Introduction
This answer is something long, because there is 3 different way on thinking: 1) perl quick or exact, 2) pure bash and 3) perl script in bash function.
That's a (common) job for perl!:
Simple and efficient:
perl -MDate::Parse -ne 'print if/^(.{15})\s/&&str2time($1)>time-600' /path/log
This version print last 10 minutes event, upto now, by using time
function.
You could test this with:
sudo cat /var/log/syslog |
perl -MDate::Parse -ne '
print if /^(\S+\s+\d+\s+\d+:\d+:\d+)\s/ && str2time($1) > time-600'
Note that first representation use only firsts 15 chars from each lines, while second construct use more detailed regexp.
As a perl script: last10m.pl
#!/usr/bin/perl -wn
use strict;
use Date::Parse;
print if /^(\S+\s+\d+\s+\d+:\d+:\d+)\s/ && str2time($1) > time-600
Strictly: extract last 10 minutes from logfile
Meaning not relative to current time, but to last entry in logfile:
There is two way for retrieving end of period:
date -r logfile +%s
tail -n1 logfile | perl -MDate::Parse -nE 'say str2time($1) if /^(.{15})/'
Where logically, last modification time of the logfile must be the time of the last entry.
So the command could become:
perl -MDate::Parse -ne 'print if/^(.{15})\s/&&str2time($1)>'$(
date -r logfile +%s)
or you could take the last entry as reference:
perl -MDate::Parse -E 'open IN,"<".$ARGV[0];seek IN,-200,2;while (<IN>) {
$ref=str2time($1) if /^(\S+\s+\d+\s+\d+:\d+:\d+)/;};seek IN,0,0;
while (<IN>) {print if /^(.{15})\s/&&str2time($1)>$ref-600}' logfile
Second version seem stronger, but access to file only once.
As a perl script, this could look like:
#!/usr/bin/perl -w
use strict;
use Date::Parse;
my $ref; # The only variable I will use in this.
open IN,"<".$ARGV[0]; # Open (READ) file submited as 1st argument
seek IN,-200,2; # Jump to 200 character before end of logfile. (This
# could not suffice if log file hold very log lines! )
while (<IN>) { # Until end of logfile...
$ref=str2time($1) if /^(\S+\s+\d+\s+\d+:\d+:\d+)/;
}; # store time into $ref variable.
seek IN,0,0; # Jump back to the begin of file
while (<IN>) {
print if /^(.{15})\s/&&str2time($1)>$ref-600;
}
But if you really wanna use bash
There is a very quick pure bash script:
Warning: This use recent bashisms, require $BASH_VERSION
4.2 or higher.
#!/bin/bash
declare -A month
for i in {1..12};do
LANG=C printf -v var "%(%b)T" $(((i-1)*31*86400))
month[$var]=$i
done
printf -v now "%(%s)T" -1
printf -v ref "%(%m%d%H%M%S)T" $((now-600))
while read line;do
printf -v crt "%02d%02d%02d%02d%02d" ${month[${line:0:3}]} \
$((10#${line:4:2})) $((10#${line:7:2})) $((10#${line:10:2})) \
$((10#${line:13:2}))
# echo " $crt < $ref ??" # Uncomment this line to print each test
[ $crt -gt $ref ] && break
done
cat
Store this script and run:
cat >last10min.sh
chmod +x last10min.sh
sudo cat /var/log/syslog | ./last10min.sh
Strictly: extract last 10 minutes from logfile
Simply replace line 10, but you have to place filename in the script and not use it as a filter:
#!/bin/bash
declare -A month
for i in {1..12};do
LANG=C printf -v var "%(%b)T" $(((i-1)*31*86400))
month[$var]=$i
done
read now < <(date -d "$(tail -n1 $1|head -c 15)" +%s)
printf -v ref "%(%m%d%H%M%S)T" $((now-600))
export -A month
{
while read line;do
printf -v crt "%02d%02d%02d%02d%02d" ${month[${line:0:3}]} \
$((10#${line:4:2})) $((10#${line:7:2})) $((10#${line:10:2})) \
$((10#${line:13:2}))
[ $crt -gt $ref ] && break
done
cat
} <$1
A perl script into a bash function
As commented by ajcg
, this could be nice to put efficient perl script into a bash function:
recentLog(){
perl -MDate::Parse -ne '
print if/^(.{'${3:-15}'})\s/ &&
str2time($1)>time-'$((
60*${2:-10}
)) ${1:-/var/log/daemon.log}
}
Usage:
recentLog [filename] [minutes] [time sting length]
filename
of log file
minutes
max before now of lines to show
time sting length
from begin of lines (default 15
).