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  • Calculate an Internet (aka IP, aka RFC791) checksum in C#

    - by Pat
    Interestingly, I can find implementations for the Internet Checksum in almost every language except C#. Does anyone have an implementation to share? Remember, the internet protocol specifies that: "The checksum field is the 16 bit one's complement of the one's complement sum of all 16 bit words in the header. For purposes of computing the checksum, the value of the checksum field is zero." More explanation can be found from Dr. Math. There are some efficiency pointers available, but that's not really a large concern for me at this point. Please include your tests! (Edit: Valid comment regarding testing someone else's code - but I am going off of the protocol and don't have test vectors of my own and would rather unit test it than put into production to see if it matches what is currently being used! ;-) Edit: Here are some unit tests that I came up with. They test an extension method which iterates through the entire byte collection. Please comment if you find fault in the tests. [TestMethod()] public void InternetChecksum_SimplestValidValue_ShouldMatch() { IEnumerable<byte> value = new byte[1]; // should work for any-length array of zeros ushort expected = 0xFFFF; ushort actual = value.InternetChecksum(); Assert.AreEqual(expected, actual); } [TestMethod()] public void InternetChecksum_ValidSingleByteExtreme_ShouldMatch() { IEnumerable<byte> value = new byte[]{0xFF}; ushort expected = 0xFF; ushort actual = value.InternetChecksum(); Assert.AreEqual(expected, actual); } [TestMethod()] public void InternetChecksum_ValidMultiByteExtrema_ShouldMatch() { IEnumerable<byte> value = new byte[] { 0x00, 0xFF }; ushort expected = 0xFF00; ushort actual = value.InternetChecksum(); Assert.AreEqual(expected, actual); }

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  • Question regarding ip checksum code

    - by looktt
    unsigned short /* this function generates header checksums */ csum (unsigned short *buf, int nwords) { unsigned long sum; for (sum = 0; nwords > 0; nwords--) // add words(16bits) together sum += *buf++; sum = (sum >> 16) + (sum & 0xffff); //add carry over sum += (sum >> 16); //what does this step do??? add possible left-over //byte? But isn't it already added in the loop (if //any)? return ((unsigned short) ~sum); } I assume nwords in the number of 16bits word, not 8bits byte (if there are odd byte, nword is rounded to next large), is it correct? The line sum = (sum 16) + (sum & 0xffff) is to add carry over to make 16bit complement sum += (sum 16); What's the purpose of this step? Add left-over byte? How? Thanks!

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  • Invalid Html Response and JS Errors when you open your Application in Visual Studio 2013

    - by imran_ku07
     I was working on an application which uses Telerik controls. The application was working fine for a while. Suddenly, the application stopped working. I mean lot of my application pages becoming very very ugly. I found JavaScript errors on every Browser's console. When I check the page view-source, the generated HTML was messy and invalid. This was only happening with my local machine. If someone else on my network accesses my application pages, he will get the correct HTML and no JavaScript errors. My mind was blowing because the same page was generating invalid HTML(and JavaScript errors) when I access the page using a local browser but generate correct HTML(and no JavaScript errors) when someone else access my application page remotely. Then I realized that I the only change I made last was opening my application in Visual Studio 2013 RTM which I installed few days ago. I closed the Visual Studio 2013, everything work like a charm. Then I became100% sure that this is only happening due to new Visual Studio 2013 feature called Browser Link. I just open the application again and add this in web.config. Everything become fine Happy coding :)   <add key="vs:EnableBrowserLink" value="false" />

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  • How to get ip address from NSNetService

    - by Vic
    When I get a NSNetService object, I try to do: NSNetService *ss=[netArray objectAtIndex:indexPath.row]; ss.delegate=self; [ss resolveWithTimeout:3.0]; Then on the delegate method: - (void)netServiceDidResolveAddress:(NSNetService *)sender { NSArray *address=sender.addresses; NSData *addressData=[NSData dataWithBytes:address length:sizeof(address)]; NSError *error; /* How? */ } Thanks.

