java.net.ConnectException: failed to connect to /192.168.253.3 (port 2468): connect failed: ECONNREFUSED (Connection refused)

Question

I want to transfer some data between PC and a mobile phone with WiFi.

This is how I get the WiFi IP address:

String ip  = String.format(                     "%d.%d.%d.%d",                     (wifiInfo.getIpAddress() & 0xff),                     (wifiInfo.getIpAddress() >> 8 & 0xff),                     (wifiInfo.getIpAddress() >> 16 & 0xff),                     (wifiInfo.getIpAddress() >> 24 & 0xff));  new Recive().execute(ip); 

This is the code about sending a message to the PC:

Socket socket = null; String message = "test\r\n"; protected Void doInBackground(String... urls) {     try {         Log.i("ip", urls[0]);         socket = new Socket(urls[0], 2468);         toserver = new DataOutputStream(socket.getOutputStream());         toserver.writeBytes(message);          toserver.flush();         toserver.close();         socket.close();         return null;     } catch (Exception e) {         Log.i("e", e.toString());         return null;     } } 

But an error occurs,

java.net.ConnectException: failed to connect to /192.168.253.3 (port 2468): connect failed: ECONNREFUSED (Connection refused)

Besides, I use a android phone to run the app.

Answer 1

A connect failed: ECONNREFUSED (Connection refused) most likely means that there is nothing listening on that port AND that IP address. Possible explanations include:

• the service has crashed or hasn't been (successfully!) started,
• your client is trying to connect using the wrong IP address or port,
• your client is trying to connect using a DNS name that resolves to the wrong IP, or
• server access is being blocked by a firewall that is "refusing" on the server/service's behalf. This is pretty unlikely given that normal practice (these days) is for firewalls to "blackhole" all unwanted connection attempts.

---

Note that while you have an array variable called urls, it cannot contain real URLs. There is no overload of the Socket constructor that takes a real URL in any form. Indeed, if you supplied a URL in string form like this:

 new Socket("http://example.com", 42) 

the result would be a different exception. Likewise, if you attempt to connect to an IP address on a network that you can't route to (e.g. "a different WiFi network"), then you will get a different exception; e.g. "host not found", "no route to host" or "no route to network".

Answer 2

A common mistake during development of an android app running on a Virtual Device on your dev machine is to forget that the virtual device is not the same host as your dev machine. So if your server is running on your dev machine you cannot use a "http://localhost/..." url as that will look for the server endpoint on the virtual device not your dev machine.