grep logs for the last X minutes every X minutes - grep

Question

I want to grep logs for the last X minutes every X minutes(ex: grep logs of last 5 min every 5 min) without grep-ing anything older. What could be the simplest way to do this?

Answer 1

A Ruby 1.9.3 solution
The second part that you require, I haven't tested it yet :(
Note: Also if you know the location of 'ruby' change the shebang (first line of the script),
for security reasons.
#! /usr/bin/env ruby
require 'date'
require 'pathname'
if ARGV.length != 4
$stderr.print "usage: #{$0} -d|-h|-m|-s time expression log_file\n"
exit 1
end
begin
total_amount = Integer ARGV[1]
rescue ArgumentError
$stderr.print "error: parameter 'time' must be an Integer\n"
$stderr.print "usage: #{$0} -d|-h|-m|-s time expression log_file\n"
end
if ARGV[0] == "-m"
gap = Rational(60, 86400)
time_str = "%b %d %H:%M"
until_next_search = total_amount * 60
elsif ARGV[0] == "-s"
gap = Rational(1, 86400)
time_str = "%b %d %H:%M:%S"
until_next_search = total_amount
elsif ARGV[0] == "-h"
gap = Rational(3600, 86400)
time_str = "%b %d %H"
until_next_search = total_amount * 3600
elsif ARGV[0] == "-d"
time_str = "%b %d"
gap = 1
until_next_search = total_amount * 86400
else
$stderr.print "usage: #{$0} -d|-h|-m|-s time expression log_file\n"
exit 1
end
pn = Pathname.new(ARGV[3])
if pn.exist?
log = (pn.directory?) ? ARGV[3] + "/*" : ARGV[3]
else
$stderr.print "error: file '" << ARGV[3] << "' does not exist\n"
$stderr.print "usage: #{$0} -d|-h|-m|-s time expression log_file\n"
end
search_str = ARGV[2]
now = DateTime.now
while true
total_amount.times do
now -= gap
system "cat " << log << " | grep '" << now.strftime(time_str) << ".*" << search_str << "'"
end
sleep until_next_search
end