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  • Mutually beneficial IP/copyright clauses for contract-based freelance work

    - by Nathan de Vries
    I have a copyright section in the contract I give to my clients stating that I retain copyright on any works produced during my work for them as an independent contractor. This is most definitely not intended to place arbitrary restrictions on my clients, but rather to maintain my ability to decide on how the software I create is licensed and distributed. Almost every project I work on results in at least one part of it being released as open source. Every project I work on makes use of third-party software released in the same fashion, so returning the favour is something I would like to continue doing. Unfortunately, the contract is not so clear when it comes to defining the rights of the client in the use of said software. I mention that the code will be licensed to them, but do not mention specifics about exclusivity, ability to produce derivatives etc. As such, a client has raised concerns about the copyright section of my contract, and has suggested that I reword it such that all copyrights are transferred entirely to the client on final payment for the project. This will almost certainly reduce my ability to distribute the software I have created; I would much prefer to find a more mutually beneficial agreement where both our concerns are appeased. Are there any tried and true approaches to licensing software in this kind of situation? To summarise: I want to maintain the ability to license (parts of) the software under my own terms, independently of my relationship with the client; with some guarantee to the client that no trade-secrets or critical business logic will be shared; giving them the ability to re-use my code in their future projects; but not necessarily letting them sell it (I'm not sure about this, though...what happens if they sell their business and the software along with it?) I realise that everyone's feedback is going to be prefixed with "IANAL", however I appreciate any thoughts you might have on the matter.

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  • Is it possible in .NET to set the local endpoint (IP address) when using webclient to consume a web

    - by Tom
    I'm wondering if it is possible using .NET to call a remote web service and in effect specify which IP the call is made on. I'm consuming a service that limits the number of calls I can make based on IP. The service costs in the 20k range after the free limit is used up. I'm very close to enough calls but not quite there using the free service. My server has 3 IP so I could in effect triple the number of calls I could make to the remote service by changing the IP.

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  • How to set up IP forwarding on Nexenta (Solaris)?

    - by Gleb
    I am trying to set up IP forwarding on my Nexenta box: root@hdd:~# uname -a SunOS hdd 5.11 NexentaOS_134f i86pc i386 i86pc Solaris The box has 2 network interfaces: root@hdd:~# ifconfig -a lo0: flags=2001000849<UP,LOOPBACK,RUNNING,MULTICAST,IPv4,VIRTUAL> mtu 8232 index 1 inet 127.0.0.1 netmask ff000000 e1000g1: flags=1001100843<UP,BROADCAST,RUNNING,MULTICAST,ROUTER,IPv4,FIXEDMTU> mtu 1500 index 2 inet 192.168.12.2 netmask ffffff00 broadcast 192.168.12.255 ether 68:5:ca:9:51:b8 myri10ge0: flags=1100843<UP,BROADCAST,RUNNING,MULTICAST,ROUTER,IPv4> mtu 9000 index 3 inet 10.10.10.10 netmask ffffff00 broadcast 10.10.10.255 ether 0:60:dd:47:87:2 lo0: flags=2002000849<UP,LOOPBACK,RUNNING,MULTICAST,IPv6,VIRTUAL> mtu 8252 index 1 inet6 ::1/128 192.168.12.0 is my normal LAN with 192.168.12.1 being the firewall/gateway 10.10.10.0 is a separate LAN for iSCSI (with no internet access) I want to set up IP forwarding so that a computer on 10.10.10.0 will be able to access the internet by using 10.10.10.10 as a gateway (I don't need any port forwarding) I have turned on IP forwarding: root@hdd:~# routeadm Configuration Current Current Option Configuration System State --------------------------------------------------------------- IPv4 routing disabled disabled IPv6 routing disabled disabled IPv4 forwarding enabled enabled IPv6 forwarding disabled disabled Routing services "route:default ripng:default" Routing daemons: STATE FMRI disabled svc:/network/routing/rdisc:default disabled svc:/network/routing/route:default disabled svc:/network/routing/legacy-routing:ipv4 disabled svc:/network/routing/legacy-routing:ipv6 disabled svc:/network/routing/ripng:default online svc:/network/routing/ndp:default But when I dry to start ipnat, I get an error: root@hdd:~# ipnat -CF -f /etc/ipf/ipnat.conf ioctl(SIOCGNATS): I/O error Here is the config: root@hdd:~# cat /etc/ipf/ipnat.conf #!/sbin/ipnat -f - # map e1000g1 10.10.10.10/24 -> 192.168.12.2/32 So the question is how to fix this.. Thanks in advance!

