Java socket connect error - Исправление ошибок и поиск оптимальных решений проблем

I’m trying to implement a TCP connection, everything works fine from the server’s side but when I run the client program (from client computer) I get the following error:

java.net.ConnectException: Connection refused
        at java.net.PlainSocketImpl.socketConnect(Native Method)
        at java.net.PlainSocketImpl.doConnect(PlainSocketImpl.java:351)
        at java.net.PlainSocketImpl.connectToAddress(PlainSocketImpl.java:213)
        at java.net.PlainSocketImpl.connect(PlainSocketImpl.java:200)
        at java.net.SocksSocketImpl.connect(SocksSocketImpl.java:432)
        at java.net.Socket.connect(Socket.java:529)
        at java.net.Socket.connect(Socket.java:478)
        at java.net.Socket.<init>(Socket.java:375)
        at java.net.Socket.<init>(Socket.java:189)
        at TCPClient.main(TCPClient.java:13)

I tried changing the socket number in case it was in use but to no avail, does anyone know what is causing this error & how to fix it.

The Server Code:

//TCPServer.java

import java.io.*;
import java.net.*;

class TCPServer {
    public static void main(String argv[]) throws Exception {
        String fromclient;
        String toclient;

        ServerSocket Server = new ServerSocket(5000);

        System.out.println("TCPServer Waiting for client on port 5000");

        while (true) {
            Socket connected = Server.accept();
            System.out.println(" THE CLIENT" + " " + connected.getInetAddress()
                    + ":" + connected.getPort() + " IS CONNECTED ");

            BufferedReader inFromUser = new BufferedReader(
                    new InputStreamReader(System.in));

            BufferedReader inFromClient = new BufferedReader(
                    new InputStreamReader(connected.getInputStream()));

            PrintWriter outToClient = new PrintWriter(
                    connected.getOutputStream(), true);

            while (true) {

                System.out.println("SEND(Type Q or q to Quit):");
                toclient = inFromUser.readLine();

                if (toclient.equals("q") || toclient.equals("Q")) {
                    outToClient.println(toclient);
                    connected.close();
                    break;
                } else {
                    outToClient.println(toclient);
                }

                fromclient = inFromClient.readLine();

                if (fromclient.equals("q") || fromclient.equals("Q")) {
                    connected.close();
                    break;
                } else {
                    System.out.println("RECIEVED:" + fromclient);
                }

            }

        }
    }
}

The Client Code:

//TCPClient.java

import java.io.*;
import java.net.*;

class TCPClient {
    public static void main(String argv[]) throws Exception {
        String FromServer;
        String ToServer;

        Socket clientSocket = new Socket("localhost", 5000);

        BufferedReader inFromUser = new BufferedReader(new InputStreamReader(
                System.in));

        PrintWriter outToServer = new PrintWriter(
                clientSocket.getOutputStream(), true);

        BufferedReader inFromServer = new BufferedReader(new InputStreamReader(
                clientSocket.getInputStream()));

        while (true) {

            FromServer = inFromServer.readLine();

            if (FromServer.equals("q") || FromServer.equals("Q")) {
                clientSocket.close();
                break;
            } else {
                System.out.println("RECIEVED:" + FromServer);
                System.out.println("SEND(Type Q or q to Quit):");

                ToServer = inFromUser.readLine();

                if (ToServer.equals("Q") || ToServer.equals("q")) {
                    outToServer.println(ToServer);
                    clientSocket.close();
                    break;
                } else {
                    outToServer.println(ToServer);
                }
            }
        }
    }
}

asked Jul 29, 2011 at 16:37

Samantha CataniaSamantha Catania

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This exception means that there is no service listening on the IP/port you are trying to connect to:

You are trying to connect to the wrong IP/Host or port.
You have not started your server.
Your server is not listening for connections.
On Windows servers, the listen backlog queue is full.

tk_

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answered Jul 29, 2011 at 16:41

Collin PriceCollin Price

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I would check:

Host name and port you’re trying to connect to
The server side has managed to start listening correctly
There’s no firewall blocking the connection

The simplest starting point is probably to try to connect manually from the client machine using telnet or Putty. If that succeeds, then the problem is in your client code. If it doesn’t, you need to work out why it hasn’t. Wireshark may help you on this front.

answered Jul 29, 2011 at 16:39

Jon SkeetJon Skeet

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You have to connect your client socket to the remote ServerSocket. Instead of

Socket clientSocket = new Socket("localhost", 5000);

Socket clientSocket = new Socket(serverName, 5000);

The client must connect to serverName which should match the name or IP of the box on which your ServerSocket was instantiated (the name must be reachable from the client machine). BTW: It’s not the name that is important, it’s all about IP addresses…

answered Jul 29, 2011 at 17:21

I had the same problem, but running the Server before running the Client fixed it.

answered Jul 25, 2012 at 18:09

Dao LamDao Lam

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One point that I would like to add to the answers above is my experience—

«I hosted on my server on localhost and was trying to connect to it through an android emulator by specifying proper URL like http://localhost/my_api/login.php . And I was getting connection refused error«

Point to note — When I just went to browser on the PC and use the same URL (http://localhost/my_api/login.php) I was getting correct response

so the Problem in my case was the term localhost which I replaced with the IP for my server (as your server is hosted on your machine) which made it reachable from my emulator on the same PC.

To get IP for your local machine, you can use ipconfig command on cmd
you will get IPv4 something like 192.68.xx.yy
Voila ..that’s your machine’s IP where you have your server hosted.
use it then instead of localhost

http://192.168.72.66/my_api/login.php

Note — you won’t be able to reach this private IP from any node outside this computer. (In case you need ,you can use Ngnix for that)

answered Feb 26, 2018 at 16:50

eRaisedToXeRaisedToX

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I had the same problem with Mqtt broker called vernemq.but solved it by adding the following.

