没有数据库选择JPA持久化

Question

我有连接到我的数据库运行。我可以执行以下没有问题：

Class.forName("com.mysql.jdbc.Driver");
Connection conn = DriverManager.getConnection("jdbc:mysql://localhost:3306/people", "root", "r00t");

PreparedStatement statement = (PreparedStatement) conn.prepareStatement("select * from users");
ResultSet result = statement.executeQuery();

然而，在设置了JPA和持久类之后，我总是得到一个“未选择数据库”错误。看起来我不需要调整我的数据库配置（MySQL连接到Glassfish 3.1），否则上面的代码将无法工作。

正在拨打的电话：

SELECT USERNAME, FIRSTNAME, LASTNAME, PASSWORD, PERMISSION FROM users

我尝试过这个调用直接在MySQL工作台和它不工作。

这一个确实有效：

SELECT USERNAME, FIRSTNAME, LASTNAME, PASSWORD, PERMISSION FROM people.users

我一直在玩游戏，似乎无法在任何地方添加数据库名称（“人”）。以下是我目前掌握的情况：

使用EclipseLink 2.0。十、

JPA实现：禁用库配置

连接：本地MySQL（我已成功连接数据库）

图式：人

从我的servlet：

package com.lowe.samples;

import java.io.IOException;
import java.io.PrintWriter;
import java.sql.Connection;
import java.sql.DriverManager;
import java.sql.ResultSet;
import java.sql.SQLException;

import javax.persistence.EntityManager;
import javax.persistence.PersistenceContext;
import javax.servlet.ServletException;
import javax.servlet.annotation.WebServlet;
import javax.servlet.http.HttpServlet;
import javax.servlet.http.HttpServletRequest;
import javax.servlet.http.HttpServletResponse;

import com.mysql.jdbc.PreparedStatement;

@WebServlet("/TestServlet")
public class TestServlet extends HttpServlet {
    private static final long serialVersionUID = 1L;

    @PersistenceContext
    private EntityManager em;

    public TestServlet() {
        super();
    }

    protected void doGet(HttpServletRequest request, HttpServletResponse response) throws ServletException, IOException {

        PrintWriter out = response.getWriter(); 
        out.println("<h1>DataBase Test:<h1>");

        try {

            Class.forName("com.mysql.jdbc.Driver");
            Connection conn = DriverManager.getConnection("jdbc:mysql://localhost:3306/people", "root", "r00t");

            PreparedStatement statement = (PreparedStatement) conn.prepareStatement("select * from users");
            ResultSet result = statement.executeQuery();

            // prep the table
            out.print("<table border=\"5\">");
            out.print("<tr>");
            out.print("<td>UserName</td>");
            out.print("<td>FirstName</td>");
            out.print("<td>LastName</td>");
            out.print("<td>Password</td>");
            out.print("<td>Permission</td>");
            out.print("</tr>");

            while(result.next()) {
                out.print("<tr>");
                out.print("<td>" + result.getString(1) + "</td>");
                out.print("<td>" + result.getString(2) + "</td>");
                out.print("<td>" + result.getString(3) + "</td>");
                out.print("<td>" + result.getString(4) + "</td>");
                out.print("<td>" + result.getString(5) + "</td>");
                out.print("</tr>");
            }

            out.print("</table>");

            User u = (User)this.em.createNamedQuery("User.findAll").getResultList();
            out.print("User Name: " + u.getFirstName());

        } catch (ClassNotFoundException e) {
            out.print("<h4>" + e.getMessage() + "</h4>");
            e.printStackTrace();
        } catch (SQLException e) {
            out.print("<h4>" + e.getMessage() + "</h4>");
            e.printStackTrace();
        }
    }

    protected void doPost(HttpServletRequest request, HttpServletResponse response) throws ServletException, IOException {
    }

}

我的坚持课：

package com.lowe.samples;

import java.io.Serializable;
import javax.persistence.*;


/**
 * The persistent class for the users database table.
 */
@Entity
@Table(name="users")
@NamedQuery(name="User.findAll", query="SELECT u FROM User u")
public class User implements Serializable {
    private static final long serialVersionUID = 1L;

    @Id
    private String userName;

    private String firstName;

    private String permission;

    private String lastName;

    private String password;

    public User() {
    }

    public String getUserName() {
        return this.userName;
    }

    public void setUserName(String userName) {
        this.userName = userName;
    }

    public String getFirstName() {
        return this.firstName;
    }

    public void setFirstName(String firstName) {
        this.firstName = firstName;
    }

    public String getPermission() {
        return this.permission;
    }

    public void setPermission(String permission) {
        this.permission = permission;
    }

    public String getLastName() {
        return this.lastName;
    }

    public void setLastName(String lastName) {
        this.lastName = lastName;
    }

    public String getPassword() {
        return this.password;
    }

    public void setPassword(String password) {
        this.password = password;
    }

}

坚持。xml

<?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="MyFriends">
        <jta-data-source>jdbc/MySQLDataSource</jta-data-source>
        <class>com.lowe.samples.User</class>
    </persistence-unit>
</persistence>

Answer 1

如果你正在使用hbm。请尝试以下方法

<class name="com.javapapers.data.user"  table ="user"  catalog
="<DBNAME.Schema>">

Answer 2

如果在指定db url时使用jta数据源或使用hibernate属性，请在db ip地址后提及db名称，如jdbc： mysql：//： 3306/。如果在您的实体中不可能这样做，您可以指定表所在的数据库@Table（name="abcd"， catalog="dbname/schema name"）如果您使用来自一台服务器的多个数据库，也可以这样做。

Answer 3

第一段代码是使用JDBC加载数据库驱动程序、连接和执行查询。它工作的事实表明，您的类路径中有MySQL驱动程序，并且您可以使用提供的URL和凭据连接到数据库。这与JPA无关。

你的坚持。xml具有名为“MyFriends”的持久化单元，并配置了一个支持JTA的数据源：

此数据源绑定到JNDI。

根据您收到的错误，连接池配置错误，可能在URL中缺少数据库名称。