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  • How to run a website domain without redirecting if IP is already used for another website? [duplicate]

    - by SSpoke
    This question already has an answer here: Hosting multiple distinct folders for distinct domains 1 answer I bought a VPS Host that gave me only 1 IP Address which I used on my first domain name and it works without any problems. Now my second domain name I can't use the same ip address as it points to the first domain name. So I figured my only option was to use a GoDaddy hosted iframe redirection which redirects to a sub folder on my first domain which worked so far. Now I'm trying to load paypal from <?php headers() ?> and I get a permission error because of that iframe Refused to display 'https://www.paypal.com/cgi-bin/webscr?notify_url=&cmd=_cart&upload=1&business=removed&address_override=1' in a frame because it set 'X-Frame-Options' to 'SAMEORIGIN'. How do I avoid the Iframe solution for my second domain while not messing up my first domain? Somebody I forgot once told me it doesn't matter if you have 1 IP Address you could host multiple websites on it? how it that possible the DNS doesn't seem to work off ports afaik, yes I could host multiple websites on different folders but that's not what I call hosting a real website it has to be pointed by a domain name, so this iframe issue doesn't happen My server configuration is httpd (apache) that comes with CentOS 6 (Linux) operating system

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  • How can I determine the IP addresses allocated by DHCP on a router that I'm connected to?

    - by user234831
    This "router" is not a typical situation. I'm using my phone as a hotspot and can only configure a select number of DHCP options. I can manage the limit on how many devices/clients can use my phone as a hotspot. I have to select from a radio-button list with the options: 2,3,4,5, or 8 I can specify the DHCP starting IP address. In this case, it begins at 192.168.6.106 When I'm connected via WIFI to my phone, an ipconfig /all command shows me that the default gateway is 192.168.6.1 and my IPv4 address is 192.168.1.148. I have the luxury of connecting another device to the phone and that device was assigned 192.168.1.121. I've tried connecting to 192.168.6.1, hoping for some sort of router setup page that I'm used to seeing, but there is no such thing or maybe it's just a matter of incompatable operating systems. In summary, the "router" (phone) has an IP address of 192.168.6.1 and a DHCP server that begins at 192.168.6.106 and allows up to 8 connections. Normally, I would assume a range of 192.168.6.106 - 192.168.6.113, but connected clients are showing otherwise. How can I figure out which IP addresses are set aside by DHCP for clients?

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  • List of all TCP/IP and WinSock Repair commands

    - by Niepojety
    I am building a C# application and I am looking for all a list of TCP/IP and WinSock Repair commands. ipconfig /flushdns netsh int reset all netsh int ipv4 reset netsh int ipv6 reset netsh int ip reset netsh int ip reset c:\ipreset.log netsh int ip reset resetlog.txt netsh int ip reset c:\resetlog.txt netsh int ip reset c:\network-connection.log netsh int 6to4 reset all netsh int httpstunnel reset all netsh int isatap reset all netsh int tcp reset all netsh int teredo reset all netsh int portproxy reset all netsh branchcache reset netsh winhttp reset netsh winsock reset c:\winsock.log netsh winsock reset netsh winsock reset all netsh winsock reset catalog

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  • routing based on source IP

    - by user1977050
    I am trying to do source-based routing, following the question http://unix.stackexchange.com/questions/131527/routing-based-on-source-ip. The source IP floating one and assigned to a cluster (consists from 2 servers). Let's say that the physical IP on server1 is 192.0.2.1, on server2 192.0.2.2, and the virtual IP is 192.0.2.3 (and this should be the source IP for outgoing traffic). How can I configure static source IP routing for this in RHEL?

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  • After Port Forwarding, how to get my external IP in Java ?

    - by Frank
    I set up a static IP and did port forwarding on my notebook, and now I have a static IP address, but it's relatively static, every time I re-start the machine, I get another address, and since I have a "static" IP I can now do Paypal IPN messaging. But how can I get this static IP from my Java program ? One way I can think of is to visit : http://portforward.com/ and on that page it tells me what my external IP is, so I can extract it with Java code, is there any other way that I can do in my Java code to get this info ?

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  • New instalation with invalid signatures: NODATA 1 NODATA 2 error

    - by marcos nobre
    I am installing 13.04 into a VirtualBox machine. After installing, I receive this error after sudo apt-get update: Ign http://archive.ubuntu.com raring Release E: Erro GPG: http://archive.ubuntu.com raring Release: the following signatures were invalid: NODATA 1 NODATA 2 I have already tried: Install minimal server version; Install 12.04 desktop version; Everything gives me the same error.

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  • CryptographicException: The handle is invalid.

    - by Wil Peck
    More than once I have come across the issue where we have had a problem using an X509Cert from the certificate store.  Everything is configured properly in the certificate store but when we attempt to create the signature we end up with a cryptographic exception for no apparent reason. See CryptographicException: The handle is invalid post by Benoit Martin explains the problem and shows how this issue can be resolved. Technorati Tags: Exceptions,Help,Cryptography

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  • invalid / unauth Product Key (10 replies)

    I recently purchased a tablet PC (Fujitsu ST5011D) on ebay. In the item description, it stated that &quot;I just installed Windows TabletXP, but it needs to be activated.&quot; Fine and dandy, valid COA on the back per auction images. Well. Upon attempting activation, I was informed that I had an &quot;invalid product key&quot;. I downloaded and ran Magical Jellybean to get the PK from within the system itself, and i...

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  • invalid / unauth Product Key (10 replies)

    I recently purchased a tablet PC (Fujitsu ST5011D) on ebay. In the item description, it stated that &quot;I just installed Windows TabletXP, but it needs to be activated.&quot; Fine and dandy, valid COA on the back per auction images. Well. Upon attempting activation, I was informed that I had an &quot;invalid product key&quot;. I downloaded and ran Magical Jellybean to get the PK from within the system itself, and i...

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  • Customer site is out of IP addresses, they want to go from /24 to /12 netmask... Bad idea?

    - by ewwhite
    One of my client sites called to ask me to change the subnet masks of the Linux servers I manage there while they re-IP/change the netmask of their network based on a 10.0.0.x scheme. "Can you change the server netmasks from 255.255.255.0 to 255.240.0.0?" You mean, 255.255.240.0? "No, 255.240.0.0." Are you sure you need that many IP addresses? "Yeah, we never want to run out of IP addresses." A quick check against the Subnet Cheat Sheet shows: a 255.255.255.0 netmask, a /24 provides 256 hosts. It's clear to see that an organization can exhaust that number of IP addresses. a 255.240.0.0 netmask, a /12 provides 1,048,576 hosts. This is a small < 200-user site. I doubt that they'd allocate more than 400 IP addresses. I suggested something that provides fewer hosts, like a /22 or /21 (1024 and 2048 hosts, respectively), but was unable to give a specific reason against using the /12 subnet. Is there anything this customer should be concerned about? Are there any specific reasons they shouldn't use such an incredibly large mask in their environment?

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  • char array split ip with strtok

    - by user1480139
    I'm trying to split a IP address like 127.0.0.1 from a file: using following C code: pch2 = strtok (ip,"."); printf("\npart 1 ip: %s",pch2); pch2 = strtok (NULL,"."); printf("\npart 2 ip: %s",pch2); And IP is a char ip[500], that containt an ip. When printing it prints 127 as part 1 but as part 2 it prints NULL? Can someone help me? EDIT: Whole function: FILE *file = fopen ("host.txt", "r"); char * pch; char * pch2; char ip[BUFFSIZE]; IPPart result; if (file != NULL) { char line [BUFFSIZE]; while(fgets(line,sizeof line,file) != NULL) { if(line[0] != '#') { //fputs(line,stdout); pch = strtok (line," "); printf ("%s\n",pch); strncpy(ip, pch, sizeof(pch)-1); ip[sizeof(pch)-1] = '\0'; //pch = strtok (line, " "); pch = strtok (NULL," "); printf("%s",pch); pch2 = strtok (ip,"."); printf("\nDeel 1 ip: %s",pch2); pch2 = strtok (NULL,"."); printf("\nDeel 2 ip: %s",pch2); //if(strcmp(pch,url) == 0) //{ // result.part1 = //} } } fclose(file); }

