使用hibernate for REST在Spring Security中进行身份验证失败（Bad Credentuals）

时间：2017-03-09 20:14:43

标签： spring hibernate rest security authentication

我已经使用spring-data-rest创建了一个spring boot应用程序。

我的Rest API工作得很好。然后我导入了弹簧安全装置。在引用了许多Web资源之后，我也完成了配置。

但是，每次发送请求时，都会收到Bad Credential Error以下是我的代码

User.java 

package com.innaun.model;

import org.springframework.data.rest.core.annotation.RestResource;

import javax.persistence.*;
import javax.validation.constraints.NotNull;
import java.util.HashSet;
import java.util.Set;

@Entity
public class User {

    @Id
    @GeneratedValue(strategy = GenerationType.AUTO)
    private Long userId;

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

    @NotNull
    @RestResource(exported = false  )
    private String password;

    @NotNull
    private boolean enabled;

    @OneToMany
    private Set<UserRole> userRoles = new HashSet<UserRole>(0);

    public User() {
    }

    public User(String username, String password, boolean enabled) {
        this.username = username;
        this.password = password;
        this.enabled = enabled;
    }

    public Set<UserRole> getUserRoles() {
        return userRoles;
    }

    public void setUserRoles(Set<UserRole> userRoles) {
        this.userRoles = userRoles;
    }

    public Long getUserId() {
        return userId;
    }

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

    public String getUsername() {
        return username;
    }

    public void setUsername(String username) {
        this.username = username;
    }

    public String getPassword() {
        return password;
    }

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

    public boolean isEnabled() {
        return enabled;
    }

    public void setEnabled(boolean enabled) {
        this.enabled = enabled;
    }
}

UserRole.java 

package com.innaun.model;

import javax.persistence.*;
import javax.validation.constraints.NotNull;

@Entity
public class UserRole {

    @Id
    @GeneratedValue(strategy = GenerationType.AUTO)
    private Long userRoleId;

    @NotNull
    private String userRole;

    @ManyToOne
    private User user;

    public UserRole() {
    }

    public UserRole(String userRole, User user) {
        this.userRole = userRole;
        this.user = user;
    }

    public Long getUserRoleId() {
        return userRoleId;
    }

    public void setUserRoleId(Long userRoleId) {
        this.userRoleId = userRoleId;
    }

    public String getUserRole() {
        return userRole;
    }

    public void setUserRole(String userRole) {
        this.userRole = userRole;
    }

    public User getUser() {
        return user;
    }

    public void setUser(User user) {
        this.user = user;
    }
}

UserRepository.java 

package com.innaun.model;

import org.springframework.data.repository.CrudRepository;
import org.springframework.data.repository.query.Param;
import org.springframework.data.rest.core.annotation.RepositoryRestResource;

@RepositoryRestResource
public interface UserRepository extends CrudRepository<User, Long>{
    User findByUsername(@Param("user") String user);
}

UserRoleRepository.java 

package com.innaun.model;

import org.springframework.data.repository.CrudRepository;
import org.springframework.data.rest.core.annotation.RepositoryRestResource;

@RepositoryRestResource
public interface UserRoleRepository extends CrudRepository<UserRole, Long> {
}

AppUserDetailsS​​ervice.java 

package com.innaun.model;

import com.innaun.model.UserRepository;
import com.innaun.model.UserRole;
import org.springframework.security.core.GrantedAuthority;
import org.springframework.security.core.authority.SimpleGrantedAuthority;
import org.springframework.security.core.userdetails.User;
import org.springframework.beans.factory.annotation.Autowired;
import org.springframework.security.core.userdetails.UserDetails;
import org.springframework.security.core.userdetails.UserDetailsService;
import org.springframework.security.core.userdetails.UsernameNotFoundException;
import org.springframework.stereotype.Service;

import javax.transaction.Transactional;
import java.util.ArrayList;
import java.util.HashSet;
import java.util.List;
import java.util.Set;

@Service("appUserDetailsService")
public class AppUserDetailsService implements UserDetailsService {

    @Autowired
    private UserRepository userRepository;


    @Transactional
    @Override
    public UserDetails loadUserByUsername(String s) throws UsernameNotFoundException {
        com.innaun.model.User user = userRepository.findByUsername(s);

