Re: HA node check

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Re: Re: HA node check by blakem (Monsignor) on Dec 13, 2001 at 00:37 UTC
Splitting on `/ /` is different from splitting on `' '`. I believe that change would break the original program.... From split: As a special case, specifying a PATTERN of space (' ') will split on white space just as `split' with no arguments does. Thus, `split(' ')' can be used to emulate awk's default behavior, whereas `split(/ /)' will give you as many null initial fields as there are leading spaces. A `split' on `/\s+/' is like a `split(' ')' except that any leading whitespace produces a null first field. A `split' with no arguments really does a `split(' ', $_)' internally. Here is an example: `#!/usr/bin/perl -wT use strict; $_ = " a b c "; print "split(): $_\n" for split(); print "split(' '): $_\n" for split(' '); print "split(/ /): $_\n" for split(/ /); __END__ =head1 OUTPUT split(): a split(): b split(): c split(' '): a split(' '): b split(' '): c split(/ /): split(/ /): split(/ /): split(/ /): a split(/ /): split(/ /): split(/ /): b split(/ /): split(/ /): split(/ /): c` [download] -Blake	[reply] [d/l] [select]
Re: Re: HA node check by coec (Chaplain) on Dec 12, 2001 at 19:48 UTC
Thats the whole point :) The netstat -i on node1 will return something like `Name Mtu Network Address Ipkts Ierrs Opkts Oerrs + Coll en0 1500 link#2 0.4.ac.3e.65.22 169651762 0 150335201 + 0 0 en0 1500 172.22.6 node1 169651762 0 150335201 + 0 0 en2 1500 link#3 0.4.ac.3e.15.54 1387003 0 1389752 +0 0 en2 1500 172.22.7 node1hb 1387003 0 1389752 0 + 0 en1 9000 link#4 0.4.ac.7c.95.ec 232558546 0 76198992 + 0 0 en1 9000 172.22.5 node1r 232558546 0 76198992 0 + 0 <loop back interfaces snipped>` [download] and a netstat -i on node2 will return something like `Name Mtu Network Address Ipkts Ierrs Opkts Oerrs + Coll en0 1500 link#2 0.4.ac.3e.65.4a 36127938 0 37211569 +0 0 en0 1500 172.22.6 node2 36127938 0 37211569 +0 0 en2 1500 link#3 0.4.ac.3e.65.c6 1412752 0 1364210 +0 0 en2 1500 172.22.7 node2hb 1412752 0 1364210 0 + 0 en1 9000 link#4 0.4.ac.7c.97.2f 76199094 0 232558588 +11 0 en1 9000 172.22.5 node2r 76199094 0 232558588 11 + 0 <loop back interfaces snipped>` [download] Where node? is the primary interface, node?hb is the heartbeat link and node?r is the redundant link that node2 will use to pretend to be node1 in the event of node1 failing, in which case interface en1 will no longer have a name of node2r but node1. Sooo, if node1 is up the test will fail on node2 (which is what we want) but if node1 is dead the test will succeed which, again, is what we want. I hope that makes more sense now.	[reply] [d/l] [select]