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  • TCP packets larger than 4 KB don't get a reply from Linux

    - by pts
    I'm running Linux 3.2.51 in a virtual machine (192.168.33.15). I'm sending Ethernet frames to it. I'm writing custom software trying to emulate a TCP peer, the other peer is Linux running in the virtual machine guest. I've noticed that TCP packets larger than about 4 KB are ignored (i.e. dropped without an ACK) by the Linux guest. If I decrease the packet size by 50 bytes, I get an ACK. I'm not sending new payload data until the Linux guest fully ACKs the previous one. I've increased ifconfig eth0 mtu 51000, and ping -c 1 -s 50000 goes through (from guest to my emulator) and the Linux guest gets a reply of the same size. I've also increased sysctl -w net.ipv4.tcp_rmem='70000 87380 87380 and tried with sysctl -w net.ipv4.tcp_mtu_probing=1 (and also =0). There is no IPv3 packet fragmentation, all packets have the DF flag set. It works the other way round: the Linux guest can send TCP packets of 6900 bytes of payload and my emulator understands them. This is very strange to me, because only TCP packets seem to be affected (large ICMP packets go through). Any idea what can be imposing this limit? Any idea how to do debug it in the Linux kernel? See the tcpdump -n -vv output below. tcpdump was run on the Linux guest. The last line is interesting: 4060 bytes of TCP payload is sent to the guest, and it doesn't get any reply packet from the Linux guest for half a minute. 14:59:32.000057 IP (tos 0x0, ttl 64, id 0, offset 0, flags [DF], proto TCP (6), length 40) 192.168.33.1.36522 > 192.168.33.15.22: Flags [S], cksum 0x8da0 (correct), seq 10000000, win 14600, length 0 14:59:32.000086 IP (tos 0x10, ttl 64, id 0, offset 0, flags [DF], proto TCP (6), length 44) 192.168.33.15.22 > 192.168.33.1.36522: Flags [S.], cksum 0xc37f (incorrect -> 0x5999), seq 1415680476, ack 10000001, win 19920, options [mss 9960], length 0 14:59:32.000218 IP (tos 0x0, ttl 64, id 0, offset 0, flags [DF], proto TCP (6), length 40) 192.168.33.1.36522 > 192.168.33.15.22: Flags [.], cksum 0xa752 (correct), ack 1, win 14600, length 0 14:59:32.000948 IP (tos 0x10, ttl 64, id 53777, offset 0, flags [DF], proto TCP (6), length 66) 192.168.33.15.22 > 192.168.33.1.36522: Flags [P.], cksum 0xc395 (incorrect -> 0xfa01), seq 1:27, ack 1, win 19920, length 26 14:59:32.001575 IP (tos 0x0, ttl 64, id 0, offset 0, flags [DF], proto TCP (6), length 40) 192.168.33.1.36522 > 192.168.33.15.22: Flags [.], cksum 0xa738 (correct), ack 27, win 14600, length 0 14:59:32.001585 IP (tos 0x0, ttl 64, id 0, offset 0, flags [DF], proto TCP (6), length 65) 192.168.33.1.36522 > 192.168.33.15.22: Flags [P.], cksum 0x48d6 (correct), seq 1:26, ack 27, win 14600, length 25 14:59:32.001589 IP (tos 0x10, ttl 64, id 53778, offset 0, flags [DF], proto TCP (6), length 40) 192.168.33.15.22 > 192.168.33.1.36522: Flags [.], cksum 0xc37b (incorrect -> 0x9257), ack 26, win 19920, length 0 14:59:32.001680 IP (tos 0x10, ttl 64, id 53779, offset 0, flags [DF], proto TCP (6), length 496) 192.168.33.15.22 > 192.168.33.1.36522: Flags [P.], seq 27:483, ack 26, win 19920, length 456 14:59:32.001784 IP (tos 0x0, ttl 64, id 0, offset 0, flags [DF], proto TCP (6), length 40) 192.168.33.1.36522 > 192.168.33.15.22: Flags [.], cksum 0xa557 (correct), ack 483, win 14600, length 0 14:59:32.006367 IP (tos 0x0, ttl 64, id 0, offset 0, flags [DF], proto TCP (6), length 1136) 192.168.33.1.36522 > 192.168.33.15.22: Flags [P.], seq 26:1122, ack 483, win 14600, length 1096 14:59:32.044150 