Question

我是GrapQL的新手。我试图用弹簧靴。我可以成功查询，它返回我需要的数据，但我现在想要使用变异。我需要在注册时向数据库添加一个用途。

这是我的schema.graphqls文件：

type Token {
    token: String
}
type Register {
    message: String
}
type User {
    username: String!
    firstName: String!
    lastName: String!
    password: String!
    role: String!
}

type Query {
    login(username: String, password: String): Token
}

type Mutation {
    register(input: RegisterUserInput!): Register
}

input RegisterUserInput {
    username: String!
    firstName: String!
    lastName: String!
    password: String!
    role: String!
}

schema {
    query: Query
    mutation: Mutation
}

因为你可以看到register是Mutation类型，它在模式中添加，就像Query一样。但由于某些原因，它似乎没有进入Mutation，它只是试图在Query中找到类型。

这是我的控制者：

@Autowired
    private UserService userService;

    /**
     * Login the user and return generated token
     * @param query
     * @return String token
     */
    @PostMapping("/login")
    public ResponseEntity<Object> login(@RequestBody String query){
        ExecutionResult executionResult = userService.getGraphQL().execute(query);

        // Check if there are errors
        if(!executionResult.getErrors().isEmpty()){
            return new ResponseEntity<>(executionResult.getErrors().get(0).getMessage(), HttpStatus.UNAUTHORIZED);
        }

        return new ResponseEntity<>(executionResult, HttpStatus.OK);
    }

    /**
 * Create new user and save him to database
 * @param mutation
 * @return String message
 */
@PostMapping("/register")
public ResponseEntity<Object> register(@RequestBody String mutation){
    ExecutionResult executionResult = userService.getGraphQL().execute(mutation);

    // Check if there are errors
    if(!executionResult.getErrors().isEmpty()){
        return new ResponseEntity<>(executionResult.getErrors().get(0).getMessage(), HttpStatus.UNAUTHORIZED);
    }

    return new ResponseEntity<>(executionResult, HttpStatus.OK);
}

正如我所说，登录工作正常，但是注册会返回我在标题中提到的错误。

我的服务类：

@Value("classpath:graphql-schema/schema.graphqls")
    Resource resource;

    private GraphQL graphQL;

    @Autowired
    private LoginDataFetcher loginDataFetcher;
    @Autowired
    private RegisterDataFetcher registerDataFetcher;

    @PostConstruct
    public void  loadSchema() throws IOException{
    // Get the schema
    File schemaFile = resource.getFile();

    // Parse schema
    TypeDefinitionRegistry typeDefinitionRegistry = new SchemaParser().parse(schemaFile);
    RuntimeWiring runtimeWiring = buildRuntimeWiring();
    GraphQLSchema graphQLSchema = new SchemaGenerator().makeExecutableSchema(typeDefinitionRegistry, runtimeWiring);
    graphQL = GraphQL.newGraphQL(graphQLSchema).build();
}

private RuntimeWiring buildRuntimeWiring() {
    return RuntimeWiring.newRuntimeWiring()
            .type("Query", typeWiring ->
                typeWiring
                    .dataFetcher("login", loginDataFetcher))
            .type("Mutation", typeWiring ->
                typeWiring
                    .dataFetcher("register", registerDataFetcher))
            .build();
}

public GraphQL getGraphQL() {
    return graphQL;
}

我的LoginDataFetcher：

@Autowired
    private AppUserRepository appUserRepository;

    private JwtGenerator jwtGenerator;

    public LoginDataFetcher(JwtGenerator jwtGenerator) {
        this.jwtGenerator = jwtGenerator;
    }

    @Override
    public TokenDAO get(DataFetchingEnvironment dataFetchingEnvironment) {
        String username = dataFetchingEnvironment.getArgument("username");
        String password = dataFetchingEnvironment.getArgument("password");

        AppUser appUser = appUserRepository.findByUsername(username);

        // If user is not foung
        if(appUser == null){
            throw new RuntimeException("Username does not exist");
        }

        // If the user is fount check passwords
        if(!appUser.getPassword().equals(password)){
            throw new RuntimeException("Incorrect password");
        }

        // Generate the token
        String token = jwtGenerator.generate(appUser);

        return new TokenDAO(token);
    }

RegisterDataFetcher：

@Autowired
    private AppUserRepository appUserRepository;

    @Override
    public RegisterDAO get(DataFetchingEnvironment dataFetchingEnvironment) {
        String username = dataFetchingEnvironment.getArgument("username");
        String firstName = dataFetchingEnvironment.getArgument("firstName");
        String lastName = dataFetchingEnvironment.getArgument("lastName");
        String password = dataFetchingEnvironment.getArgument("password");
        String role = dataFetchingEnvironment.getArgument("role");

        AppUser appUser = appUserRepository.findByUsername(username);

        // Check if username exists
        if(appUser != null){
            throw new RuntimeException("Username already taken");
        }

        AppUser newAppUser = new AppUser(username, password, role, firstName, lastName);

        // Save new user
        appUserRepository.save(newAppUser);

        return new RegisterDAO("You have successfully registered");
    }

我在控制台中收到的错误：

graphql.GraphQL                          : Query failed to validate : '{
    register(username: "user", firstName: "Bla", lastName: "Blabla", password: "password", role: "DEVELOPER") {
        message
    }
}'

感谢您的帮助。

更新

根据我得到的答案，我改变了我的架构文件：

query UserQuery{
    login(username: String, password: String){
        token
    }
}

mutation UserMutation{
    register(input: RegisterUserInput) {
        message
    }
}

input RegisterUserInput {
    username: String!
    firstName: String!
    lastName: String!
    password: String!
    role: String!
}

schema {
    query: UserQuery
    mutation: UserMutation
}

但现在我收到了这个错误：

解析类型'query'时，操作类型'UserQuery'不存在解析类型'mutation'时，操作类型'UserMutation'不存在

那么现在的问题是什么？我怎样才能做到这一点？

Answer 1

您告诉GraphQL您正在请求查询，而实际上register是一个Mutation。在编写GraphQL请求时，语法通常遵循以下查询格式：

query someOperationName {
  login {
    # other fields
  }
}

在编写变异时，您只需指定：

mutation someOperationName {
  register {
    # other fields
  }
}

您可以保留操作名称，尽管包含它是一种很好的做法。您可能会看到以下格式的示例：

{
  someQuery {
    # other fields
  }
}

在这种情况下，操作名称和操作类型（查询与变异）都保持不变。这仍然是一个有效的请求，因为当您不使用操作类型时，GraphQL会假设您的意思是query。来自规范：

如果一个文档只包含一个操作，那么该操作可以是未命名的，也可以用速记形式表示，这省略了查询关键字和操作名称。

因此，在您的请求中，GraphQL假设register是一个查询，而实际上它是一个变异，并且它返回了一个错误。

同样，在编写请求时，总是同时包含操作名称和查询/变异关键字是一种很好的做法。

FieldUndefined类型的验证错误：类型'Query'中的字段'register'未定义

更新

1 个答案: