send method

@override Future<StreamedResponse> send(BaseRequest request)

Sends an HTTP request and asynchronously returns the response.

Implementers should call BaseRequest.finalize to get the body of the request as a ByteStream. They shouldn't make any assumptions about the state of the stream; it could have data written to it asynchronously at a later point, or it could already be closed when it's returned. Any internal HTTP errors should be wrapped as ClientExceptions.

Implementation

@override
Future<http.StreamedResponse> send(http.BaseRequest request) async {
  String ts = DateTime.now().toUtc().toIso8601String();

  request = trace ? await _logRequest(ts, request) : request;

  http.StreamedResponse response;

  try {
    final ex = forceException;
    if (ex != null) {
      forceException = null;
      throw ex;
    }
    if (forceThrottleDelay != null) {
      final delay = forceThrottleDelay!.inMilliseconds;
      forceThrottleDelay = null;
      response = http.StreamedResponse(
        Stream<List<int>>.fromIterable([]),
        HttpStatusCode.tooManyRequests,
        reasonPhrase: 'Forced Throttle',
        headers: {HttpHeader.msRetryAfterMs: delay.toString()},
      );
    } else if (forceForbidden) {
      response = http.StreamedResponse(
        Stream<List<int>>.fromIterable([]),
        HttpStatusCode.forbidden,
        reasonPhrase: 'Forced Forbidden',
      );
    } else {
      response = await _http.send(request);
    }
    response = trace ? await _logResponse(ts, response) : response;
  } catch (ex) {
    if (trace) {
      _logException(ts, ex);
    }
    rethrow;
  }

  return response;
}