How to get external IP successfully

Question

After reading: Getting the 'external' IP address in Java

code:

public static void main(String[] args) throws IOException
{
    URL whatismyip = new URL("http://automation.whatismyip.com/n09230945.asp");
    BufferedReader in = new BufferedReader(new InputStreamReader(whatismyip.openStream()));

    String ip = in.readLine(); //you get the IP as a String
    System.out.println(ip);
}

I thought I was a winner but I get the following error

Exception in thread "main" java.io.IOException: Server returned HTTP response code: 403 for URL: http://automation.whatismyip.com/n09230945.asp
at sun.net.www.protocol.http.HttpURLConnection.getInputStream(Unknown Source)
at java.net.URL.openStream(Unknown Source)
at getIP.main(getIP.java:12)

I think this is because the server isnt responding quick enough, is there anyway to ensure that it will get the external ip?

EDIT: okay so its getting rejected, anyone else know of another site that can do the same function

the 403 error code for HTTP means Forbidden. I however can go to that website in my browser with no problem. — twain249, Apr 25 '12 at 19:35

score 12 · Answer 1 · answered Sep 26 '13 at 16:22

    public static void main(String[] args) throws IOException 
    {
    URL connection = new URL("http://checkip.amazonaws.com/");
    URLConnection con = connection.openConnection();
    String str = null;
    BufferedReader reader = new BufferedReader(new InputStreamReader(con.getInputStream()));
    str = reader.readLine();
    System.out.println(str);
     }

score 8 · Accepted Answer · edited Jun 14 '12 at 10:49

8

Before you run the following code take a look at this: http://www.whatismyip.com/faq/automation.asp

public static void main(String[] args) throws Exception {

    URL whatismyip = new URL("http://automation.whatismyip.com/n09230945.asp");
    URLConnection connection = whatismyip.openConnection();
    connection.addRequestProperty("Protocol", "Http/1.1");
    connection.addRequestProperty("Connection", "keep-alive");
    connection.addRequestProperty("Keep-Alive", "1000");
    connection.addRequestProperty("User-Agent", "Web-Agent");

    BufferedReader in = 
        new BufferedReader(new InputStreamReader(connection.getInputStream()));

    String ip = in.readLine(); //you get the IP as a String
    System.out.println(ip);
}

edited Jun 14 '12 at 10:49

dacwe

43,066
12
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140

answered Apr 25 '12 at 20:39

Justin

436
1
3
9

1

Down voted as this comment is a bit obsolete and from now on it's not as easy to check ip on this site. What is more above comment is no longer working. – Drachenfels Feb 16 '13 at 20:04
http://checkip.amazonaws.com worked for me instead of URL("http://automation.whatismyip.com/n09230945.asp"); – iltaf khalid Mar 25 '17 at 16:21

Adam Gent · Answer 3 · 2012-04-25T21:09:26.613

While playing with Go I saw your question. I made a quick App on Google App Engine using Go:

Hit this URL:

http://agentgatech.appspot.com/

Java code:

new BufferedReader(new InputStreamReader(new URL('http://agentgatech.appspot.com').openStream())).readLine()

Go code for the app which you can copy and make your own app:

package hello

import (
    "fmt"
    "net/http"
)

func init() {
    http.HandleFunc("/", handler)
}

func handler(w http.ResponseWriter, r *http.Request) {
    fmt.Fprint(w, r.RemoteAddr)
}

score 4 · Answer 4 · answered Apr 25 '12 at 19:45

Some servers has triggers that blocks access from "non-browsers". They understand that you are some kind of automatic app that can do a DOS attack. To avoid this, you can try to use a lib to access the resource and set the "browser" header.

wget works in this way:

 wget  -r -p -U Mozilla http://www.site.com/resource.html

Using Java, you can use the HttpClient lib and set the "User-Agent" header. Look the topic 5 of "Things To Try" section.

