postUrl method

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

Opens an HTTP connection using the POST method.

The URL to use is specified in url.

See openUrl for details.

Implementation

@override
Future<HttpClientRequest> postUrl(Uri uri) {
  checkOverride(uri);
  if (shouldOverride) {
    return fakeHttpClient.postUrl(uri);
  }

  return realHttpClient.postUrl(uri);
}