        List<GrantedAuthority> authorities = buildUserAuthority(user.getUserRoles());

        return buildUserForAuthentication(user, authorities);
    }

    private User buildUserForAuthentication(com.innaun.model.User user, List<GrantedAuthority> authorities){
                return new User(user.getUsername(), user.getPassword(), user.isEnabled(), true, true, true, authorities);
    }

    private List<GrantedAuthority> buildUserAuthority(Set<UserRole> userRoles){
        Set<GrantedAuthority> setAuths = new HashSet<GrantedAuthority>();

        for (UserRole userRole : userRoles){
            setAuths.add(new SimpleGrantedAuthority(userRole.getUserRole()));
        }

        List<GrantedAuthority> result = new ArrayList<GrantedAuthority>(setAuths);

        return result;
    }


}

ApplicationRESTSecurity.java 

package com.innaun;

import org.springframework.beans.factory.annotation.Autowired;
import org.springframework.beans.factory.annotation.Qualifier;
import org.springframework.context.annotation.Bean;
import org.springframework.context.annotation.Configuration;
import org.springframework.security.config.annotation.authentication.builders.AuthenticationManagerBuilder;
import org.springframework.security.config.annotation.web.builders.HttpSecurity;
import org.springframework.security.config.annotation.web.configuration.EnableWebSecurity;
import org.springframework.security.config.annotation.web.configuration.WebSecurityConfigurerAdapter;
import org.springframework.security.core.userdetails.UserDetailsService;
import org.springframework.security.crypto.bcrypt.BCryptPasswordEncoder;
import org.springframework.security.crypto.password.PasswordEncoder;


@Configuration
@EnableWebSecurity
public class ApplicationRESTSecurity extends WebSecurityConfigurerAdapter {

    @Qualifier("appUserDetailsService")
    @Autowired
    UserDetailsService userDetailsService;

    @Autowired
    public void configureGlobal(AuthenticationManagerBuilder auth) throws Exception{
        auth.userDetailsService(userDetailsService).passwordEncoder(passwordEncoder());
    }

    @Bean
    public PasswordEncoder passwordEncoder() {
        PasswordEncoder encoder = new BCryptPasswordEncoder();
        return encoder;
    }

    @Override
    protected void configure(HttpSecurity http) throws Exception {
        http
                .authorizeRequests()
                .anyRequest().fullyAuthenticated()
                .and().httpBasic()
                .and().csrf()
                .disable();
    }
}

另外，我已添加以下内容将测试用户添加到数据库

package com.innaun;

import com.innaun.model.User;
import com.innaun.model.UserRepository;
import org.springframework.boot.CommandLineRunner;
import org.springframework.boot.SpringApplication;
import org.springframework.boot.autoconfigure.SpringBootApplication;
import org.springframework.context.annotation.Bean;

@SpringBootApplication
public class PitchuApplication {

    public static void main(String[] args) {
        SpringApplication.run(PitchuApplication.class, args);
    }

    @Bean
    CommandLineRunner init(UserRepository userRepository) {

        return (args) -> {
            userRepository.save(new User("myuser", "mypassword", true));
        };
    }

}

正如我想的那样，数据库现在拥有上述用户并启用了用户。

Screenshot of the User data table

所有其他表格都是空白的。

然而，当我尝试卷曲时

curl -u myuser:mypassword localhost:8080

它返回了

{"timestamp":1489090315435,"status":401,"error":"Unauthorized","message":"Bad credentials","path":"/"}

任何人都可以解释我哪里出错了。

2 个答案:

答案 0 :(得分：0)

您的配置对我来说很好。所以，我最好的猜测是数据库中的password列的长度小于60，这是散列BCrypt将产生的长度。

答案 1 :(得分：0)

我明白了。这是如此简单的错误。我用来在userRepository中保存新用户的密码是raw而不是加密的。我想通了：

//Create a new password encoder
private PasswordEncoder passwordEncoder = new BCryptPasswordEncoder();

//Encode the password
String password = "mypassword";
String hashedPassword = passwordEncoder.encode(password);

//Create the user with the encoded password
User user = new User(myuser, hashedPassword, true);

//then persist
userRepository.save(user);
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