postRequest method

Future postRequest(
  1. UrlRequest request
)

Sends an HTTP POST request with the UrlRequest object waiting for a 200, 202 or 204 status code.

Implementation

Future<dynamic> postRequest(UrlRequest request) async {
  final response = await http.post(request.url,
      headers: request.headers,
      body: request.body,
      encoding: request.encoding);

  if (response.statusCode == 200) {
    try {
      return jsonDecode(response.body);
    } catch (error) {
      return Future.error(error);
    }
  } else {
    return Future.error("The server did not return a 200 code");
  }
}