【问题标题】:Cannot make a request to the GraphQL with the wrapper in QueryVariables无法使用 QueryVariables 中的包装器向 GraphQL 发出请求
【发布时间】:2021-11-07 23:51:26
【问题描述】:

我正在尝试向 GraphQl 发出请求。我可以在操场上做。 Request in the playground

当我尝试使用颤振发出请求时,我无法在查询变量中创建包装器登录。 这是我来自 Flutter 的代码:

InviteCodeRepository.dart

import 'package:fauna/data/api/graphql_api_client.dart';
import 'package:get/get.dart';
import 'package:graphql_flutter/graphql_flutter.dart';

abstract class InviteCodeRepository {
Future<void> inviteCode(String inviteCode, String phone);
}

class InviteRepositoryImplementation extends GetxController
  implements InviteCodeRepository {
String login = r'''
 mutation login($Login: Login){
login(login: $Login) {
  token
}
}

final _gqlClient = Get.put(GraphQLApiClient(), permanent: true);

@override
Future<void> inviteCode(String inviteCode, String phone) async {
  final variables = {'phone': phone, 'accessCode': inviteCode};
  final QueryResult result = await _gqlClient.mutation(
    login,
    variables: variables,
  );
  print(result.data);
}
}

【问题讨论】:

    标签: flutter graphql mutation query-variables


    【解决方案1】:

    这对我有用: InviteCodeRepository.dart

    import 'package:fauna/data/api/graphql_api_client.dart';
    import 'package:fauna/data/api/graphql_configuration.dart';
    import 'package:get/get.dart';
    import 'package:graphql_flutter/graphql_flutter.dart';
    
    class InviteRepositoryImplementation extends GetxController
        implements InviteCodeRepository {
      String _loginError = "";
      String login = r'''
      mutation login($Login: Login){
      login(login: $Login) {
        token
      }
    }
      ''';
    
    final _gqlClient = Get.put(GraphQLApiClient(), permanent: true);
    
    Future<QueryResult> _logIn(String accessCode, String phone) async {
        final loginData = {'phone': phone, 'accessCode': accessCode};
        final variables = {'Login': loginData};
        final QueryResult result = await _gqlClient.mutation(
          login,
          variables: variables,
        );
        return result;
      }
    
    

    【讨论】:

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