Question

我正在尝试实现DAO以在Hibernate / JPA2中使用Spring Security数据库身份验证。 Spring使用以下关系和关联来表示用户和＆amp;作用：

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作为postgresql创建查询重复：

CREATE TABLE users
(
  username character varying(50) NOT NULL,
  "password" character varying(50) NOT NULL,
  enabled boolean NOT NULL,
  CONSTRAINT users_pkey PRIMARY KEY (username)
);
CREATE TABLE authorities
(
  username character varying(50) NOT NULL,
  authority character varying(50) NOT NULL,
  CONSTRAINT fk_authorities_users FOREIGN KEY (username)
      REFERENCES users (username) MATCH SIMPLE
      ON UPDATE NO ACTION ON DELETE NO ACTION
);

使用GrantedAuthorities，UserDetailsService和UserDetailsmanager的板载实现，一切都很好。但是，我对Spring的JDBC实现不满意，并且想编写自己的实现。为此，我尝试通过遵循业务对象来创建关系的表示：

用户实体：

@Entity
@Table(name = "users", uniqueConstraints = {@UniqueConstraint(columnNames = {"username"})})
public class AppUser implements UserDetails, CredentialsContainer {

    private static final long serialVersionUID = -8275492272371421013L;

    @Id
    @Column(name = "username", nullable = false, unique = true)
    private String username;

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

    @OneToMany(
            fetch = FetchType.EAGER, cascade = CascadeType.ALL,
            mappedBy = "appUser"
    )
    private Set<AppAuthority> appAuthorities;

    @Column(name = "accountNonExpired")
    private Boolean accountNonExpired;

    @Column(name = "accountNonLocked")
    private Boolean accountNonLocked;

    @Column(name = "credentialsNonExpired")
    private Boolean credentialsNonExpired;

    @OneToOne(fetch = FetchType.EAGER, cascade = CascadeType.ALL)
    @JoinColumn(name = "personalinformation_fk", nullable = true)
    @JsonIgnore
    private PersonalInformation personalInformation;

    @Column(name = "enabled", nullable = false)
    @NotNull
    private Boolean enabled;

    public AppUser(
            String username,
            String password,
            boolean enabled,
            boolean accountNonExpired,
            boolean credentialsNonExpired,
            boolean accountNonLocked,
            Collection<? extends AppAuthority> authorities,
            PersonalInformation personalInformation
    ) {
        if (((username == null) || "".equals(username)) || (password == null)) {
            throw new IllegalArgumentException("Cannot pass null or empty values to constructor");
        }

        this.username = username;
        this.password = password;
        this.enabled = enabled;
        this.accountNonExpired = accountNonExpired;
        this.credentialsNonExpired = credentialsNonExpired;
        this.accountNonLocked = accountNonLocked;
        this.appAuthorities = Collections.unmodifiableSet(sortAuthorities(authorities));
        this.personalInformation = personalInformation;
    }

    public AppUser() {
    }

    @JsonIgnore
    public PersonalInformation getPersonalInformation() {
        return personalInformation;
    }

    @JsonIgnore
    public void setPersonalInformation(PersonalInformation personalInformation) {
        this.personalInformation = personalInformation;
    }

    // Getters, setters 'n other stuff

作为GrantedAuthorities的实现的权威实体：

@Entity
@Table(name = "authorities", uniqueConstraints = {@UniqueConstraint(columnNames = {"id"})})
public class AppAuthority implements GrantedAuthority, Serializable {
    //~ Instance fields ================================================================================================

    private static final long serialVersionUID = 1L;

    @Id
    @GeneratedValue(strategy = GenerationType.TABLE)
    @Column(name = "id", nullable = false)
    private Integer id;

    @Column(name = "username", nullable = false)
    private String username;

    @Column(name = "authority", nullable = false)
    private String authority;

    // Here comes the buggy attribute. It is supposed to repesent the
    // association username<->username, but I just don't know how to
    // implement it 
    @ManyToOne(fetch = FetchType.EAGER, cascade = CascadeType.ALL)
    @JoinColumn(name = "appuser_fk")
    private AppUser appUser;

    //~ Constructors ===================================================================================================

    public AppAuthority(String username, String authority) {
        Assert.hasText(authority,
                "A granted authority textual representation is required");
        this.username = username;
        this.authority = authority;
    }

    public AppAuthority() {
    }

    // Getters 'n setters 'n other stuff

我的问题是@ManyToOne assoc。 AppAuthorities：它应该是“用户名”，但尝试并这样做会引发错误，因为我必须将该属性表示为String ...而Hibernate期望关联的实体。所以我尝试的实际上是提供正确的实体并通过@JoinColumn(name = "appuser_fk")创建关联。这当然是垃圾，因为为了加载用户，我将在username中使用外键，而Hibernate在appuser_fk中搜索它，它将始终为空。

所以这是我的问题：关于如何修改上述代码以获得数据模型的正确JPA2实现的任何建议？

由于

Answer 1

AppAuthority根本不需要username。 Spring Security不能依赖它，因为它取决于没有任何方法来访问用户名的GrantedAuthority interface。

但更好的做法是将您的域模型与Spring Security分离。如果您有自定义UserDetailsService，则无需模拟Spring Security的默认数据库架构及其对象模型。您的UserDetailsService可以加载您自己的AppUser和AppAuthority，然后根据它们创建UserDetails和GrantedAuthority。这样可以实现更清晰的设计，更好地分离关注点。

Answer 2

这看起来像使用特定于域的密钥的经典Hibernate问题。可能的解决方法是创建一个新的主键字段;例如对于userId int和Users个实体/表格Authorities，请移除Authorities.userName，然后将Users.userName更改为唯一的辅助密钥。

Answer 3

还有一种方法可以将UserDetailsService与JPA / Hibernate分离。

您可以根据需要为User和Authority类建模，并在配置中定义userDetailsService时使用它： -

<sec:jdbc-user-service data-source-ref="webDS"
                id="userDetailsService"
                users-by-username-query="SELECT USER_NAME,PASSWORD,TRUE FROM CUSTOMER WHERE USER_NAME=?"
                authorities-by-username-query="SELECT c.USER_NAME,r.ROLE_NAME from CUSTOMER c 
                                          JOIN CUSTOMER_ROLE cr ON c.CUSTOMER_ID = cr.CUSTOMER_ID 
                                          JOIN ROLE r ON cr.ROLE_ID = r.ROLE_ID 
                                          WHERE USER_NAME=?" />

通过这种方式，您可以定义微调的SQL查询，以从数据库中获取用户和角色。您只需要处理表和列名称。

如何使用Hibernate / JPA2实现Spring Security用户/权限？

3 个答案: