java - 如何拆分此函数以降低认知复杂度？

Question

我是 Java 新手，这个菜鸟功能一直困扰着我。我已经下载了 sonarLint 并发现它把这个问题抛在了我的脸上方法的认知复杂性不应该太高。我知道它看起来很难看，但任何人都可以指出我如何重新格式化以具有干燥的概念，并且不具有 SonarLint 提到的高复杂性。

@PostMapping("/adduser")
//  @PreAuthorize("hasRole('ADMIN')")
    public ResponseEntity<MessageResponse> registerUser(@Valid @RequestBody SignupRequest signUpRequest) {
        /*
         * This controller Creates new user based on all the entities for the user
         * 
         */
         if(dbValue)
         {
             if (repository.existsByUsername(signUpRequest.getUsername())) {
                    return ResponseEntity
                            .badRequest()
                            .body(new MessageResponse("Error: Username is already taken!"));
                }

                if (repository.existsByEmail(signUpRequest.getEmail())) {
                    return ResponseEntity
                            .badRequest()
                            .body(new MessageResponse("Error: Email is already in use!"));
                }

                // Create new user's account
                User2 user = new User2(signUpRequest.getUsername(), 
                                     signUpRequest.getEmail(),
                                     signUpRequest.getCustomername(),
                                     signUpRequest.getCustomerid(),
                                     signUpRequest.getDescription(),
                                     encoder.encode(signUpRequest.getPassword()));              
                
                Set<String> strRoles = signUpRequest.getRoles();                
                Set<Role2> roles = new HashSet<>();
                Role2 e = new Role2();
                e.setName("ROLE_ADMIN");
                roles.add(e);               
                user.setRoles(roles);
                repository.save(user);      
                
         }
         
         else {
             
             if (repository.existsByUsername(signUpRequest.getUsername())) {
                    return ResponseEntity
                            .badRequest()
                            .body(new MessageResponse("Error: Username is already taken!"));
                }

                if (repository.existsByEmail(signUpRequest.getEmail())) {
                    return ResponseEntity
                            .badRequest()
                            .body(new MessageResponse("Error: Email is already in use!"));
                }

                // Create new user's account
                User1 user = new User1(signUpRequest.getUsername(), 
                                     signUpRequest.getEmail(),
                                     signUpRequest.getCustomername(),
                                     signUpRequest.getCustomerid(),
                                     signUpRequest.getDescription(),
                                     encoder.encode(signUpRequest.getPassword()));
                

                Set<String> strRoles = signUpRequest.getRoles();
                Set<Role> roles = new HashSet<>();
              
                if (strRoles == null) {
                    Role userRole = roleRepository1.findByName(URole.ROLE_USER)
                            .orElseThrow(() -> new RuntimeException("Error: Role is not found."));
                    roles.add(userRole);
                } else {
                    strRoles.forEach(role -> {
                        switch (role) {
                        case "admin":
                            Role adminRole = roleRepository1.findByName(URole.ROLE_ADMIN)
                                    .orElseThrow(() -> new RuntimeException("Error: Role is not found."));                          
                            roles.add(adminRole);

                            break;
                        
                        default:
                            Role userRole = roleRepository1.findByName(URole.ROLE_USER)
                                    .orElseThrow(() -> new RuntimeException("Error: Role is not found."));
                            roles.add(userRole);
                        }
                    });
                }
                user.setRoles(roles);
                repository.save(user);
         }      
         return ResponseEntity.ok(new MessageResponse("User Added successfully!"));     
    }   

感谢批评和任何帮助。提前致谢。

Answer 1

1. 在 Controller 中，您需要使用服务，而不是存储库。在那里创建用户创建方法。
2. 使用方法创建 UserValidator 接口(interface)(existsByUsername、existsByEmail 实现谓词)可选检查(谓词条件、字符串消息)或者将此方法委托(delegate)给新的验证实用程序类。
3. 您有重复，可以将其移出条件

if (repository.existsByUsername(signUpRequest.getUsername())) 
    return ResponseEntity
        .badRequest() // below, you always create a new string and a new message response, use a constant as not to clog the memory
        .body(new MessageResponse("Error: Username is already taken!"));
if (repository.existsByEmail(signUpRequest.getEmail())) 
    return ResponseEntity
        .badRequest() // here too
        .body(new MessageResponse("Error: Email is already in use!"));

signUpRequest.getRoles() 可能会返回一个可选值，或者 Collections.emptyList()(如果为空，而不是 null)

通过 .valueOf() 获取角色，它会使 .forEach() 条件不被使用，并且也会切换情况。

替换这个

Role userRole = roleRepository1.findByName(URole.ROLE_USER)
   .orElseThrow(() -> new RuntimeException("Error: Role is not found."));
roles.add(userRole);

与

roleRepository1.findByName(URole.ROLE_USER).ifPresent(roles::add);