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

1 周，4 日 Questions & Answers 45

当从邮递员发送GET请求以测试我的端点时，我遇到了一个问题。相同名称的查询以前在Derby中只使用一个模式，因此无需进行区分。我现在已经改为在本地Docker实例中运行DB2，并使用Maven运行OpenLiberty框架来创建端点。我收到以下错误。我需要创建一个orm吗。使用实体映射的xml文件，或者执行其他操作来解决此问题？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

邮递员：

Error 500: java.lang.NullPointerException: Cannot invoke >"javax.persistence.EntityManager.createNamedQuery(String, java.lang.Class)" because "this.em" is null

马文：

[INFO] [ERROR ] CWWJP0015E: An error occurred in the org.eclipse.persistence.jpa.PersistenceProvider persistence provider when it attempted to create the container entity manager factory for the jpa-unit persistence unit. The following error occurred: Exception [EclipseLink-28019] (Eclipse Persistence Services - 2.7.9.v20210604-2c549e2208): org.eclipse.persistence.exceptions.EntityManagerSetupException

[INFO] Exception Description: Deployment of PersistenceUnit [jpa-unit] failed. Close all factories for this PersistenceUnit.
[INFO] Internal Exception: Exception [EclipseLink-0] (Eclipse Persistence Services - 2.7.9.v20210604-2c549e2208): org.eclipse.persistence.exceptions.JPQLException
[INFO] Exception Description: Problem compiling [SELECT u FROM Sankofa.Users u].
[INFO] [14, 29] 'Sankofa.Users u' cannot be the first declaration of the FROM clause.

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;

坚持。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>

服务器。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：

码头工人：20.10.8，建造3967b7d
DB2:ibm/DB2 docker映像版本11.5.6
马文：3.8.3
Java:JDK14.0.2

如果需要更多的细节，我很乐意提供。谢谢，迪伦

# 1 楼答案

How to specify tables from different DB2 schemas using @NamedQuery

好的，您不能在查询级别配置模式值。在上的实体下定义的所有命名查询都将针对同一架构执行

可以通过orm在持久性单元级别设置数据库模式。xml映射文件：

奥姆。xml

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

坚持。xml

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

对于EclipseLink，可以配置SessionCustomizer类

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

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

坚持。xml

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

@Table注释有一个“schema”元素，用于在实体级别配置模式
```
@Table(name = "Users", schema = "Sankofa")
```

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

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