I am creating an API to simulate a vending machine and I'm using Spring, JPA and MySQL. I have an endpoint for a POST request that allows for new user creation by entering the user's details into a table called users. I have a check to see if the username already exists in the table and if it does, I want to return a message that says the creation of a new user was unsuccessful because that username is already in use. If the user name is not in the table, I want to return the User object.

How do I return this error message? From what I have found so far, suggestions include the usage of ResponseEntity or creating a custom exception handler which all seemed overly complicated for something that I feel is quite straightforward. Is there an easier way to do this?

So this is what I want the response to look like when a new user has been created successfully (which I have currently managed to get working successfully):

{
    "username": "user",
    "password": "password",
    "role": "BUYER",
    "deposit": 0.0,
    "id": 12
}

And if it fails, I want it to return something that looks along the lines of:

Error: username already in use.

or:

{
  "error" : "Username already in use"
}

User object:

@Entity
@Table(name = "users")
public class User
{
    @Id
    @GeneratedValue(strategy = GenerationType.IDENTITY)
    @Column(name = "id")
    private int userId;

    @NotNull
    @Column(name = "username", unique = true)
    private String username;

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

    @NotNull
    @Column(name = "role")
    private Role role;

    @NotNull
    @Column(name = "deposit")
    private BigDecimal deposit;

    public User() {}

    public User(String username, String password, Role role, BigDecimal deposit)
    {
        this.username = username;
        this.password = password;
        this.role = role;
        this.deposit = deposit;
    }

    public int getId()
    {
        return 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 Role getRole()
    {
        return role;
    }

    public void setRole(String role)
    {
        this.role = Role.valueOf(role);
    }

    public void setRole(Role role) {
        this.role = role;
    }

    public BigDecimal getDeposit()
    {
        return deposit;
    }

    public void setDeposit(BigDecimal deposit)
    {
        this.deposit = deposit;
    }
}

Method in the controller that is being called:

    @PostMapping(value = "/create-user", consumes = {"application/json"})
    public User createUser(@RequestBody User user)
    {
        return userService.createUser(user);
    }

createUser method in UserService:

    public User createUser(User user)
    {
//        if(userRepository.userExists(user.getUsername()) == 0)
        return userRepository.save(user);
    }

UserRepository is an interface that extends JpaRepository.

If I have missed some information or have worded the question incorrectly, please let me know and I'll update accordingly.

CodePudding user response：

This can be achieved in multiple ways.

1. You can use JpaRepository's existsById() or similar "exists" methods. This would return a boolean. If it's true, that is if the entry already exists for a given id, you can send an error response. Otherwise, new user object. For that, you can have a model class something like UserOrError which will either hold a user object or an error string at a time. Use @JsonView annotation from Jackson library to show/ hide the fields. Read more
2. Based on the result of the above "exists" method, wrap it with ResponseEntity<T>. The return type of your controller method should be ResponseEntity<T>. Here's the link. The advantage of this method is that you can send different HTTP status codes
3. Throw a custom RuntimeException in your service layer and use Spring's exception handling like the RestControllerAdvice annotation