Apache Monitoring Script

[mmidgett]

Using this method 2 I have it working for all of my servers but one.

I am using this article http://www.zabbix.com/wiki/templates/apache to monitor Apache on a Ubuntu Server.

I have one special server that I have to run the server-status page on a non standard port of 8082

I can get the status page to display in my browser so I know the page is on the non standard port.

I have created a copy of the template to only be assigned to this one server.

In Zabbix I need to have it do the web check on this port of 8082. Can someone help me with this?

[raoul]

Hi,

You should use a Web test (Configuration -> Web), declaring in the Step your test URL (http://your_server:8082/your_page.html)

Then you can setup a trigger based on the result of this web test.

(see WEB Monitoring in Zabbix Manual)

Cheers,

y/

[mmidgett]

I maybe no understanding why I'm setting up a trigger to see if apache is running. I don't want to see if apache is running. I want to get the statistics from www.url.domain.tld:8082/server-status into my graphs. The default connector for method 2 is over 80. I have port 80 doing a redirect to 443 on the base url of the server. So I can't let the script go to url.domain.tld/server-status

Can't I just change # Setup connection string
ConnectionString = ("http://%s:%s%s") % (WebServer, Port, URL)
to

# Setup connection string
ConnectionString = ("http://url.domain.tld:8082") % (WebServer, Port, URL)

############### Function to fire off an HTTP request to the web server, throw an exception gracefully
############### and print FAIL if a connection can't be made
def getURL(WebServer,Port,URL):
try:
# Setup connection string
ConnectionString = ("http://%s:%s%s") % (WebServer, Port, URL)

conn = urllib.urlopen(ConnectionString)
URLresponse = conn.read()

# Clean up the connection
conn.close()

# The response to the function is the output of the URL called
return URLresponse

# Catch all exceptions
except:
print "Error getting URL"

[mmidgett]

can somebody help?