Spring Boot Data Rest JPA - Entity custom create (User)

Question

I am trying to learn Spring. I created a project with Spring Boot using the following tools:

Spring Data JPA
Spring Data REST
Spring HATEOAS
Spring Security

I am trying to create a User entity. I want the user to have an encrypted password (+ salt).

When i do POST to /api/users i successfully create a new user.

{
"firstname":"John",
"lastname":"Doe",
"email":"[email protected]",
"password":"12345678"
}

But i have 2 problems:

the password is saved in clear-text
the salt is null

+----+---------------------+-----------+----------+----------+------+
| id |        email        | firstname | lastname | password | salt |
+----+---------------------+-----------+----------+----------+------+
|  1 | [email protected] | John      | Doe      | 12345678 | NULL |
+----+---------------------+-----------+----------+----------+------+

The problem i think is that the default constructor is used and not the other one i have created. I am new to Spring and JPA so i must be missing something. Here is my code.

User.java

@Entity
@Table(name = "users")
public class User{

    @Id
    @GeneratedValue
    private Long id;

    @Column(nullable = false)
    public String firstname;

    @Column(nullable = false)
    public String lastname;

    @Column(nullable = false, unique = true)
    public String email;

    @JsonIgnore
    @Column(nullable = false)
    public String password;

    @JsonIgnore
    @Column
    private String salt;

    public User() {}

    public User(String email, String firstname, String lastname, String password) {
        this.email = email;
        this.firstname = firstname;
        this.lastname = lastname;
        this.salt = UUID.randomUUID().toString();
        this.password = new BCryptPasswordEncoder().encode(password + this.salt);
    }


    @JsonIgnore
    public String getSalt() {
        return salt;
    }

    @JsonProperty
    public void setSalt(String salt) {
        this.salt = salt;
    }

    public String getEmail() {
        return email;
    }

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

    public String getFirstname() {
        return firstname;
    }

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

    public Long getId() {
        return id;
    }

    public void setId(Long id) {
        this.id = id;
    }

    public String getLastname() {
        return lastname;
    }

    public void setLastname(String lastname) {
        this.lastname = lastname;
    }

    @JsonIgnore
    public String getPassword() {
        return password;
    }

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

}

UserRepository.java

public interface UserRepository extends JpaRepository<User, Long> {

    public User findByEmail(String email);

    public User findByEmailAndPassword(String email, String password);
}

Application.java

@SpringBootApplication
public class Application {


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

}

Also if someone finds what i did wrong, i would like to point me where/how i should put the user login code (decryption).

Thanks.

@Makoton i don't have one. Spring Data Rest automatically exposes my entities in a RESTful API. Do i need one in this case? — Christos Baziotis
– Christos Baziotis, Commented Jul 14, 2015 at 18:09
and did you try to create a UserService and : public void registerUser(final User user) { final String encodedPassword = passwordEncoder.encode(user.getPassword()); user.setPassword(encodedPassword); userRepo.save(user); } — Makoton
– Makoton, Commented Jul 14, 2015 at 18:20
@agnostos Did you solve your problem? If not, did you try the solution in my answer? — hzpz
– hzpz, Commented Jul 17, 2015 at 15:13

Christos Baziotis · Accepted Answer · 2015-07-25 12:05:08Z

So, here is how i solved my problem: i created a Controller as my custom endpoint and then i created a service in which i placed the logic i wanted for the creation of the user. Here is the code:

UserController.java

@Controller
public class UserController {

    @Autowired
    private UserService userService;

    @RequestMapping("/api/register")
    @ResponseBody
    public Long register(@RequestBody User user) {
        return userService.registerUser(user);
    }

    ...

}

UserService .java

@Service
public class UserService {

    @Autowired
    private UserRepository userRepository;

    public Long registerUser(User user) {
        user.setPassword(new BCryptPasswordEncoder().encode(password));
        userRepository.save(user);
        return user.getId();
    }

        ...

}

so by doing a POST with

{
"firstname":"John",
"lastname":"Doe",
"email":"[email protected]",
"password":"12345678"
}

in /api/register, i can now create a user with a hashed password.

hzpz · Accepted Answer · 2015-07-14 18:53:43Z

2

If you want Spring to use your constructor, you need to

remove the no-argument constructor
annotate every parameter in the other constructor with @JsonProperty like this

public User(@JsonProperty("email") String email, 
            @JsonProperty("firstname") String firstname, 
            @JsonProperty("lastname") String lastname, 
            @JsonProperty("password") String password) {
    this.email = email;
    this.firstname = firstname;
    this.lastname = lastname;
    this.password = new BCryptPasswordEncoder().encode(password);
}

You don't need to provide a salt value to the BCryptPasswordEncoder because it already salts passwords by itself.

answered Jul 14, 2015 at 18:53

hzpz

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1 Comment

aycanadal Over a year ago

How do you do when one of the arguments is a reference to another entity? Can @JsonProperty be used then?

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