Spring Security Login Form Example with Database Authentication

In this Spring Security tutorial, we will learn how to implement a custom login form with database authentication using Spring Security and the MySQL database.

In the case of database authentication, the User enters login credentials like username and password in a Login form and submits the form. Next, we will validate User entered login credentials with the username and password in a database table.

Database Set up

Add Maven Dependencies

Add below Maven dependencies to your Spring Boot project:
		<dependency>
			<groupId>org.springframework.boot</groupId>
			<artifactId>spring-boot-starter-data-jpa</artifactId>
		</dependency>
		<dependency>
			<groupId>org.springframework.boot</groupId>
			<artifactId>spring-boot-starter-web</artifactId>
		</dependency>
		<dependency>
			<groupId>com.mysql</groupId>
			<artifactId>mysql-connector-j</artifactId>
			<scope>runtime</scope>
		</dependency>
		<dependency>
			<groupId>org.projectlombok</groupId>
			<artifactId>lombok</artifactId>
			<optional>true</optional>
		</dependency>
		<dependency>
			<groupId>org.springframework.boot</groupId>
			<artifactId>spring-boot-starter-security</artifactId>
		</dependency>

		<dependency>
			<groupId>org.springframework.boot</groupId>
			<artifactId>spring-boot-starter-thymeleaf</artifactId>
		</dependency>

Configure MySQL Database

Let's first create a database in MySQL server using the below command:
create database login_system
Since we’re using MySQL as our database, we need to configure the database URLusername, and password so that Spring can establish a connection with the database on startup. Open the src/main/resources/application.properties file and add the following properties to it:
spring.datasource.url = jdbc:mysql://localhost:3306/login_system
spring.datasource.username = root
spring.datasource.password = root

# Hibernate ddl auto (create, create-drop, validate, update)
spring.jpa.hibernate.ddl-auto = update

logging.level.org.springframework.security=DEBUG

Model Layer - Create JPA Entities 

In this step, we will create User and Role JPA entities and establish MANY-to-MANY relationships between them. Let's use JPA annotations to establish MANY-to-MANY relationships between User and Role entities.

User

import jakarta.persistence.*;
import lombok.AllArgsConstructor;
import lombok.Getter;
import lombok.NoArgsConstructor;
import lombok.Setter;

import java.util.Set;

@Setter
@Getter
@NoArgsConstructor
@AllArgsConstructor
@Entity
@Table(name = "users")
public class User {

    @Id
    @GeneratedValue(strategy = GenerationType.IDENTITY)
    private Long id;
    private String name;
    @Column(nullable = false, unique = true)
    private String username;
    @Column(nullable = false, unique = true)
    private String email;
    @Column(nullable = false)
    private String password;

    @ManyToMany(fetch = FetchType.EAGER, cascade = CascadeType.ALL)
    @JoinTable(name = "users_roles",
        joinColumns = @JoinColumn(name = "user_id", referencedColumnName = "id"),
            inverseJoinColumns = @JoinColumn(name = "role_id", referencedColumnName = "id")
    )
    private Set<Role> roles;
}

Role

import jakarta.persistence.*;
import lombok.AllArgsConstructor;
import lombok.Getter;
import lombok.NoArgsConstructor;
import lombok.Setter;

@Getter
@Setter
@NoArgsConstructor
@AllArgsConstructor
@Entity
@Table(name = "roles")
public class Role {

    @Id
    @GeneratedValue(strategy = GenerationType.IDENTITY)
    private Long id;
    private String name;
}

Repository Layer

UserRepository

import net.javaguides.todo.entity.User;
import org.springframework.data.jpa.repository.JpaRepository;

import java.util.Optional;

public interface UserRepository extends JpaRepository<User, Long> {

    Optional<User> findByUsername(String username);

    Boolean existsByEmail(String email);

    Optional<User> findByUsernameOrEmail(String username, String email);

    boolean existsByUsername(String username);
}

RoleRepository

import net.javaguides.todo.entity.Role;
import org.springframework.data.jpa.repository.JpaRepository;

import java.util.Map;
import java.util.Optional;

public interface RoleRepository extends JpaRepository<Role, Long> {
    Optional<Role> findByName(String name);
}

Service Layer - CustomUserDetailsService

Let's write a logic to load user details by name or email from the database. 

