类 java.util.Optional 中的方法 flatMap<T> 不能应用于给定的类型

Question

我的代码在flatmap中给出错误。

public Optional<User> getUserWithAuthorities() {
return SecurityUtils.getCurrentUserLogin().flatMap(UserDao::findOneWithAuthoritiesByLogin);
}

下面是错误:

controller/AuthenticationController.java:[77,51] method flatMap in class java.util.Optional<T> cannot be applied to given types;
[ERROR]   required: java.util.function.Function<? super java.lang.String,java.util.Optional<U>>
[ERROR]   found: UserDao::f[...]Login
[ERROR]   reason: cannot infer type-variable(s) U
[ERROR]     (argument mismatch; invalid method reference
[ERROR]       cannot find symbol
[ERROR]         symbol:   method findOneWithAuthoritiesByLogin(java.lang.String)
[ERROR]         location: interface net.javaguides.springboot.dao.UserDao)

UserDao代码:

@Repository
public interface UserDao extends JpaRepository<User, Long> {
User findByUsername(String username);
Optional<User> findOneWithAuthoritiesByLogin(String login);
}

这是securitytils的代码:

public final class SecurityUtils {
private SecurityUtils() {}
/**
* Get the login of the current user.
*
* @return the login of the current user.
*/
public static Optional<String> getCurrentUserLogin() {
SecurityContext securityContext = SecurityContextHolder.getContext();
return Optional.ofNullable(extractPrincipal(securityContext.getAuthentication()));
}
private static String extractPrincipal(Authentication authentication) {
if (authentication == null) {
return null;
} else if (authentication.getPrincipal() instanceof UserDetails) {
UserDetails springSecurityUser = (UserDetails) authentication.getPrincipal();
return springSecurityUser.getUsername();
} else if (authentication.getPrincipal() instanceof String) {
return (String) authentication.getPrincipal();
}
return null;
}

有什么问题吗?

Answer 1

引用接口类UserDao::findOneWithAuthoritiesByLogin导致错误。Java不能单独使用接口调用该方法。

要解决这个问题，可以引用UserDao类型的实例化对象。

下面是接口的一个示例实现:

class UserDaoImplementation implements UserDao {
@Override
public Optional<User> findOneWithAuthoritiesByLogin(String login) {
User user = User.builder().name("Tom").build();
return Optional.of(user);
}
}

然后创建一个实例并引用它而不是接口类:

public static Optional<User> getUserWithAuthorities() {
UserDao userDao = new UserDaoImplementation();
return SecurityUtils.getCurrentUserLogin().flatMap(userDao::findOneWithAuthoritiesByLogin);
}

在这种情况下，getUserWithAuthorities方法现在返回一个包含User对象的Optional。