Hallo all, I’m a Comodo newby and would need some help on configuring CPF Firewall. My System is WinXP/SP2 with an Apache HTTP Server 2.2.3 and PHP 5.2.0 for developing reasons. Afer reading manuals and faq’s for 2 hours I’m still not able to create a network rule which lets pass through the apache’s traffic. Maybe anyone could help?
Ports are properly forwarded through my router to the virtual server [my local machine]. Apache is reachable within LAN on 127.0.0.1 and on WAN on its static IP if CPF’s security level is switched to ‘allow all’, but this does not appear to be the solution. (:NRD)
When setting security level to ‘custom’ and trying to connect to 127.0.0.1 I receive a browser’s error ‘connection refused’. I get a browser ‘timeout’ when attempting to connect the webserver on it’s static WAN IP.
My application settings are:
httpd.exe and ApacheMonitor.exe both on Destination 192.168.0.2 [my machine’s LAN IP] with Port ‘Any’ and Protocol ‘TCP In/Out’ Permission Allow.
My network settings are default config:
Rule ID 5: Block & Log Protocol IP In/Out Source Any Dest. Any where IPROTO IS ANY
Other Rule ID’s above are also set to allow by default settings.
Knowing that the prob may be much easier solved when knowing about the CPF’s logfiles I will publish the concerning parts as text-only [upload folder is full, sorry
]:
Comodo Firewall Logs:
Date/Time: 2006-12-17 14:07:37 Severity: Medium Reporter: Network Monitor Description: Inbound Policy Violation (Access Denied, IP = 192.168.0.2, Port = nbname(137)) Protocol: UDP Incoming Source: 192.168.0.1:2360 Destination: 192.168.0.2:nbname(137) Reason: Network Control Rule ID = 5
Date/Time: 2006-12-17 14:05:50 Severity: Medium Reporter: Application Monitor Description: Application Access Denied (httpd.exe:127.0.0.1:http(80)) Application: C:\Programme\Apache2.2\bin\httpd.exe Parent: C:\WINDOWS\system32\services.exe Protocol: TCP In Destination: 127.0.0.1:http(80)
Date/Time: 2006-12-17 14:00:43 Severity: Medium Reporter: Application Monitor Description: Application Access Denied (httpd.exe:127.0.0.1:http(80)) Application: C:\Programme\Apache2.2\bin\httpd.exe Parent: C:\WINDOWS\system32\services.exe Protocol: TCP In Destination: 127.0.0.1:http(80)
Date/Time: 2006-12-17 13:43:17 Severity: Medium Reporter: Network Monitor Description: Inbound Policy Violation (Access Denied, IP = 192.168.0.2, Port = http(80)) Protocol: TCP Incoming Source: 88.217.225.12:1317 Destination: 192.168.0.2:http(80) TCP Flags: SYN Reason: Network Control Rule ID = 5
Date/Time: 2006-12-17 12:51:01 Severity: Medium Reporter: Network Monitor Description: Inbound Policy Violation (Access Denied, IP = 192.168.0.2, Port = nbname(137)) Protocol: UDP Incoming Source: 192.168.0.1:2329 Destination: 192.168.0.2:nbname(137) Reason: Network Control Rule ID = 5
Date/Time: 2006-12-17 12:49:57 Severity: Medium Reporter: Application Monitor Description: Application Access Denied (httpd.exe:127.0.0.1:http(80)) Application: C:\Programme\Apache2.2\bin\httpd.exe Parent: C:\WINDOWS\system32\services.exe Protocol: TCP In Destination: 127.0.0.1:http(80)
Date/Time: 2006-12-17 12:49:36 Severity: Medium Reporter: Network Monitor Description: Inbound Policy Violation (Access Denied, IP = 192.168.0.2, Port = http(80)) Protocol: TCP Incoming Source: 88.217.225.12:1186 Destination: 192.168.0.2:http(80) TCP Flags: SYN Reason: Network Control Rule ID = 5
Date/Time: 2006-12-17 12:49:26 Severity: Medium Reporter: Network Monitor Description: Inbound Policy Violation (Access Denied, IP = 192.168.0.2, Port = http(80)) Protocol: TCP Incoming Source: 88.217.225.12:1186 Destination: 192.168.0.2:http(80) TCP Flags: SYN Reason: Network Control Rule ID = 5
I saw any connection attempt to my local machine 192.168.0.2 blocked by network rule ID 5 and also sometimes blocked incoming traffic to Apache’s httpd.exe on local IP 127.0.0.1 with htt-protocol on port 80 blocked by the Application Monitor
Thank you very much for any help. When reading this forum for the last two hours I saw many user problems solved due the comodo community also hopefully my small prob. will be solved too…
Greetings from Munich, dah