up vote 0 down vote favorite
share [g+] share [fb]

Is there a tool in Windows XP that allows me to see what remote hosts my machine connects to in real-time? It would be even better if it could tell me how much data is being exchanged.

link|improve this question
57% accept rate
feedback

2 Answers

up vote 1 down vote

Sysinternals TCPView

TCPView is a Windows program that will show you detailed listings of all TCP and UDP endpoints on your system, including the local and remote addresses and state of TCP connections.

By default, TCPView updates every second, but you can use the Options|Refresh Rate menu item to change the rate. Endpoints that change state from one update to the next are highlighted in yellow; those that are deleted are shown in red, and new endpoints are shown in green.

If by "in Windows XP" you mean "part of a standard Windows XP installation and not some sort of downloaded application", you have netstat 1 in a Command-Prompt window, but it isn't pretty.

link|improve this answer
hah netstat 1 (I see too now in -?) I wouldn't have believed it! – barlop Sep 29 '11 at 20:35
of course, netstat doesn't show amount of data exchanged, as op wanted, but does tcpview? – barlop Sep 29 '11 at 20:37
@barlop: No you are right, TCPView doesn't show amount of data exchanged. For Linux there is nettop and nethogs but I don't know of a Windows port of those. – RedGrittyBrick Sep 29 '11 at 22:46
feedback
up vote 0 down vote

Net Balancer can do it, it shows incoming and outgoing traffic for any PID, keeps totals as well as live data, Highlight any process and you can see the remote ip in the lower right pane. I use the "Show Online Processes Only" button to make viewing of traffic easy, this software is more for managing network traffic but shows the data you are looking for, I don't think it logs anything.

http://seriousbit.com/netbalancer/

link|improve this answer
feedback

Your Answer

 
or
required, but never shown
discard

Not the answer you're looking for? Browse other questions tagged or ask your own question.