IP (tos 0x10, ttl 64, id 53780, offset 0, flags [DF], proto TCP (6), length 40) 192.168.33.15.22 > 192.168.33.1.36522: Flags [.], cksum 0xc37b (incorrect -> 0x8c47), ack 1122, win 19920, length 0 14:59:32.045310 IP (tos 0x0, ttl 64, id 0, offset 0, flags [DF], proto TCP (6), length 312) 192.168.33.1.36522 > 192.168.33.15.22: Flags [P.], seq 1122:1394, ack 483, win 14600, length 272 14:59:32.045322 IP (tos 0x10, ttl 64, id 53781, offset 0, flags [DF], proto TCP (6), length 40) 192.168.33.15.22 > 192.168.33.1.36522: Flags [.], cksum 0xc37b (incorrect -> 0x8b37), ack 1394, win 19920, length 0 14:59:32.925726 IP (tos 0x10, ttl 64, id 53782, offset 0, flags [DF], proto TCP (6), length 1112) 192.168.33.15.22 > 192.168.33.1.36522: Flags [.], seq 483:1555, ack 1394, win 19920, length 1072 14:59:32.925750 IP (tos 0x10, ttl 64, id 53784, offset 0, flags [DF], proto TCP (6), length 312) 192.168.33.15.22 > 192.168.33.1.36522: Flags [P.], seq 1555:1827, ack 1394, win 19920, length 272 14:59:32.927131 IP (tos 0x0, ttl 64, id 0, offset 0, flags [DF], proto TCP (6), length 40) 192.168.33.1.36522 > 192.168.33.15.22: Flags [.], cksum 0x9bcf (correct), ack 1555, win 14600, length 0 14:59:32.927148 IP (tos 0x0, ttl 64, id 0, offset 0, flags [DF], proto TCP (6), length 40) 192.168.33.1.36522 > 192.168.33.15.22: Flags [.], cksum 0x9abf (correct), ack 1827, win 14600, length 0 14:59:32.932248 IP (tos 0x10, ttl 64, id 53785, offset 0, flags [DF], proto TCP (6), length 56) 192.168.33.15.22 > 192.168.33.1.36522: Flags [P.], cksum 0xc38b (incorrect -> 0xd247), seq 1827:1843, ack 1394, win 19920, length 16 14:59:32.932366 IP (tos 0x0, ttl 64, id 0, offset 0, flags [DF], proto TCP (6), length 40) 192.168.33.1.36522 > 192.168.33.15.22: Flags [.], cksum 0x9aaf (correct), ack 1843, win 14600, length 0 14:59:32.964295 IP (tos 0x0, ttl 64, id 0, offset 0, flags [DF], proto TCP (6), length 104) 192.168.33.1.36522 > 192.168.33.15.22: Flags [P.], seq 1394:1458, ack 1843, win 14600, length 64 14:59:32.964310 IP (tos 0x10, ttl 64, id 53786, offset 0, flags [DF], proto TCP (6), length 40) 192.168.33.15.22 > 192.168.33.1.36522: Flags [.], cksum 0xc37b (incorrect -> 0x85a7), ack 1458, win 19920, length 0 14:59:32.964561 IP (tos 0x10, ttl 64, id 53787, offset 0, flags [DF], proto TCP (6), length 88) 192.168.33.15.22 > 192.168.33.1.36522: Flags [P.], seq 1843:1891, ack 1458, win 19920, length 48 14:59:32.965185 IP (tos 0x0, ttl 64, id 0, offset 0, flags [DF], proto TCP (6), length 40) 192.168.33.1.36522 > 192.168.33.15.22: Flags [.], cksum 0x9a3f (correct), ack 1891, win 14600, length 0 14:59:32.965196 IP (tos 0x0, ttl 64, id 0, offset 0, flags [DF], proto TCP (6), length 104) 192.168.33.1.36522 > 192.168.33.15.22: Flags [P.], seq 1458:1522, ack 1891, win 14600, length 64 14:59:32.965233 IP (tos 0x10, ttl 64, id 53788, offset 0, flags [DF], proto TCP (6), length 88) 192.168.33.15.22 > 192.168.33.1.36522: Flags [P.], seq 1891:1939, ack 1522, win 19920, length 48 14:59:32.965970 IP (tos 0x0, ttl 64, id 0, offset 0, flags [DF], proto TCP (6), length 40) 192.168.33.1.36522 > 192.168.33.15.22: Flags [.], cksum 0x99cf (correct), ack 1939, win 14600, length 0 14:59:32.965979 IP (tos 0x0, ttl 64, id 0, offset 0, flags [DF], proto TCP (6), length 568) 192.168.33.1.36522 > 192.168.33.15.22: Flags [P.], seq 1522:2050, ack 1939, win 14600, length 528 14:59:32.966112 IP (tos 0x10, ttl 64, id 53789, offset 0, flags [DF], proto TCP (6), length 520) 192.168.33.15.22 > 192.168.33.1.36522: Flags [P.], seq 1939:2419, ack 2050, win 19920, length 480 14:59:32.970059 