如何成功获取外部IP

o f*_*o f 3 java ip

阅读后:用Java获取"外部"IP地址

码:

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);
}
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我以为我是胜利者,但我得到以下错误

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)
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我认为这是因为服务器没有足够快的响应,无论如何确保它将获得外部IP?

编辑:好吧所以它被拒绝,其他人都知道另一个可以做同样功能的网站

Sou*_*301 11

    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);
     }
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Jus*_*tin 8

在运行以下代码之前,请查看以下内容: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);
}
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Ada*_*ent 8

在玩Go时,我看到了你的问题.我使用Go在Google App Engine上制作了一个快速应用程序:

点击此网址:

http://agentgatech.appspot.com/

Java代码:

new BufferedReader(new InputStreamReader(new URL('http://agentgatech.appspot.com').openStream())).readLine()
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转到应用程序的代码,您可以复制并制作自己的应用程序:

package hello

import (
    "fmt"
    "net/http"
)

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

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