Hi there
I came across this script online to convert Epoch time to proper date format, but I am receiving the following error
Also, I have HISTTIMEFORMAT set in user's .profile so that their history output shows time stamps. Additionally I have changed their .history location to a dedicated directory in /var/log. Example as below:
export HISTFILE=/var/log/user_history/JP1_history
export HISTFILESIZE=20000
export HISTSIZE=5000
export HISTTIMEFORMAT="%F %T "
This is the error message from the script:
[root@H99A100 user_history]# ./test
Month '23' out of range 0..11 at -e line 2
This is the script I am using:
#!/bin/bash
# YYYY MM DD HH MM SS
# perl autosplits the string and uses timelocal to return
# the number of seconds from the Epoch
# No error checking!
function seconds_from_epoch {
echo $*| perl -MTime::Local -ane '
my $epochseconds = timelocal($F[5], $F[4], $F[3], $F[2], $F[1], $F[0]);
print "$epochseconds\n"; '
}
# parse file s32adm_bash_history date and time
set - $(ls -l s32adm_bash_history)
fdate=$6
ftime=$7
# parse the year, month, and day
set - $(IFS="-"; echo $fdate)
fyear=$1
fmonth=$2
fday=$3
# parse the hours and minutes
set - $(IFS=":"; echo $ftime)
fhour=$1
fmin=$2
totsecs=$(seconds_from_epoch "$fyear $fmonth $fday $fhour $fmin 0")
echo $totsecs
Can someone please point out what I am doing wrong?
echo $*
you are not passing any input arguments for your program.
Hi itkamaraj
I added echo $epochseconds| perl -MTime::Local -ane '
But i am getting this error now
[root@H99A100 user_history]# ./test
Day '' out of range 1..31 at -e line 2
@itkamaraj: $* is used in a function and it receives from the 2nd last line of the original script in post #1.
@hedkandi: Your original script in post #1 works for me without errors. Though, I didn't validate if the time in seconds from epoch is correct or not.
Couple of things that come to mind
- The parameters required for the perl timelocal function seem to be reversed (Time::Local - perldoc.perl.org)
- The timelocal function uses months 0-11 instead of 1-12
- Using ls -l like that is unreliable unless you control the format. Also, some ls switch to another format after 6 months and/or a year
- You need the echo $* because only one parameter gets past to the function instead of 6 (lose the double quotes around the parameters to make it 6)
- use set -- instead of set -
Hi Scrutinizer, thanks for your advise. I amended the script but its still not working
#!/bin/bash
# YYYY MM DD HH MM SS
# perl autosplits the string and uses timelocal to return
# the number of seconds from the Epoch
# No error checking!
function seconds_from_epoch {
echo $*| perl -MTime::Local -ane '
my $epochseconds = timegm($F[5], $F[4], $F[3], $F[2], $F[1], $F[0]);
print "$epochseconds\n"; '
}
# parse file s32adm_bash_history date and time
set -- $(ls -l s12adm_history)
fdate=$6
ftime=$7
# parse the year, month, and day
set -- $(IFS="-"; echo $fdate)
fyear=$0
fmonth=$1
fday=$2
# parse the hours and minutes
set -- $(IFS=":"; echo $ftime)
fhour=$3
fmin=$4
fsecs=$5
totsecs=$($fsecs $fmin $fhour $fday $fmonth $fyear)
echo $totsecs
I am getting this error now
[root@H99A100 user_history]# ./test
./test: line 29: Mar: command not found
Sorry, I'm not really good with scripts 
What is the output of ls -l s12adm_history ? It seems to me that "Mar" stands for "March", and the is hardly the input that you should feed into timegm or timelocal.
Hi Scrutinizer
This is my output
[root@H99 user_history]# ls -l s12adm_history
-rw-r--r-- 1 s12adm root 1890 Mar 27 12:23 s12adm_history
So if we look at this part of your script, we see that $fdate gets set to "Mar" and $ftime gets set to "27", which I don't think is as intended..
Hi Scrutinizer
I give up, I am using this one-liner added to cron that runs once a month at the end of each month and it worked. Only thing is that I have to set a cron for each users on the system
cat /var/log/user_history/nttd_jp1_history | while read line ; do if [[ $line =~ '^#' ]]; then date -d "@$(echo $line | cut -c2-)
"; else echo $line ; fi; done