApplication 
        location="hyperledger-api.war" 
        contextRoot="${app.context.root}"
    />

    <!-- DB2 Library Configuration -->
    <library id="DB2JCCLib">
        <file name="${shared.resource.dir}/jcc-11.5.6.0" />
    </library>

    <dataSource id="DefaultDataSource" jndiName="jdbc/db2">
        <jdbcDriver libraryRef="jdbcLib"/>
        <properties.db2.jcc
            databaseName="testdb"
            serverName="localhost" 
            portNumber="50000"
            user="****" password="****"
        />
    </dataSource>
</server>

版本:

  • 扩展坞:20.10.8,内部版本号3967 b7 d
  • DB2:ibm/db2 docker映像版本11.5.6
  • 美芬:3.8.3
  • Java语言:JDK 14.0.2

如果需要更多的细节,我很乐意提供。谢谢,迪伦

r6vfmomb

r6vfmomb1#

如何使用@NamedQuery指定不同DB2模式中的表
AFAIK,您无法在查询级别配置架构值。在的实体下定义的所有命名查询都应针对同一架构执行。
1.可以通过orm.xmlMap文件在持久性单元级别设置数据库模式:
orm.xml

<entity-mappings ... >
     <persistence-unit-metadata>
         <persistence-unit-defaults>
             <schema>Sankofa</schema>
         </persistence-unit-defaults>
     </persistence-unit-metadata>   
     . . .
 </entity-mappings>

persistence.xml

<persistence ... >
    <persistence-unit name="foo">
        <mapping-file>orm.xml</mapping-file>
    </persistence-unit>
</persistence>

1.对于EclipseLink,您可以配置一个SessionCustomizer类

public class FooSessionCustomizer 
     implements org.eclipse.persistence.config.SessionCustomizer {

     @Override
     public void customize(Session session) throws Exception {
         session.getLogin().setTableQualifier("Sankofa");
     }
 }

persistence.xml

<persistence ... >
    <persistence-unit name="foo">
        <properties>
            <property name="eclipselink.session.customizer" value="foo.customizer.FooSessionCustomizer" />
        </properties>
    </persistence-unit>
</persistence>

1.@Table注解具有一个“schema”元素,用于在实体级别配置架构

@Table(name = "Users", schema = "Sankofa")

相关问题