IP (tos 0x0, ttl 64, id 0, offset 0, flags [DF], proto TCP (6), length 40) 192.168.33.1.36522 > 192.168.33.15.22: Flags [.], cksum 0x95df (correct), ack 2419, win 14600, length 0 14:59:32.970089 IP (tos 0x0, ttl 64, id 0, offset 0, flags [DF], proto TCP (6), length 616) 192.168.33.1.36522 > 192.168.33.15.22: Flags [P.], seq 2050:2626, ack 2419, win 14600, length 576 14:59:32.981159 IP (tos 0x10, ttl 64, id 53790, offset 0, flags [DF], proto TCP (6), length 72) 192.168.33.15.22 > 192.168.33.1.36522: Flags [P.], cksum 0xc39b (incorrect -> 0xa84f), seq 2419:2451, ack 2626, win 19920, length 32 14:59:32.982347 IP (tos 0x0, ttl 64, id 0, offset 0, flags [DF], proto TCP (6), length 40) 192.168.33.1.36522 > 192.168.33.15.22: Flags [.], cksum 0x937f (correct), ack 2451, win 14600, length 0 14:59:32.982357 IP (tos 0x0, ttl 64, id 0, offset 0, flags [DF], proto TCP (6), length 104) 192.168.33.1.36522 > 192.168.33.15.22: Flags [P.], seq 2626:2690, ack 2451, win 14600, length 64 14:59:32.982401 IP (tos 0x10, ttl 64, id 53791, offset 0, flags [DF], proto TCP (6), length 88) 192.168.33.15.22 > 192.168.33.1.36522: Flags [P.], seq 2451:2499, ack 2690, win 19920, length 48 14:59:32.982570 IP (tos 0x0, ttl 64, id 0, offset 0, flags [DF], proto TCP (6), length 40) 192.168.33.1.36522 > 192.168.33.15.22: Flags [.], cksum 0x930f (correct), ack 2499, win 14600, length 0 14:59:32.982702 IP (tos 0x0, ttl 64, id 0, offset 0, flags [DF], proto TCP (6), length 104) 192.168.33.1.36522 > 192.168.33.15.22: Flags [P.], seq 2690:2754, ack 2499, win 14600, length 64 14:59:33.020066 IP (tos 0x10, ttl 64, id 53792, offset 0, flags [DF], proto TCP (6), length 40) 192.168.33.15.22 > 192.168.33.1.36522: Flags [.], cksum 0xc37b (incorrect -> 0x7e07), ack 2754, win 19920, length 0 14:59:33.983503 IP (tos 0x10, ttl 64, id 53793, offset 0, flags [DF], proto TCP (6), length 72) 192.168.33.15.22 > 192.168.33.1.36522: Flags [P.], cksum 0xc39b (incorrect -> 0x2aa7), seq 2499:2531, ack 2754, win 19920, length 32 14:59:33.983810 IP (tos 0x10, ttl 64, id 53794, offset 0, flags [DF], proto TCP (6), length 88) 192.168.33.15.22 > 192.168.33.1.36522: Flags [P.], seq 2531:2579, ack 2754, win 19920, length 48 14:59:33.984100 IP (tos 0x0, ttl 64, id 0, offset 0, flags [DF], proto TCP (6), length 40) 192.168.33.1.36522 > 192.168.33.15.22: Flags [.], cksum 0x92af (correct), ack 2531, win 14600, length 0 14:59:33.984139 IP (tos 0x0, ttl 64, id 0, offset 0, flags [DF], proto TCP (6), length 40) 192.168.33.1.36522 > 192.168.33.15.22: Flags [.], cksum 0x927f (correct), ack 2579, win 14600, length 0 14:59:34.022914 IP (tos 0x0, ttl 64, id 0, offset 0, flags [DF], proto TCP (6), length 104) 192.168.33.1.36522 > 192.168.33.15.22: Flags [P.], seq 2754:2818, ack 2579, win 14600, length 64 14:59:34.022939 IP (tos 0x10, ttl 64, id 53795, offset 0, flags [DF], proto TCP (6), length 40) 192.168.33.15.22 > 192.168.33.1.36522: Flags [.], cksum 0xc37b (incorrect -> 0x7d77), ack 2818, win 19920, length 0 14:59:34.023554 IP (tos 0x10, ttl 64, id 53796, offset 0, flags [DF], proto TCP (6), length 88) 192.168.33.15.22 > 192.168.33.1.36522: Flags [P.], seq 2579:2627, ack 2818, win 19920, length 48 14:59:34.027571 IP (tos 0x0, ttl 64, id 0, offset 0, flags [DF], proto TCP (6), length 40) 192.168.33.1.36522 > 192.168.33.15.22: Flags [.], cksum 0x920f (correct), ack 2627, win 14600, length 0 14:59:34.027603 IP (tos 0x0, ttl 64, id 0, offset 0, flags [DF], proto TCP (6), length 4100) 192.168.33.1.36522 > 192.168.33.15.22: Flags [P.], seq 2818:6878, ack 2627, win 14600, length 4060

