How can I ignore a future exception of a http.post call? In the following code I want to send a message to the server, but I don't care about the response:

                      http.post('http:/10.0.2.2/log-message',
                          body: json.encode({
                            'message': 'Event occurred',
                          }));

If the server at that URL is not running, this exception is thrown:

SocketException (SocketException: Connection refused (OS Error: Connection refused, errno = 111), address = 10.0.2.2, port = 41874)

I can prevent this exception from being thrown by doing await on the above call and wrapping that around a try-catch, but I don't want to block. The following code still results in the above exception:

                      http.post('http:/10.0.2.2/log-message',
                          body: json.encode({
                            'message': 'Event occurred',
                          }));
                        .catchError((_) => http.Response('Logging message failed', 404));

ignore() and onError() have the same result.

I want ignore whatever exception http.post could throw without having to do an await, which blocks the code.

CodePudding user response：

If you just print the error that you catched, it should't block the app.

http.post('http:/10.0.2.2/log-message',
                      body: json.encode({
                        'message': 'Event occurred',
                      }));
                    .catchError((_) => print('Logging message failed'));

CodePudding user response：

There is no way to catch a SocketException without waiting for the result of the http.post operation, because it occurs later than the request is sent. Whether you use async/await with try/catch or .then and .catchError is your choice, but the preferred way is async/await:

Future<Response> myFunction async {
  try {
    return http.post(...);
  } on SocketException {
    // handle the exception here as you like
  }
}