如何將 multiPartRequest 的響應轉換為 flutter 中的正常 http 響應？

Question

我必須發送 api 請求才能將文件上傳到服務器。

模式：repo，提供者服務

由於 multipartRequest 發送的響應，我關注失敗

如何將請求轉換為正常的獲取/發布請求？

Answer 1

     Future mutliPartPostRequest(
          {required String method,
          required String url,
          Map<String, String>? headers,
          File? file,
          String? document,
          Map? data}) async {
        Map<String, dynamic> jsonResponse;
        try {
          var request = http.MultipartRequest(method, Uri.parse(url));
          data!.forEach((key, value) {
            request.fields[key] = value.toString();
          });
          request.headers['Authorization'] = headers!['authorization']!;
    
          /// [FILTERING] : the null image if there are no photos in the request it will skip adding the photo in the request
          if (file != null) {
            var picture = await http.MultipartFile.fromPath("photo", file.path);
            request.files.add(picture); //adds the photo to the request
          }
    
          // Checking the document if it is empty, if it is empty it will skip to add the document in the request
          if (document != null) {
            var doc = await http.MultipartFile.fromPath('document', document);
            request.files.add(doc); // adds the document to the request 
          }

///Our case starts from this line, while sending multipart request, the response we get is quite different than the normal json data response. 


// response => stores the response got
          var response = await request.send().timeout(const Duration(seconds: 10)); //sends the request body to the server


// result => coverts the multipart response got from the server to the normal request
          var result = await http.Response.fromStream(response);
          jsonResponse = returnResponse(result); /// returns the response according to the status code from [returnResponse] function
        } on SocketException {
          throw FetchDataException({"detail": "No Internet Connection"});
        }
        return jsonResponse;
      }