Flutter 如何在颤振post请求中传递Map作为键的值?

Flutter 如何在颤振post请求中传递Map作为键的值?,flutter,dart,flutter-dependencies,Flutter,Dart,Flutter Dependencies,代码 Future editResume()异步{ var url='1〕https://example.com", String body = jsonEncode({ "resume".toString(): { "id".toString(): 22720456, "name".toString(): "Accounting-Accountant-2020-11-05".toString()

代码 Future editResume()异步{ var url='1〕https://example.com",

String body = jsonEncode({
  "resume".toString(): {
    "id".toString(): 22720456,
    "name".toString(): "Accounting-Accountant-2020-11-05".toString(),
    "years".toString(): 2,
    "functional_area_id".toString(): 1,
    "sub_functional_area_id".toString(): 1,
    "industry_id".toString(): null,
    "user_id".toString(): 191451,
    "resume_type_id".toString(): 29,
    "created_at".toString(): "2020-11-05T11:03:52.846+05:30".toString(),
    "updated_at".toString(): "2021-04-27T14:05:09.457+05:30".toString(),
    "hide_employer_name".toString(): false,
    "hide_educational_marks".toString(): false,
    "hide_email".toString(): false
  }
});

try {
  var dio = Dio();

  print(body);
  final response = await dio.post(
    url,
    data: body,
    options: Options(
      contentType: Headers.jsonContentType,
      responseType: ResponseType.json,
    ),
  );

  final responseData = json.decode(response.data);
  if (responseData["message"] == "All Good") {
    throw HttpException('Something went wrong');
  } else {
    throw HttpException('Something went wrong');
  }
} catch (error) {
  print(error.toString());
  // return error;
}
} 我正在将此代码放入providers.dart文件中,一旦我点击请求,它就会向我提供以下响应: I/flatter(9781):DioError[DioErrorType.response]:Http状态错误[404] I/flatter(9781):#0 DioMixin.assureDioError(包:dio/src/dio_mixin.dart:819:20) I/flatter(9781):#1 DioMixin.(发送请求#文件包:dio/src/dio_mixin.dart:678:13) I/颤振(9781): 我/弗利特(9781):#2迪奥米辛。拿。。(包装:dio/src/dio_mixin.dart) I/颤振(9781):

最糟糕的是,如果我在《邮递员》上做了同样的事情,用同样的数据点击请求,我会得到正确/预期的响应