Login with SpringBoot - spring-boot

Question

I'm taking the first steps with SpringBoot ... I want to make a service Log in where you put a user and a password and if it is correct it returns the user's data.
At the moment, with what I can do the user's data according to, but I can solve what I want, I would appreciate it.
The problem that I have 2. I do not know how to send 2 parameters "username", "password" for a moment Send 1.
Repository:
import org.springframework.data.repository.CrudRepository;
import prueba.Usuario;
public interface UsuRepository extends CrudRepository<Usuario,String>{
}
usuarioClass:
#Entity
#Table(name="usuario")
#JsonIgnoreProperties({"hibernateLazyInitializer", "handler"})
public class Usuario {
#Id
#Column(name="id")
private int id;
#Column(name="name")
private String name;
#Column(name="email")
private String email;
public Usuario() {
}
public Usuario(int id, String name, String email) {
this.id = id;
this.name = name;
this.email = email;
}
public int getId() {
return id;
}
public void setId(int id) {
this.id = id;
}
public String getName() {
return name;
}
public void setName(String name) {
this.name = name;
}
public String getEmail() {
return email;
}
public void setEmail(String email) {
this.email = email;
}
#Override
public String toString() {
return "User[id=" + id + ", name=" + name + ", email=" + email + "]";
}
Controller
#RestController
#RequestMapping(value={"/usu"})
public class UsuController {
#Autowired
UsuService usuRepository;
#RequestMapping("/")
String home() {
return "Servicio Usuario!";
}
#RequestMapping(method=RequestMethod.GET, path = "/Login/{userName}" , produces = MediaType.APPLICATION_JSON_VALUE)
public Usuario Login(#PathVariable("userName") String userName) {
Usuario tasks = usuRepository.getUsu(userName);
return tasks;
}
Interface:
public interface UsuService {
public List<Usuario> getUsus() ;
public Usuario getUsu(String name);
}
Service:
#Override
public Usuario getUsu(String user) {
Usuario usu;
return usu = usuRepository.findOne(user);
}
run:
http://localhost:8080/usu/Login/edu
Error:
Whitelabel Error Page
This application has no explicit mapping for /error, so you are seeing this as a fallback.
Mon Jul 02 10:31:34 CEST 2018
There was an unexpected error (type=Internal Server Error, status=500).
Provided id of the wrong type for class prueba.Usuario. Expected: class java.lang.Integer, got class java.lang.String; nested exception is java.lang.IllegalArgumentException: Provided id of the wrong type for class prueba.Usuario. Expected: class java.lang.Integer, got class java.lang.String
I already did a test returning all the users and it works.
At the moment I want to have my username, I return my data, and then I will do the Log in with password. Thanks.
I saw others tutorials The Login with SpringBoot, but this tutorials has "Security, encriptyng , etc " but I am new With this, I need service Login easy...

Answer 1

I think you're mixing some concepts here. See you have this:
1) Class Usuario with an id of type int:
public class Usuario {
#Id
#Column(name="id")
private int id;
...
}
2) A request mapping receiving an user name of type String:
#RequestMapping(method=RequestMethod.GET, path = "/Login/{userName}" , produces = MediaType.APPLICATION_JSON_VALUE)
public Usuario Login(#PathVariable("userName") String userName) { ...
3) A repository configured with String as the id type of the Usuario entity:
public interface UsuRepository extends CrudRepository<Usuario,String>
Conclussion:
When you do:
public Usuario getUsu(**String** user) {
...
return usu = usuRepository.findOne(**user**);
}
You're asking the repository to find an Usuario entity with a String ID but Usuario has an int as an ID.
Therefore the error:
Provided id of the wrong type for class prueba.Usuario. **Expected**: class java.lang.Integer, **got** class java.lang.String; nested exception is java.lang.IllegalArgumentException: Provided id of the wrong type for class prueba.Usuario. Expected: class java.lang.Integer, got class java.lang.String
You need something like this:
public interface UsuRepository extends CrudRepository<Usuario,Integer> {
Usuario getByName(String username);
}
Hope I helped. :-)