When I run program JPA does not create table in MySQL

问题

Here is my entity named 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;   
}

And here is my persistance.xml file:

<?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>

I have created connection pool named jdbc/BankService and mySQL scheme named cards. But when I deploy and run program it does not create table in that scheme. Please help what have I done wrong here.

回答1:

I suppose that in your case you don't initialize the EntityManagerFactory which means that your eclipselink does not receive a command to create a tables and even connect to the DB.

In your case you can try to use ServletContextListener which have to be regestered withing your web.xml file.

Quick example:

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>

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

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();
    }
}

So your persistence.xml is absolutely valid. Adapted to my domain model file looks like:

RESOURCE_LOCAL

You can configure your connection through the RESOURCE_LOCAL.

<?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-data-source

In this case all configurations will be taken from Resource block from your tomcat:

<?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>

Output of the tomcat start:

But I also suggest you to look into direction of spring. Using it initializer approach you can initialize EntityManagerFactory directly from configuration file.

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>

Hope this will help.

回答2:

If your DB already contains card table then just drop it first. Use

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

来源：https://stackoverflow.com/questions/46618802/when-i-run-program-jpa-does-not-create-table-in-mysql