Let's create a CustomUserDetailsService which implements the UserDetailsService interface ( Spring security in-build interface) and provides an implementation for the loadUserByUername() method:
import lombok.AllArgsConstructor;
import net.javaguides.todo.entity.User;
import net.javaguides.todo.repository.UserRepository;
import org.springframework.security.core.GrantedAuthority;
import org.springframework.security.core.authority.SimpleGrantedAuthority;
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 java.util.Set;
import java.util.stream.Collectors;

@Service
@AllArgsConstructor
public class CustomUserDetailsService implements UserDetailsService {

    private UserRepository userRepository;

    @Override
    public UserDetails loadUserByUsername(String usernameOrEmail) throws UsernameNotFoundException {

        User user = userRepository.findByUsernameOrEmail(usernameOrEmail, usernameOrEmail)
                .orElseThrow(() -> new UsernameNotFoundException("User not exists by Username or Email"));

        Set<GrantedAuthority> authorities = user.getRoles().stream()
                .map((role) -> new SimpleGrantedAuthority(role.getName()))
                .collect(Collectors.toSet());

        return new org.springframework.security.core.userdetails.User(
                usernameOrEmail,
                user.getPassword(),
                authorities
        );
    }
}

Spring Security uses the UserDetailsService interface, which contains the loadUserByUsername(String username) method to look up UserDetails for a given username

The UserDetails interface represents an authenticated user object and Spring Security provides an out-of-the-box implementation of org.springframework.security.core.userdetails.User

Spring Security Configuration

Let's create a class SpringSecurityConfig and add the following configuration to it:
import lombok.AllArgsConstructor;
import org.springframework.context.annotation.Bean;
import org.springframework.context.annotation.Configuration;
import org.springframework.security.authentication.AuthenticationManager;
import org.springframework.security.config.annotation.authentication.configuration.AuthenticationConfiguration;
import org.springframework.security.config.annotation.web.builders.HttpSecurity;
import org.springframework.security.core.userdetails.UserDetailsService;
import org.springframework.security.crypto.bcrypt.BCryptPasswordEncoder;
import org.springframework.security.crypto.password.PasswordEncoder;
import org.springframework.security.web.SecurityFilterChain;
import org.springframework.security.web.util.matcher.AntPathRequestMatcher;

@Configuration
@AllArgsConstructor
public class SpringSecurityConfig {

    private UserDetailsService userDetailsService;
    @Bean
    public static PasswordEncoder passwordEncoder(){
        return new BCryptPasswordEncoder();
    }

    @Bean
    public SecurityFilterChain filterChain(HttpSecurity http) throws Exception {
        http.csrf().disable()
                .authorizeHttpRequests((authorize) ->
                        authorize.anyRequest().authenticated()
                ).formLogin(
                        form -> form
                                .loginPage("/login")
                                .loginProcessingUrl("/login")
                                .defaultSuccessUrl("/welcome")
                                .permitAll()
                ).logout(
                        logout -> logout
                                .logoutRequestMatcher(new AntPathRequestMatcher("/logout"))
                                .permitAll()
                );
        return http.build();
    }

    @Bean
    public AuthenticationManager authenticationManager(AuthenticationConfiguration configuration) throws Exception {
        return configuration.getAuthenticationManager();
    }
}
When the login page is specified in the Spring Security configuration, you are responsible for rendering the page.

We are using the Spring security provided BCryptPasswordEncoder class to encrypt the passwords.

Thymeleaf Template - Custom Login Page

The following Thymeleaf template produces an HTML login form that complies with a login page of /login.

Login Form - src/main/resources/templates/login.html

<!DOCTYPE html>
<html lang="en"
      xmlns:th="http://www.thymeleaf.org"
>
<head>
    <meta charset="UTF-8">
    <title>Login System</title>
    <link href="https://cdn.jsdelivr.net/npm/[email protected]/dist/css/bootstrap.min.css"
          rel="stylesheet"
          integrity="sha384-EVSTQN3/azprG1Anm3QDgpJLIm9Nao0Yz1ztcQTwFspd3yD65VohhpuuCOmLASjC"
          crossorigin="anonymous">
</head>
<body>
<nav class="navbar navbar-expand-lg navbar-dark bg-dark">
    <div class="container-fluid">
        <a class="navbar-brand" th:href="@{/index}">Spring Security Custom Login Example</a>
        <button class="navbar-toggler" type="button" data-bs-toggle="collapse" data-bs-target="#navbarSupportedContent" aria-controls="navbarSupportedContent" aria-expanded="false" aria-label="Toggle navigation">
            <span class="navbar-toggler-icon"></span>
        </button>
    </div>
</nav>
<br /><br />
<div class="container">
    <div class="row">
        <div class="col-md-6 offset-md-3">

            <div th:if="${param.error}">
                <div class="alert alert-danger">Invalid Email or Password</div>
            </div>
            <div th:if="${param.logout}">
                <div class="alert alert-success"> You have been logged out.</div>
            </div>