$ sudo vmq-admin listener show

to show the list o allowed ips and ports for vernemq

$ sudo vmq-admin listener start port=1885 -a 0.0.0.0 --mountpoint /appname --nr_of_acceptors=10 --max_connections=20000

to add any ip and your new port. now u should be able to connect without any problem.

Hope it solves your problem.

answered Apr 5, 2016 at 7:58

MrOnyanchaMrOnyancha

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Hope my experience may be useful to someone. I faced the problem with the same exception stack trace and I couldn’t understand what the issue was. The Database server which I was trying to connect was running and the port was open and was accepting connections.

The issue was with internet connection. The internet connection that I was using was not allowed to connect to the corresponding server. When I changed the connection details, the issue got resolved.

answered Nov 17, 2014 at 12:01

phoenixphoenix

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In my case, I gave the socket the name of the server (in my case «raspberrypi»), and instead an IPv4 address made it, or to specify, IPv6 was broken (the name resolved to an IPv6)

answered Dec 21, 2016 at 18:20

ZhyanoZhyano

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In my case, I had to put a check mark near Expose daemon on tcp://localhost:2375 without TLS in docker setting (on the right side of the task bar, right click on docker, select setting)

answered Aug 23, 2017 at 13:42

user1419243user1419243

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i got this error because I closed ServerSocket inside a for loop that try to accept number of clients inside it (I did not finished accepting all clints)

so be careful where to close your Socket

answered May 21, 2016 at 21:07

I had same problem and the problem was that I was not closing socket object.After using socket.close(); problem solved.
This code works for me.

ClientDemo.java

public class ClientDemo {
    public static void main(String[] args) throws UnknownHostException,
            IOException {
        Socket socket = new Socket("127.0.0.1", 55286);
        OutputStreamWriter os = new OutputStreamWriter(socket.getOutputStream());
        os.write("Santosh Karna");
        os.flush();
        socket.close();
    }
}

and
ServerDemo.java

public class ServerDemo {
    public static void main(String[] args) throws IOException {
        System.out.println("server is started");
        ServerSocket serverSocket= new ServerSocket(55286);
        System.out.println("server is waiting");
        Socket socket=serverSocket.accept();
        System.out.println("Client connected");
        BufferedReader reader=new BufferedReader(new InputStreamReader(socket.getInputStream()));
        String str=reader.readLine();
        System.out.println("Client data: "+str);
        socket.close();
        serverSocket.close();

    }
}

answered May 14, 2017 at 17:13

Santosh KarnaSantosh Karna

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I changed my DNS network and it fixed the problem

answered Nov 23, 2019 at 9:51

You probably didn’t initialize the server or client is trying to connect to wrong ip/port.

answered Aug 10, 2020 at 16:21

Change local host to your ip address

 localhost
 //to you local ip
 192.168.xxxx

answered Apr 28, 2021 at 10:46

Gabriel RogathGabriel Rogath

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I saw the same error message «»java.net.ConnectException: Connection refused» in SQuirreLSQL when it was trying to connect to a postgresql database through an ssh tunnel.

Example of opening tunel:

Example of error in Squirrel with Postgresql:

It was trying to connect to the wrong port. After entering the correct port, the process execution was successful.

See more options to fix this error at: https://stackoverflow.com/a/6876306/5857023

answered Jan 6, 2022 at 19:57

GenivanGenivan

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In my case, with server written in c# and client written in Java, I resolved it by specifying hostname as ‘localhost‘ in the server, and ‘[::1]‘ in the client. I don’t know why that is, but specifying ‘localhost’ in the client did not work.
Supposedly these are synonyms in many ways, but apparently, not not a 100% match. Hope it helps someone avoid a headache.

answered Aug 12, 2022 at 15:07

AlexeiOstAlexeiOst

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For those who are experiencing the same problem and use Spring framework, I would suggest to check an http connection provider configuration. I mean RestTemplate, WebClient, etc.

In my case there was a problem with configured RestTemplate (it’s just an example):

public RestTemplate localRestTemplate() {
   Proxy proxy = new Proxy(Proxy.Type.HTTP, new InetSocketAddress("localhost", <some port>));
   SimpleClientHttpRequestFactory clientHttpReq = new SimpleClientHttpRequestFactory();
   clientHttpReq.setProxy(proxy);
   return new RestTemplate(clientHttpReq);
}

I just simplified configuration to:

public RestTemplate restTemplate() {
   return new RestTemplate(new SimpleClientHttpRequestFactory());
}

And it started to work properly.

answered Jan 24 at 17:45

There is a service called MySQL80 that should be running to connect to the database

for windows you can access it by searching for services than look for MySQL80 service and make sure it is running

answered Jul 16, 2021 at 19:50

It could be that there is a previous instance of the client still running and listening on port 5000.

answered Jan 10, 2013 at 18:27

Michael MunseyMichael Munsey

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Источник

Methods declared in class java.lang.Object

Constructor Detail

Socket

If there is a security manager, its checkConnect method is called with the proxy host address and port number as its arguments. This could result in a SecurityException.

Examples:

Socket s = new Socket(Proxy.NO_PROXY); will create a plain socket ignoring any other proxy configuration.
Socket s = new Socket(new Proxy(Proxy.Type.SOCKS, new InetSocketAddress(«socks.mydom.com», 1080))); will create a socket connecting through the specified SOCKS proxy server.

Socket

If the specified host is null it is the equivalent of specifying the address as InetAddress.getByName (null) . In other words, it is equivalent to specifying an address of the loopback interface.

If the application has specified a server socket factory, that factory’s createSocketImpl method is called to create the actual socket implementation. Otherwise a «plain» socket is created.