Formatting epoch time - solaris

Question

I have this command, which I found here:
/usr/bin/truss /usr/bin/date 2>&1 |
nawk -F= '/^time\(\)/ {gsub(/" "/,"",$2);printf "0t%d=Y\n", $2-30*86400}' |
adb
which prints 2014 Jan 11 09:48:54. Does anyone know how to reformat the output to YYYYMMDDHH?
I have to use ksh on Solaris. I can't use perl.

Answer 1

A couple of comments:
the first command:
/usr/bin/truss /usr/bin/date 2>&1 |
nawk -F= '/^time\(\)/ {gsub(/" "/,"",$2);printf "0t%d=Y\n", $2-30*86400}'
might be replaced by the much lighter and simpler
nawk 'BEGIN {printf("0t%d=Y\n",srand()-30*86400)}'`
I'm not sure adb is understanding the 0txx=Y syntax but mdb definitely does.
That said, here is something that should fit your needs:
nawk 'BEGIN {printf("0t%d=Y\n",srand()-30*86400)}' | mdb | nawk '
function m2m(m)
{
if(m=="Jan")return 1
if(m=="Feb")return 2
if(m=="Mar")return 3
if(m=="Apr")return 4
if(m=="May")return 5
if(m=="Jun")return 6
if(m=="Jul")return 7
if(m=="Aug")return 8
if(m=="Sep")return 9
if(m=="Oct")return 10
if(m=="Nov")return 11
if(m=="Dec")return 12
}
{
y=$1
m=$2
d=$3
split($4,t,":")
h=t[1]
min=t[2]
printf("%s%02d%02d%02d%02d\n",y,m2m(m),d,h,min)
}'
Notes:
srand (not to be confused with rand) is returning the previous seed when called.
When no argument is passed, srand is using the "time of the day" as seed.
As srand is initialized internally under Solaris nawk, the first srand() call does return the time of the day, i.e. the number of seconds elapsed since the epoch.
If you want a strictly portable way, you need to run a first srand() call before getting its return value but this is not required under Solaris.
This is portable (i.e. POSIX) unlike using truss.

Answer 2

If you have Solaris 11, the answer is easy, because its ksh is ksh93. The Solaris 11 ksh man page has this info about printf:
A %(date-format)T format can be use to treat an argument as a date/time string
and to format the date/time according to the date-format as defined for the
date(1) command.
The allowable date/time string formats aren't specified in the ksh man page, but are (somewhat) mentioned in the AST tm(3) man page. A wide variety are accepted. In particular, this will do what you want:
$ printf "%(%Y%m%d%H)T\n"
2014021118
$ printf "%(%Y%m%d%H)T\n" now
2014021118
$ printf "%(%Y%m%d%H)T\n" "30 days ago"
2014011218

