Flutter 颤振通用RESTAPI调用函数

我正在开发我的第一个颤振应用程序。调用rest api并返回结果所需的应用程序。我希望创建

generic

函数来调用RESTAPI。我已经写了下面的代码，但我并没有轻描淡写，我如何才能解码特定模型中的api响应

Future<T> apiRequest<T>(
    String endPoint,
    RequestMethod method, {
    String body = '',
    String token = '',
  }) async {
    http.Response resp;
    final String url = LocalConstants.apiBaseUrl + endPoint;
    final Map<String, String> headers = new Map<String, String>();
    headers.putIfAbsent(
        HttpHeaders.contentTypeHeader, () => 'application/json');
    if (token != null && token.isNotEmpty) {
      headers.putIfAbsent(
          HttpHeaders.authorizationHeader, () => 'Bearer ' + token);
    }
    try {
      if (method == RequestMethod.get) {
        resp = await http.get(
          url,
          headers: headers,
        );
      } else if (method == RequestMethod.put) {
        resp = await http.put(
          url,
          headers: headers,
          body: body,
        );
      } else if (method == RequestMethod.post) {
        resp = await http.post(
          url,
          headers: headers,
          body: body,
        );
      } else if (method == RequestMethod.delete) {
        resp = await http.delete(
          url,
          headers: headers,
        );
      }
      if (resp != null && this.validateResponse(resp)) {
        return json.decode(resp.body);
      }
      // else {
      //   Response resp = new Response();
      //   resp.respMsg = LocalConstants.genericError;
      //   resp.respCode = LocalConstants.resp_failure;
      //   Response.
      // }
    } on TimeoutException catch (e) {
      //handleTimeout();
    } on SocketException catch (e) {
      print('Socket Error: $e');
      //handleTimeout();
    } on Error catch (e) {
      print('General Error: $e');
      //showError();
    }
  }

---

如何解码body

json.decode（resp.body）

到类型为

T

？

的

GenericResp

，您可以添加一个泛型参数，将json数据反序列化到

GenericResp

。诸如此类：

Future<T> apiRequest<T>(
    String endPoint,
    RequestMethod method, T Function(Object json) fromJson, {
      String body = '',
      String token = '',
    }) async { ... }

然后一个电话会是这样的：

await ApiService.newInstance(context).apiRequest<GenericResp>('/api/people',
    RequestMethod.get, (json) => GenericResp.fromJson(json));

wait-ApiService.newInstance（context）.apirest（'/api/people'，
RequestMethod.get，（json）=>GenericResp.fromJson（json））；

无法理解您对t类型的GenericResp所说的话，GenericResp已经是一个

类了

？

Future<T> apiRequest<T>(
    String endPoint,
    RequestMethod method, T Function(Object json) fromJson, {
      String body = '',
      String token = '',
    }) async { ... }

if (resp != null && this.validateResponse(resp)) {
  return fromJson(json.decode(resp.body));
}

await ApiService.newInstance(context).apiRequest<GenericResp>('/api/people',
    RequestMethod.get, (json) => GenericResp.fromJson(json));