ip

What is the correct way to convert my IP Packet data into a String?

二次信任 提交于 2020-01-06 05:47:14
问题 When I receive an IP Packet with my Android VpnService I read the headers first (as here), then I try to print the received data as follows: int lengthRemaining = packet.remaining(); if (lengthRemaining > 0) { byte[] data = new byte[lengthRemaining]; packet.get(data, packet.arrayOffset(), lengthRemaining); Log.d(TAG, "Packet-Data: " + new String(data, Charset.forName("UTF-8"))); } This results in out put like the following: Packet-Data: ����5��(��������������������www�google�com������ Or

Dynamic loose source routing in IP networks

时光毁灭记忆、已成空白 提交于 2020-01-06 03:11:11
问题 Okay guys, I have this idea to recreate a program wich could find the best rout in IP networks. I found one guy on the internet who did it, and he shared some code..but whatever I'm doing with it, I can't make it work. I'm using Visual Studio 2012 with latest BOOST libs, C++. The IP protocol has a possibility of data transmission by the non-standard routes. therefore, at the IP-packege header, you can specify specific source routing option of IP-addresses list network routers through which

When a socket is disconnected in Go, can I know if it was a good or bad close?

断了今生、忘了曾经 提交于 2020-01-05 11:26:35
问题 Say I open a connection to a server like: conn, _ := net.DialTimeout("tcp", "127.0.0.1:2121", 10000000) and I do some stuff and then call conn.Close() that's option 1. But what if I don't call Close on the conn and just call os.Exit(1) leaving the connection still open. That's option 2. i.e. I'm trying to simulate a user uploading a file and either reaching the ending and really Closing the connection... or ending the stream being writen to the socket but never officially closing it. Back on

Java TCP/IP Server - Client Communication over the internet

和自甴很熟 提交于 2020-01-05 10:37:20
问题 I'd like to get a tcp/ip connection working over the internet. I already have two classes, one tcpserver which handles requests and a tcpclient which connects, sends and receives data. Until now I had it working perfectly on local networks. The usual 127.0.0.1 / localhost and my 192.168.xxx.xxx adresses are not giving any problems. My question is, what do I have to do to make it work over the internet. As I might want to do some research involving a simple http/server I need this to work. I

Java TCP/IP Server - Client Communication over the internet

筅森魡賤 提交于 2020-01-05 10:35:53
问题 I'd like to get a tcp/ip connection working over the internet. I already have two classes, one tcpserver which handles requests and a tcpclient which connects, sends and receives data. Until now I had it working perfectly on local networks. The usual 127.0.0.1 / localhost and my 192.168.xxx.xxx adresses are not giving any problems. My question is, what do I have to do to make it work over the internet. As I might want to do some research involving a simple http/server I need this to work. I

Remote IP Address for HttpServletRequest Object being obscured by Apache Proxy

杀马特。学长 韩版系。学妹 提交于 2020-01-05 05:54:27
问题 I am writing an application in Java that serves a web page with data from an underlying database. I am limiting access to the web page based on some IP restrictions. Basically any IP that falls into the 'Accepted' range will be allowed to access the web page and any IP outside of this range will be redirected to an error page. To get the IP address of the user attempting to access the page I am using the following: String userIPAddress = request.getRemoteAddr(); Where 'request' is my

Remote IP Address for HttpServletRequest Object being obscured by Apache Proxy

 ̄綄美尐妖づ 提交于 2020-01-05 05:54:09
问题 I am writing an application in Java that serves a web page with data from an underlying database. I am limiting access to the web page based on some IP restrictions. Basically any IP that falls into the 'Accepted' range will be allowed to access the web page and any IP outside of this range will be redirected to an error page. To get the IP address of the user attempting to access the page I am using the following: String userIPAddress = request.getRemoteAddr(); Where 'request' is my

Replace IP Address from file

假装没事ソ 提交于 2020-01-04 07:09:30
问题 Sample.txt file: ......................... some log file entries some log file entries some log file entries some log file entries This system ip is not found some log file entries some log file entries some log file entries This system IP is 172.16.80.10 some log file entries some log file entries This system IP:172.16.80.10 some log file entries some log file entries some log file entries Hostname::ip-172.16.80.10.ec2.internal some log file entries some log file entries ....................

Why should I store both REMOTE and FORWARDED as user's IP?

老子叫甜甜 提交于 2020-01-04 05:20:28
问题 Currently I get usr's IP like this: if ( isset($_SERVER['HTTP_X_FORWARDED_FOR']) ){ $ip = $_SERVER['HTTP_X_FORWARDED_FOR']; } elseif ( isset($_SERVER['REMOTE_ADDR']) ) { $ip = $_SERVER['REMOTE_ADDR']; } // IPs +----+----------------+-------------+ | id | user_ip | date_time | +----+----------------+-------------+ | 1 | 43.12.9.9 | 1468070172 | | 2 | 173.3.0.1 | 1468070667 | +----+----------------+-------------+ But now, I read this in here: if you are going to save the $_SERVER['HTTP_X

Linux C: Get default interface's IP address

♀尐吖头ヾ 提交于 2020-01-03 17:48:30
问题 My question is about the following code (in this link): #include <stdio.h> #include <sys/types.h> #include <ifaddrs.h> #include <netinet/in.h> #include <string.h> #include <arpa/inet.h> int main (int argc, const char * argv[]) { struct ifaddrs * ifAddrStruct = NULL; struct ifaddrs * ifa = NULL; void * tmpAddrPtr = NULL; getifaddrs(&ifAddrStruct); for (ifa = ifAddrStruct; ifa != NULL; ifa = ifa->ifa_next) { if (ifa ->ifa_addr->sa_family==AF_INET) { // Check it is // a valid IPv4 address