类型子选择权的验证误差:字段类型的子选择所需的子选择

Question

我使用graphql-spqr库问题是"类型subselectionRequired类型的验证错误：类型Timestamp所需的子选择"也许在查询时间戳中有表达或实体中的格式

{"query":
"{findUserPointByUserId(userId:73){rowNum userAccountPointUserId totalPoint pointTypeDescription point userAccountCreatedDate} findUserAccountImgByUserId(userId:73){imageId,userId,presentImgNum}}"

}

错误

{
    "errors": [
        {
            "message": "Validation error of type SubSelectionRequired: Sub selection required for type Timestamp of field userAccountCreatedDate",
            "locations": [
                {
                    "line": 1,
                    "column": 103
                }
            ]
        }
    ]
}

实体

@NoArgsConstructor
@AllArgsConstructor
@Getter
@Setter
@Entity
@Table(name = "view_user_account_point", schema = "public", catalog = "corus")
public class ViewUserAccountPoint {
    @Id
    @Basic
    @GraphQLQuery(name = "rowNum")
    @Column(name = "row_num", nullable = true)
    private Long rowNum;
    @Basic
    @Column(name = "user_account_point_userid", nullable = true)
    @GraphQLQuery(name = "userAccountPointUserId")
    private Integer userAccountPointUserId;
    @Basic
    @Column(name = "subject_id", nullable = true)
    @GraphQLQuery(name = "subjectId")
    private Integer subjectId;
    @Basic
    @Column(name = "point", nullable = true)
    @GraphQLQuery(name = "point")
    private Integer point;
    @Basic
    @Column(name = "user_account_point_typeid", nullable = true)
    @GraphQLQuery(name = "userAccountPointTypeId")
    private Integer userAccountPointTypeId;
    @Basic
    @Column(name = "date_created", nullable = true)
    @GraphQLQuery(name = "userAccountCreatedDate")
    private Timestamp userAccountCreatedDate;

服务

public List<ViewUserAccountPoint> findUserPointByUserId(@GraphQLArgument(name = "userId") Integer userId){
        return viewUserAccountPointRepository.findByUserAccountPointUserIdOrderByUserAccountCreatedDateDesc(userId);
    }

控制器

 private final GraphQL graphQL;
    public UserController(UserAccountService userAccountService) {
        GraphQLSchema schema = new GraphQLSchemaGenerator()
                .withResolverBuilders(
                        //Resolve by annotations
                        new AnnotatedResolverBuilder())
                .withOperationsFromSingleton(userAccountService,UserAccountService.class)
                .withValueMapperFactory(new JacksonValueMapperFactory())
                .generate();
        graphQL = GraphQL.newGraphQL(schema).build();
    }
    @PostMapping(value = "/graphql", consumes = MediaType.APPLICATION_JSON_UTF8_VALUE, produces = MediaType.APPLICATION_JSON_UTF8_VALUE)
    @ResponseBody
    public Map<String, Object> graphql(@RequestBody Map<String, String> request, HttpServletRequest raw) {
        ExecutionResult executionResult = graphQL.execute(ExecutionInput.newExecutionInput()
                .query(request.get("query"))
                .operationName(request.get("operationName"))
                .context(raw)
                .build());
        return executionResult.toSpecification();
    }

我搜索所有查询时间戳格式但是，我找不到我希望听到解决方案。谢谢

Answer 1

出于某种原因，Timestamp的映射不正确。它最终是一个对象，而不是标量。如您打开的问题所述，尚不清楚代码中的Timestamp来自哪里。

java.sql.Timestamp在最近版本的GraphQL SPQR中受支持，因此您可能会使用较旧的版本。

如果不是这种情况，则意味着Timestamp除了java.sql.Timestamp之外，您需要为其注册一个自定义映射器。

public class TimestampMapper implements TypeMapper {
    // Define the scalar as needed, see io.leangen.graphql.util.Scalars for inspiration
    private static final GraphQLScalarType TIMESTAMP = ...;
    @Override
    public GraphQLOutputType toGraphQLType(AnnotatedType javaType, OperationMapper operationMapper, Set<Class<? extends TypeMapper>> mappersToSkip, BuildContext buildContext) {
        return TIMESTAMP; //it's important to always return the same instance
    }
    @Override
    public GraphQLInputType toGraphQLInputType(AnnotatedType javaType, OperationMapper operationMapper, Set<Class<? extends TypeMapper>> mappersToSkip, BuildContext buildContext) {
        return TIMESTAMP; //same as above
    }
    @Override
    public boolean supports(AnnotatedType type) {
        return ClassUtils.isSuperClass(Timestamp.class, type);
    }
}

然后注册您的映射器：

generator.withTypeMappers(new TimestampMapper())

Answer 2

对我的情况来说是不正确的查询体，请确保您有正确的查询体。