Answer 3

EDIT: Originally I had formed a response that works nicely on Linux, but not on Solaris. This is much longer now, with the original answer below.
To do this purely in the shell, on Solaris, using only the date function, is possible. Kind of. You would have to create some shell functions as well to help out with the subtraction of days from a given date.
Using the functions below, you could do the following to get the desired results:
$ mydate="$(date '+%Y%m%d')"
$ hours="$(date '+%H')"
$ echo "$(date_subtract "${mydate}" 30)${hours}" | adb
or something like that...
On Solaris, the date command only provides Gregorian dates. This is harder to work with for the purpose of this question, so I created a shell function, gregorian_to_julian which can convert a date in 'YYYYMMDD' format to a Julian date of 'YYYYDDD'.
function gregorian_to_julian {
# "Expecting input as YYYYMMDD."
typeset gregorian_date=$1
typeset _year=$(echo "${gregorian_date}" | cut -c1-4)
typeset _month=$(echo "${gregorian_date}" | cut -c5-6)
typeset _day=$(echo "${gregorian_date}" | cut -c7-8)
typeset days
days[01]=31
days[02]=$(( days[01] + 28 ))
if (( _year % 4 == 0 )) && (( _year % 100 != 0 )) || (( _year % 400 == 0 )); then
days[02]=$(( days[01] + 29 ))
fi
days[03]=$(( 31 + days[02] ))
days[04]=$(( 30 + days[03] ))
days[05]=$(( 31 + days[04] ))
days[06]=$(( 30 + days[05] ))
days[07]=$(( 31 + days[06] ))
days[08]=$(( 31 + days[07] ))
days[09]=$(( 30 + days[08] ))
days[10]=$(( 31 + days[09] ))
days[11]=$(( 30 + days[10] ))
typeset julian_date=0
case "${_month}" in
"01") julian_date=$_day ;;
"02") julian_date=$(( days[01] + _day )) ;;
"03") julian_date=$(( days[02] + _day )) ;;
"04") julian_date=$(( days[03] + _day )) ;;
"05") julian_date=$(( days[04] + _day )) ;;
"06") julian_date=$(( days[05] + _day )) ;;
"07") julian_date=$(( days[06] + _day )) ;;
"08") julian_date=$(( days[07] + _day )) ;;
"09") julian_date=$(( days[08] + _day )) ;;
"10") julian_date=$(( days[09] + _day )) ;;
"11") julian_date=$(( days[10] + _day )) ;;
"12") julian_date=$(( days[11] + _day )) ;;
esac
julian_date="${_year}${julian_date}"
echo "${julian_date}"
}
Understanding that the OP wants the date formatted as a Gregorian date, I created a second function, julian_to_gregorian which converts the Julian date back.
function julian_to_gregorian {
# "Expecting input as YYYYDDD."
#set -x
typeset julian_date=$1
typeset _year="$(echo "${julian_date}" | cut -c1-4)"
typeset _month=""
typeset julian_day="$(echo "${julian_date}" | cut -c5-7)"
typeset -RZ2 _day=0
typeset days
days[01]=31
days[02]=$(( days[01] + 28 ))
if (( _year % 4 == 0 )) && (( _year % 100 != 0 )) || (( _year % 400 == 0 )); then
days[02]=$(( days[01] + 29 ))
fi
days[03]=$(( 31 + days[02] ))
days[04]=$(( 30 + days[03] ))
days[05]=$(( 31 + days[04] ))
days[06]=$(( 30 + days[05] ))
days[07]=$(( 31 + days[06] ))
days[08]=$(( 31 + days[07] ))
days[09]=$(( 30 + days[08] ))
days[10]=$(( 31 + days[09] ))
days[11]=$(( 30 + days[10] ))
if (( days[11] < julian_day )); then _month="12"; _day=$(( julian_day - days[11] ));
elif (( days[10] < julian_day )); then _month="11"; _day=$(( julian_day - days[10] ));
elif (( days[09] < julian_day )); then _month="10"; _day=$(( julian_day - days[09] ));
elif (( days[08] < julian_day )); then _month="09"; _day=$(( julian_day - days[08] ));
elif (( days[07] < julian_day )); then _month="08"; _day=$(( julian_day - days[07] ));
elif (( days[06] < julian_day )); then _month="07"; _day=$(( julian_day - days[06] ));
elif (( days[05] < julian_day )); then _month="06"; _day=$(( julian_day - days[05] ));
elif (( days[04] < julian_day )); then _month="05"; _day=$(( julian_day - days[04] ));
elif (( days[03] < julian_day )); then _month="04"; _day=$(( julian_day - days[03] ));
elif (( days[02] < julian_day )); then _month="03"; _day=$(( julian_day - days[02] ));
elif (( days[01] < julian_day )); then _month="02"; _day=$(( julian_day - days[01] ));
else
_month="01"; _day=${julian_day};
fi
echo "${_year}${_month}${_day}"
}
A third function, date_subtract, takes a Gregorian date, converts it to Julian, does the date math, and converts back to Gregorian.
function date_subtract {
typeset julian_from_date=$(gregorian_to_julian "$1")
typeset number_of_days=$2
typeset julian_year=$(echo "${julian_from_date}" | cut -c1-4)
typeset julian_days=$(echo "${julian_from_date}" | cut -c5-7)
typeset leap_year="FALSE"
if (( julian_days - number_of_days > 0 )); then
(( julian_days -= number_of_days ))
else
(( julian_year -= 1 ))
if (( julian_year % 4 == 0 )) && (( julian_year % 100 != 0 )) || (( julian_year % 400 == 0 )); then
leap_year="TRUE"
fi
if [[ "${leap_year}" == "TRUE" ]]; then
(( julian_days = julian_days + 366 - number_of_days ))
else
(( julian_days = julian_days + 365 - number_of_days ))
fi
fi
typeset gregorian_date=$(julian_to_gregorian "${julian_year}${julian_days}")
echo "${gregorian_date}"
}
EDIT: The following works nicely on Linux. And is much simpler.
If you need the date formatted as YYMMDDHH then that is simple enough, as #FrankH. pointed out in his comment:
date +"%Y%m%d%H"
The trickier part is calculating the date 30 days ago, which seems to be your intent. The date command can take the seconds since the epoch and convert that to a date string. This is done with the -d flag. e.g.
$ date -d #123456789
Thu Nov 29 15:33:09 CST 1973
So, if you need the date for thirty days ago, we can take this principle and do a little math to get the needed value.
Note that date '+%s' gives us seconds since the epoch.
$ date -d #$(( $(date '+%s') - $((60 * 60 * 24 * 30 )) )) '+%Y%m%d%H'
2014011211
If you need to pipe that date string to adb then you can surround the command with $(...).
echo $( date -d #$(( $(date '+%s') - $((60 * 60 * 24 * 30 )) )) '+%Y%m%d%H' ) | adb