Java ee，JPA 2.1 返回 null

Question

首先很抱歉我的英语不好。我正在尝试从一个表中显示数据库（Postgres）行，并且它总是返回 null。我正在使用 JPA hibernate 在 Java EE 中做项目。

<?xml version="1.0" encoding="UTF-8"?>
<persistence xmlns="http://java.sun.com/xml/ns/persistence" version="2.0">

    <persistence-unit name="mesPU">

        <class>pl.mes.model.Users</class>
        <properties>
            <property name="javax.persistence.jdbc.driver" value="org.postgresql.Driver" />
            <property name="javax.persistence.jdbc.url" value="jdbc:postgresql://localhost:5432/test" />
            <property name="javax.persistence.jdbc.user" value="postgres" />
            <property name="javax.persistence.jdbc.password" value="xxxxxx" />
            <property name="hibernate.show_sql" value="true" />
            <property name="hibernate.hbm2ddl.auto" value="update" />
            <property name="hibernate.connection.url" value="jdbc:postgresql://localhost:5432/test"/>
            <property name="hibernate.connection.driver_class" value="org.postgresql.Driver"/>
            <property name="hibernate.dialect" value="org.hibernate.dialect.PostgreSQL9Dialect"/>
        </properties>
    </persistence-unit>
</persistence>

用户实体

package pl.mes.model;

import javax.persistence.Basic;
import javax.persistence.Column;
import javax.persistence.Entity;
import javax.persistence.Id;

@Entity
public class Users {
    private Integer userId;
    private String firstname;
    private String secondname;
    private String email;

    @Id
    @Column(name = "userId", nullable = false)
    public Integer getUserId() {
        return userId;
    }

    public void setUserId(Integer userId) {
        this.userId = userId;
    }

    @Basic
    @Column(name = "firstname", nullable = true)
    public String getFirstname() {
        return firstname;
    }

    public void setFirstname(String firstname) {
        this.firstname = firstname;
    }

    @Basic
    @Column(name = "secondname", nullable = true)
    public String getSecondname() {
        return secondname;
    }

    public void setSecondname(String secondname) {
        this.secondname = secondname;
    }

    @Basic
    @Column(name = "email", nullable = true)
    public String getEmail() {
        return email;
    }

    public void setEmail(String email) {
        this.email = email;
    }

    @Override
    public boolean equals(Object o) {
        if (this == o) return true;
        if (o == null || getClass() != o.getClass()) return false;

        Users users = (Users) o;

        if (userId != null ? !userId.equals(users.userId) : users.userId != null) return false;
        if (firstname != null ? !firstname.equals(users.firstname) : users.firstname != null) return false;
        if (secondname != null ? !secondname.equals(users.secondname) : users.secondname != null) return false;
        if (email != null ? !email.equals(users.email) : users.email != null) return false;

        return true;
    }

    @Override
    public int hashCode() {
        int result = userId != null ? userId.hashCode() : 0;
        result = 31 * result + (firstname != null ? firstname.hashCode() : 0);
        result = 31 * result + (secondname != null ? secondname.hashCode() : 0);
        result = 31 * result + (email != null ? email.hashCode() : 0);
        return result;
    }
}

当我使用这个时

从用户中选择你

在控制台输出看起来像这样： output

DbSOURCE 如果有人可以帮助我，我将不胜感激:)

Answer 1

您已为属性 javax.persistence.jdbc.url 指定 URL jdbc:postgresql://localhost:5432/test。但是当我查看第二个屏幕截图时，我看不到名为 test 的数据库架构。

我猜，您连接到错误的数据库。也许第二个实例在不同的端口监听？

还请删除 JPA 标准属性已经指定的 Hibernate 特定属性（如 Billy Frost 在 cmets 中所述）。