java - Spring JDBC 中 Hibernate 模型的等价物是什么

Question

我正在尝试修改此this使用 Spring Security 和 jdbc 创建用户并以权限登录的示例我现在已经思考并搜索了如何将示例中的一些模型转换为 JDBC。所以我问 JDBC 中的以下 Hibernate 模型的等价物是什么

import org.hibernate.validator.constraints.NotEmpty;

@Entity
@Table(name="APP_USER")
public class User implements Serializable{

    @Id @GeneratedValue(strategy=GenerationType.IDENTITY)
    private Integer id;

    @NotEmpty
    @Column(name="SSO_ID", unique=true, nullable=false)
    private String ssoId;

    @NotEmpty
    @Column(name="PASSWORD", nullable=false)
    private String password;

    @NotEmpty
    @Column(name="FIRST_NAME", nullable=false)
    private String firstName;

    @NotEmpty
    @Column(name="LAST_NAME", nullable=false)
    private String lastName;

    @NotEmpty
    @Column(name="EMAIL", nullable=false)
    private String email;

    @NotEmpty
    @ManyToMany(fetch = FetchType.LAZY)
    @JoinTable(name = "APP_USER_USER_PROFILE", 
             joinColumns = { @JoinColumn(name = "USER_ID") }, 
             inverseJoinColumns = { @JoinColumn(name = "USER_PROFILE_ID") })
    private Set<UserProfile> userProfiles = new HashSet<UserProfile>();
    public Integer getId() {
        return id;
    }
    public void setId(Integer id) {
        this.id = id;
    }
    public String getSsoId() {
        return ssoId;
    }
    public void setSsoId(String ssoId) {
        this.ssoId = ssoId;
    }
    public String getPassword() {
        return password;
    }
    public void setPassword(String password) {
        this.password = password;
    }
    public String getFirstName() {
        return firstName;
    }
    public void setFirstName(String firstName) {
        this.firstName = firstName;
    }
    public String getLastName() {
        return lastName;
    }
    public void setLastName(String lastName) {
        this.lastName = lastName;
    }
    public String getEmail() {
        return email;
    }
    public void setEmail(String email) {
        this.email = email;
    }
    public Set<UserProfile> getUserProfiles() {
        return userProfiles;
    }
    public void setUserProfiles(Set<UserProfile> userProfiles) {
        this.userProfiles = userProfiles;
    }
    @Override
    public int hashCode() {
        final int prime = 31;
        int result = 1;
        result = prime * result + ((id == null) ? 0 : id.hashCode());
        result = prime * result + ((ssoId == null) ? 0 : ssoId.hashCode());
        return result;
    }
    @Override
    public boolean equals(Object obj) {
        if (this == obj)
            return true;
        if (obj == null)
            return false;
        if (!(obj instanceof User))
            return false;
        User other = (User) obj;
        if (id == null) {
            if (other.id != null)
                return false;
        } else if (!id.equals(other.id))
            return false;
        if (ssoId == null) {
            if (other.ssoId != null)
                return false;
        } else if (!ssoId.equals(other.ssoId))
            return false;
        return true;
    }

    /*
     * DO-NOT-INCLUDE passwords in toString function.
     * It is done here just for convenience purpose.
     */
    @Override
    public String toString() {
        return "User [id=" + id + ", ssoId=" + ssoId + ", password=" + password
                + ", firstName=" + firstName + ", lastName=" + lastName
                + ", email=" + email + "]";
    }    
}

@Entity
@Table(name="USER_PROFILE")
public class UserProfile implements Serializable{
    @Id @GeneratedValue(strategy=GenerationType.IDENTITY)
    private Integer id; 
    @Column(name="TYPE", length=15, unique=true, nullable=false)
    private String type = UserProfileType.USER.getUserProfileType(); 
    public Integer getId() {
        return id;
    }
    public void setId(Integer id) {
        this.id = id;
    }
    public String getType() {
        return type;
    }
    public void setType(String type) {
        this.type = type;
    }
    @Override
    public int hashCode() {
        final int prime = 31;
        int result = 1;
        result = prime * result + ((id == null) ? 0 : id.hashCode());
        result = prime * result + ((type == null) ? 0 : type.hashCode());
        return result;
    }
    @Override
    public boolean equals(Object obj) {
        if (this == obj)
            return true;
        if (obj == null)
            return false;
        if (!(obj instanceof UserProfile))
            return false;
        UserProfile other = (UserProfile) obj;
        if (id == null) {
            if (other.id != null)
                return false;
        } else if (!id.equals(other.id))
            return false;
        if (type == null) {
            if (other.type != null)
                return false;
        } else if (!type.equals(other.type))
            return false;
        return true;
    }
    @Override
    public String toString() {
        return "UserProfile [id=" + id + ", type=" + type + "]";
    }
}

我仍在学习，我想了解 @ManyToMany 和 @JoinTable 注释，我想知道如何在 JDBC 中完成它

Answer 1

以下是Spring支持的数据访问技术，有多种选择。

• Spring-DAO ( http://docs.spring.io/spring/docs/2.0.8/reference/dao.html )
• Spring-ORM ( http://docs.spring.io/spring/docs/3.0.x/spring-framework-reference/html/orm.html )
• Spring-JDBC ( http://docs.spring.io/spring/docs/3.0.x/spring-framework-reference/html/jdbc.html )

Spring JDBC 提供了模板，用于减少通过简单的旧方式访问数据库的样板代码 - 编写您自己的 SQL 查询。

Spring-ORM 提供了通过 ORM 技术访问数据库的简化模板，例如 Hibernate、开放 JPA 等。

Spring-DAO:

Spring 中的数据访问对象 (DAO) 支持旨在以一致的方式轻松使用 JDBC、Hibernate 或 JDO 等数据访问技术