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  • How to lookup an IP address in an Excel spreadsheet?

    - by Kevin Williams
    I am working with a decent sized spreadsheet of domains and server names. Another user of the spreadsheet needs the IP address for each of the DNS entries on the worksheet. Instead of manually adding and then having to maintain this list I was hoping there was an easy way to do an IPAddress lookup to display the IP address in a cell. I've seen some VBScripts that call gethostbyname, e.g.: Declare Function GetHostByName Lib "wsock32.dll" Alias "gethostbyname" (ByVal Host As String) As Long But I'm not a VB expert so I'm not sure if this is the right way to go. Any advice/links would be appreciated! also if this is a question better suited for Stack Overflow - let me know, I'm new here.

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  • Using IP Restrictions with URL Rewrite-Week 25

    - by OWScott
    URL Rewrite offers tremendous flexibility for customizing rules to your environment. One area of functionality that is often desired for URL Rewrite is to allow a large list of approved or denied IP addresses and subnet ranges. IIS’s original IP Restrictions is helpful for fully blocking an IP address, but it doesn’t offer the flexibility that URL Rewrite does. An example where URL Rewrite is helpful is where you want to allow only authorized IPs to access staging.yoursite.com, but where staging.yoursite.com is part of the same site as www.yoursite.com. This requires conditional logic for the user’s IP. This lesson covers this unique situation while also introducing Rewrite Maps, server variables, and pairing rules to add more flexibility. This is week 25 of a 52 week series for the Web Pro. Past and future videos can be found here: http://dotnetslackers.com/projects/LearnIIS7/ You can find this week’s video here.

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  • Are there any negative impact running all services on a single static IP?

    - by Jake
    I need to setup a VPN trunk (with remote branch office location), Video Conference and Web Server on-premise in the office. I am used to ISP providing blocks of 8 or 16 IPs. But I have a new ISP which says they only can provide a single IP. Are there any issues with running all services on a single IP? I don't think this has any thing to do with bandwidth..? I'm not using SSL certificates... I can do port forwarding to different machines... What else...? Disclaimer: I am a programmer by training. Sorry for noob question.

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