当我运行程序时，JPA 不会在 MySQL 中创建表

Question

这是我的名为 Class 的实体：

package az.bank.entities;
import java.io.Serializable;
import javax.persistence.Entity;
import javax.persistence.GeneratedValue;
import javax.persistence.GenerationType;
import javax.persistence.Id;
import javax.persistence.Table;

@Entity
@Table (name = "cards")
public class Card implements Serializable{

    @Id
    @GeneratedValue(strategy = GenerationType.IDENTITY)
    private Long id;
    private String cardHolder;
    private String cardNumber;
    private String cardPassword;
    private String expiryYear;
    private String expiryMonth;
    private String cardType;
    private double cardBalance;   
}

这是我的persistance.xml 文件：

<?xml version="1.0" encoding="UTF-8"?>
<persistence version="2.1" 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_1.xsd">
    <persistence-unit name="BankServicePU" transaction-type="JTA">
        <jta-data-source>jdbc/BankService</jta-data-source>
        <class>az.bank.entities.Card</class>
        <exclude-unlisted-classes>true</exclude-unlisted-classes>
        <properties>
            <property name="javax.persistence.jdbc.driver" value="com.mysql.jdbc.Driver" />
            <property name="javax.persistence.jdbc.url" value="jdbc:mysql://localhost:3306/cards" />
            <property name="javax.persistence.jdbc.user" value="root" />
            <property name="javax.persistence.jdbc.password" value="root" />
            <property name="javax.persistence.schema-generation.database.action" value="create"/>
        </properties>
    </persistence-unit>
</persistence>

我创建了名为 jdbc/BankService 的连接池和名为 cards 的 mySQL 方案。但是当我部署和运行程序时，它不会在该方案中创建表。请帮助我在这里做错了什么。

Answer 1

我想在您的情况下，您没有初始化 EntityManagerFactory，这意味着您的 eclipselink 没有收到创建表甚至连接到数据库的命令。

在您的情况下，您可以尝试使用ServletContextListener，它必须在您的web.xml 文件中注册。

快速示例：

web.xml

<?xml version="1.0" encoding="UTF-8"?>
<web-app xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
         xmlns="http://java.sun.com/xml/ns/javaee"
         xmlns:web="http://java.sun.com/xml/ns/javaee/web-app_3_0.xsd"
         xsi:schemaLocation="http://java.sun.com/xml/ns/javaee http://java.sun.com/xml/ns/javaee/web-app_3_0.xsd"
         version="3.0">
    <listener>
        <listener-class>
            com.mberazouski.stackoverflow.AppServletContextListener
        </listener-class>
    </listener>
</web-app>

启动 Servlet 3.0 你可以只使用@WebListener 注解而不是在web.xml 中注册。

AppServletContextListener.java

package com.mberazouski.stackoverflow;

import javax.persistence.EntityManager;
import javax.persistence.EntityManagerFactory;
import javax.persistence.Persistence;
import javax.servlet.ServletContextEvent;
import javax.servlet.ServletContextListener;

public class AppServletContextListener implements ServletContextListener {
    private static EntityManagerFactory emf;

    public void contextInitialized(ServletContextEvent event) {
        emf = Persistence.createEntityManagerFactory("default");
        createEntityManager();
    }

    public void contextDestroyed(ServletContextEvent event) {
        emf.close();
    }

    public static EntityManager createEntityManager() {
        if (emf == null) {
            throw new IllegalStateException("Context is not initialized yet.");
        }

        return emf.createEntityManager();
    }
}

所以你的persistence.xml 是绝对有效的。适应我的域模型文件看起来像：

RESOURCE_LOCAL

您可以通过RESOURCE_LOCAL 配置您的连接。

<?xml version="1.0" encoding="UTF-8"?>
<persistence version="2.0" xmlns="http://java.sun.com/xml/ns/persistence"
             xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
             xsi:schemaLocation="http://java.sun.com/xml/ns/persistence http://java.sun.com/xml/ns/persistence/persistence_2_0.xsd">
    <persistence-unit name="default" transaction-type="RESOURCE_LOCAL">
        <class>com.mberazouski.stackoverflow.domain.Cards</class>
        <properties>
            <property name="javax.persistence.jdbc.driver" value="com.mysql.jdbc.Driver"/>
            <property name="javax.persistence.jdbc.url" value="jdbc:mysql://localhost:3306/rsreu"/>
            <property name="javax.persistence.jdbc.user" value="root"/>
            <property name="javax.persistence.jdbc.password" value="root"/>
            <property name="javax.persistence.schema-generation.database.action" value="create"/>
        </properties>
    </persistence-unit>
</persistence>

jta-数据源

在这种情况下，所有配置都将从您的 tomcat 的 Resource 块中获取：

<?xml version="1.0" encoding="UTF-8"?>
<persistence version="2.0" xmlns="http://java.sun.com/xml/ns/persistence"
             xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
             xsi:schemaLocation="http://java.sun.com/xml/ns/persistence http://java.sun.com/xml/ns/persistence/persistence_2_0.xsd">
    <persistence-unit name="default" transaction-type="JTA">
        <provider>org.eclipse.persistence.jpa.PersistenceProvider</provider>
        <jta-data-source>java:comp/env/jdbc/EclispeLinkDB</jta-data-source>
        <class>com.mberazouski.stackoverflow.domain.Cards</class>
        <properties>
            <property name="javax.persistence.schema-generation.database.action" value="create"/>
        </properties>
    </persistence-unit>
</persistence>

tomcat 启动的输出：

但我也建议你看看spring的方向。使用它的初始化方法，您可以直接从配置文件初始化EntityManagerFactory。

applicationContext.xml

<?xml version="1.0" encoding="UTF-8"?>
<beans xmlns="http://www.springframework.org/schema/beans" xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
       xsi:schemaLocation="http://www.springframework.org/schema/beans http://www.springframework.org/schema/beans/spring-beans-3.0.xsd">

    <bean id="emf" class="org.springframework.orm.jpa.LocalEntityManagerFactoryBean ">
        <property name="persistenceUnitName" value="default"/>
    </bean>
</beans>

希望这会有所帮助。

Answer 2

如果您的数据库已经包含卡表，则只需先将其删除。使用

 <property name="javax.persistence.schema-generation.database.action" value="drop-and-create"/>