Hope this can help you.

score 3 · Answer 5 · answered Apr 25 '12 at 19:35

3

A 403 response indicates that the server is explicitly rejecting your request for some reason. Contact the operator of WhatIsMyIP for details.

answered Apr 25 '12 at 19:35

2

Possibly related to the terms of service at: http://www.whatismyip.com/faq/automation.asp – Flexo Apr 25 '12 at 19:38

score 3 · Answer 6 · edited Apr 01 '13 at 11:11

3

We've set up CloudFlare and as designed they're challenging unfamiliar useragents. If you can set your UA to something common, you should be able to gain access.

edited Apr 01 '13 at 11:11

Rais Alam

6,970
12
53
84

answered May 01 '12 at 14:10

Brian

31
1

score 2 · Answer 7 · answered Apr 25 '12 at 19:36

2

You can use another web service like this; http://freegeoip.net/static/index.html

answered Apr 25 '12 at 19:36

Baran

1,388
6
16
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score 2 · Answer 8 · answered May 26 '16 at 17:35

Using the Check IP address link on AWS worked for me.Please note that MalformedURLException,IOException are to be added as well

public String getPublicIpAddress() throws MalformedURLException,IOException {

    URL connection = new URL("http://checkip.amazonaws.com/");
    URLConnection con = connection.openConnection();
    String str = null;
    BufferedReader reader = new BufferedReader(new InputStreamReader(con.getInputStream()));
    str = reader.readLine();


    return str;
}

Someone Somewhere · Answer 9 · 2018-05-08T22:02:36.817

This is how I do it with rxJava2 and Butterknife. You'll want to run the networking code in another thread because you'll get an exception for running network code on the main thread! I use rxJava instead of AsyncTask because the rxJava cleans up nicely when the user moves on to the next UI before the thread is finished. (this is super useful for very busy UI's)

public class ConfigurationActivity extends AppCompatActivity {

    // VIEWS
    @BindView(R.id.externalip) TextInputEditText externalIp;//this could be TextView, etc.

    // rxJava - note: I have this line in the base class - for demo purposes it's here
    private CompositeDisposable compositeSubscription = new CompositeDisposable();


    @Override
    public void onCreate(Bundle savedInstanceState) {
        super.onCreate(savedInstanceState);
        setContentView(R.layout.my_wonderful_layout);

        ButterKnife.bind(this);

        getExternalIpAsync();
    }

    // note: I have this code in the base class - for demo purposes it's here
    @Override
    protected void onStop() {
        super.onStop();
        clearRxSubscriptions();
    }

    // note: I have this code in the base class - for demo purposes it's here
    protected void addRxSubscription(Disposable subscription) {
        if (compositeSubscription != null) compositeSubscription.add(subscription);
    }

    // note: I have this code in the base class - for demo purposes it's here
    private void clearRxSubscriptions() {
        if (compositeSubscription != null) compositeSubscription.clear();
    }

    private void getExternalIpAsync() {
        addRxSubscription(
                Observable.just("")
                        .map(s -> getExternalIp())
                        .subscribeOn(Schedulers.io())
                        .observeOn(AndroidSchedulers.mainThread())
                        .subscribe((String ip) -> {
                            if (ip != null) {
                                externalIp.setText(ip);
                            }
                        })
        );
    }

    private String getExternalIp() {
        String externIp = null;
        try {
            URL connection = new URL("http://checkip.amazonaws.com/");
            URLConnection con = connection.openConnection(Proxy.NO_PROXY);
            con.setConnectTimeout(1000);//low value for quicker result (otherwise takes about 20secs)
            con.setReadTimeout(5000);
            BufferedReader reader = new BufferedReader(new InputStreamReader(con.getInputStream()));
            externIp = reader.readLine();
        } catch (Exception e) {
            e.printStackTrace();
        }
        return externIp;
    }
}

UPDATE - I've found that URLConnection is really quite shit; it'll take a long time to get a result, not really time out very well, etc. The code below improves the situation with OKhttp

private String getExternalIp() {
    String externIp = "no connection";
    OkHttpClient client = new OkHttpClient();//should have this as a member variable
    try {
        String url = "http://checkip.amazonaws.com/";
        Request request = new Request.Builder().url(url).build();
        Response response = client.newCall(request).execute();
        ResponseBody responseBody = response.body();
        if (responseBody != null) externIp = responseBody.string();
    } catch (IOException e) {
        e.printStackTrace();
    }
    return externIp;
}

How to get external IP successfully

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