当我 运行 程序 JPA 不在 MySQL 中创建 table

When I run program JPA does not create table in MySQL

这是我的实体 Class:

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 table。请帮助我在这里做错了什么。

如果您的数据库中已经包含卡table,那么先将其删除。使用

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

我想在你的情况下你没有初始化 EntityManagerFactory 这意味着你的 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,它必须与您的 web.xml 文件一起注册。

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.Whosebug.AppServletContextListener
        </listener-class>
    </listener>
</web-app>

Starting Servlet 3.0 you can just use @WebListener annotation instead registration in web.xml.

AppServletContextListener.java

package com.mberazouski.Whosebug;

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如下所示:

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.Whosebug.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:

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.Whosebug.domain.Cards</class>
        <properties>
            <property name="javax.persistence.schema-generation.database.action" value="create"/>
        </properties>
    </persistence-unit>
</persistence>

tomcat 开始的输出:

不过我也建议你看看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。