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I am using Windows XP pro, and I need to know if something is registered on a port.

If so how can I tell what is on the port?

EDIT

What I mean by registered is that I am trying to test a .NET remoting application, and I need to see if the application is running or registered on a given port.

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4 Answers 4

Reset to default
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netstat -a -b will show all listening ports and the executable name (rather than just the PID).

If you prefer a graphical version, Microsoft's TCPView will show you the same information, updating in real-time.

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  • 2
    What about searching for particular port is in use or not
    – FindOutIslamNow
    Commented Oct 12, 2016 at 10:19
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    @Kilanny The findstr command finds lines that match some pattern, so pipe the netstat output into it: netstat -a | findstr 8888
    – Stephen Jennings
    Commented Oct 15, 2016 at 18:30
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    @Kilanny Or, redirect the output to a file and search it with your text editor: netstat -a -b > output.txt
    – Stephen Jennings
    Commented Oct 15, 2016 at 18:31
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TCPview from the Sysinternals Suite does it in a nice, GUI way.

It includes a configurable update interval, address resolving and difference highlighting. And it is a lot faster than netstat -a -b.

Also available as a command line version (tcpvcon.exe). Both are free.

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  • This tool just made my day. Once again sysinternals to the rescue. Never had a need for this before, it won't be forgotten!
    – Chris Marisic
    Commented May 23, 2017 at 19:34
  • So nice to see that an eight-year-old answer still helps people. Thanks for the feedback!
    – Tomalak
    Commented May 23, 2017 at 20:00
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Run netstat from the command line and it will show you all of the current connections. netstat -b will show you what program is using the port.

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How do you define "registered"?

Anyway, netstat -o will list the currently used and listening ports with PIDs and tasklist (or whatever process id lister is available) would sort the process name and executable out I guess? ^^

edit: Oh nice, didn't notice the netstat -b option - that's obviously the better option :)

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