            <div class="card">
                <div class="card-header">
                    <h2 class="text-center">Login Form</h2>
                </div>
                <div class="card-body">
                    <form
                            method="post"
                            role="form"
                            th:action="@{/login}"
                            class="form-horizontal"
                    >
                        <div class="form-group mb-3">
                            <label class="control-label"> Email</label>
                            <input
                                    type="text"
                                    id="username"
                                    name="username"
                                    class="form-control"
                                    placeholder="Enter email address"
                            />
                        </div>

                        <div class="form-group mb-3">
                            <label class="control-label"> Password</label>
                            <input
                                    type="password"
                                    id="password"
                                    name="password"
                                    class="form-control"
                                    placeholder="Enter password"
                            />
                        </div>
                        <div class="form-group mb-3">
                            <button type="submit" class="btn btn-primary" >Submit</button>
                        </div>
                    </form>
                </div>
            </div>
        </div>
    </div>
</div>
</body>
</html>
There are a few key points about the custom Login HTML form:
  • The form should perform a post to /login.
  • The form should specify the username in a parameter named username.
  • The form should specify the password in a parameter named password.
  • If the HTTP parameter named error is found, it indicates the user failed to provide a valid username or password.
  • If the HTTP parameter named logout is found, it indicates the user has logged out successfully.
  • Many users do not need much more than to customize the login page. However, if needed, you can customize everything shown earlier with additional configuration.

Spring MVC Controller

Let's create a handler method in the Spring MVC controller that maps GET /login to the login template we created:

import org.springframework.stereotype.Controller;
import org.springframework.web.bind.annotation.GetMapping;

@Controller
public class WelComeController {

    @GetMapping("/welcome")
    public String greeting() {
        return "welcome";
    }

    @GetMapping("/login")
    public String login(){
        return "login";
    }
}

Note that we have also created a welcome handler method to return the welcome Thymeleaf template page.

Thymeleaf Template - welcome.html

Once the user login success then this welcome page will be displayed:
<!DOCTYPE html>
<html lang="en"
      xmlns:th="http://www.thymeleaf.org"
>
<head>
    <meta charset="UTF-8">
    <title>Registration and Login System</title>
    <link href="https://cdn.jsdelivr.net/npm/[email protected]/dist/css/bootstrap.min.css"
          rel="stylesheet"
          integrity="sha384-EVSTQN3/azprG1Anm3QDgpJLIm9Nao0Yz1ztcQTwFspd3yD65VohhpuuCOmLASjC"
          crossorigin="anonymous">
</head>
<body>
<nav class="navbar navbar-expand-lg navbar-dark bg-dark">
    <div class="container-fluid">
        <a class="navbar-brand" th:href="@{/index}">Spring Security Custom Login Example</a>
        <button class="navbar-toggler" type="button" data-bs-toggle="collapse" data-bs-target="#navbarSupportedContent" aria-controls="navbarSupportedContent" aria-expanded="false" aria-label="Toggle navigation">
            <span class="navbar-toggler-icon"></span>
        </button>
        <div class="collapse navbar-collapse" id="navbarSupportedContent">
            <ul class="navbar-nav me-auto mb-2 mb-lg-0">
                <li class="nav-item">
                    <a class="nav-link active" aria-current="page" th:href="@{/logout}">Logout</a>
                </li>
            </ul>
        </div>
    </div>
</nav>
<br /><br />

<body>
<div class="container">
    <div class="row">
        <h1> Welcome to Spring Security world!</h1>
    </div>
</div>

</body>
</html>

Insert SQL Scripts

Before testing Spring security, make sure that you use below SQL scripts to insert the database into respective tables:
INSERT INTO `users` VALUES
(1,'[email protected]','ramesh','$2a$10$5PiyN0MsG0y886d8xWXtwuLXK0Y7zZwcN5xm82b4oDSVr7yF0O6em','ramesh'),
(2,'[email protected]','admin','$2a$10$gqHrslMttQWSsDSVRTK1OehkkBiXsJ/a4z2OURU./dizwOQu5Lovu','admin');

INSERT INTO `roles` VALUES (1,'ROLE_ADMIN'),(2,'ROLE_USER');

INSERT INTO `users_roles` VALUES (2,1),(1,2);
Hibernate will automatically create the database tables so you don't need to create the tables manually.

Test Login Form with Database Authentication using Browser

Enter http://localhost:8080 URL in the browser and it will navigate to the login page. Next, enter a username as admin, password as admin, and click on the Sign-in button:

After successful login, you will see the below web page:
Next, click on the Logout button in the application to logout from the application:

Conclusion

In this Spring Security tutorial, we learned how to implement a custom login form with database authentication using Spring Security and the MySQL database.

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