postUrl method

  1. @override
Future<HttpClientRequest> postUrl(
  1. Uri url
)
override

Opens a HTTP connection using the POST method.

The URL to use is specified in url.

See openUrl for details.

Implementation

@override
Future<HttpClientRequest> postUrl(Uri url) async {
  await this.resolveProxies(url);
  return await _delegate.postUrl(url);
}