Question

我正在尝试将此this示例修改为创建用户并使用Spring Security和jdbc以权限登录我现在认为并且正在搜索如何将示例中的一些模型转换为JDBC。所以我问在JDBC中等效的以下Hibernate模型

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 + "]";
    }
}

我还在学习我想了解@ManyToMany和@JoinTable注释我想知道如何在JDBC中完成

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，用于通过简单的方式减少用于访问数据库的样板代码 - 编写自己的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，open JPA等。

Spring-DAO：

Spring中的数据访问对象（DAO）支持旨在使您能够以一致的方式轻松使用JDBC，Hibernate或JDO等数据访问技术

在Spring JDBC中Hibernate模型的等价物是什么

1 个答案: