date manipulation

[csunix]

I am calculating "yesterdays date" but am losing the leading zero aon the day and month.

 

[PHV]

Take a look here: faq822-4802

Hope This Help, PH.
Want to get great answers to your Tek-Tips questions? Have a look at FAQ219-2884

 

[rkrishnan]

The following function library might be of some help to you. I got the outline logic for this from somewhere in the internet and added my improvements to these functions.

*********** START LIBRARY SCRIPT **********
#!/bin/ksh

#=========================================================================
dateAdd() # Adds number of days to a given date
#=========================================================================
# ARG_1 -> The date to add days DDMMYYYY format
# ARG_2 -> The number of days to add
# ARG_3 -> The date is returned in which format?
# 1 -> dd/mm/yyyy
# 2 -> mm/dd/yyyy
# 3 -> dd/mm/yy
# 4 -> mm/dd/yy
{

# Check that the input is valid.
# There should be exactly 3 arguments
if [ $# -ne 3 ]; then
return 1
fi

# The argument should be an integer.
n=`expr $2 + 0 2> /dev/null`
if [ $? -ne 0 ]; then
qnbad=0
elif [ $n -lt 0 ]; then
qnbad=0
else
qnbad=1
fi

if [ $qnbad -eq 0 ]; then
return 1
fi

# The "local" defnitions help to trickily
# return the values to the caller
local day=''
local month=''
local year=''

# Set the month day and year.
day=`echo $1 | cut -c1-2`
month=`echo $1 | cut -c3-4`
year=`echo $1 | cut -c5-8`

# Add 0 to month. This is a
# trick to make month an unpadded integer.
month=`expr $month + 0`

# Add n to the current day.
day=`expr $day + $n`

# Add months and subtract days until
# the future date is reached.
while : # Indefinite TRUE loop
do
# Determine the number of days in the
# current month.
case $month in
1|3|5|7|8|10|12) curMonthDay=31;;
4|6|9|11) curMonthDay=30;;
2)
if [ `expr $year % 4` -eq 0 ]; then
if [ `expr $year % 400` -eq 0 ]; then
curMonthDay=29
elif [ `expr $year % 100` -eq 0 ]; then
curMonthDay=28
else
curMonthDay=29
fi
else
curMonthDay=28
fi
;;
esac

# If the day is greater than the number
# of days in the current month then
# month is incremented; otherwise we are done.
if [ $day -gt $curMonthDay ]; then
day=`expr $day - $curMonthDay`
month=`expr $month + 1`

# If the month is 13 then it is jan of
# next year.
if [ $month -eq 13 ]; then
month=1
year=`expr $year + 1`
fi
else
break
fi
done

# Pad the integers before sending
[ $day -lt 10 ] && day="0$day"
[ $month -lt 10 ] && month="0$month"

# Print the month day and year according to the format
case "$3" in
1) echo $day'/'$month'/'$year;;
2) echo $month'/'$day'/'$year;;
3) echo $day'/'$month'/'`echo $year | cut -c3-4`;;
4) echo $month'/'$day'/'`echo $year | cut -c3-4`;;
esac

exit 0
}

#=========================================================================
dateSub() # Subtracts number of days to a given date
#=========================================================================
# ARG_1 -> The date to subtract days DDMMYYYY format
# ARG_2 -> The number of days to subtract
# ARG_3 -> The date is returned in which format?
# 1 -> dd/mm/yyyy
# 2 -> mm/dd/yyyy
# 3 -> dd/mm/yy
# 4 -> mm/dd/yy
{
# Check that the input is valid.
# There should be exactly 3 arguments.
if [ $# -ne 3 ]; then
return 1
fi

# The argument should be an integer.
n=`expr $2 + 0 2> /dev/null`
if [ $? -ne 0 ]; then
qnbad=0
elif [ $n -lt 0 ]; then
qnbad=0
else
qnbad=1
fi

if [ $qnbad -eq 0 ]; then
return 1
fi

# The "local" defnitions help to trickily
# return the values to the caller
local day=''
local month=''
local year=''

# Set the month day and year.
day=`echo $1 | cut -c1-2`
month=`echo $1 | cut -c3-4`
year=`echo $1 | cut -c5-8`

# Add 0 to month. This is a
# trick to make month an unpadded integer.
month=`expr $month + 0`

# Subtrace n from the current day.
day=`expr $day - $n`

# While the day is less than or equal to
# 0, deincrement the month.
while [ $day -le 0 ]
do
month=`expr $month - 1`

# If month is 0 then it is Dec of last year.
if [ $month -eq 0 ]; then
year=`expr $year - 1`
month=12
fi

# Add the number of days appropriate to the
# month.
case $month in
1|3|5|7|8|10|12) day=`expr $day + 31`;;
4|6|9|11) day=`expr $day + 30`;;
2)
if [ `expr $year % 4` -eq 0 ]; then
if [ `expr $year % 400` -eq 0 ]; then
day=`expr $day + 29`
elif [ `expr $year % 100` -eq 0 ]; then
day=`expr $day + 28`
else
day=`expr $day + 29`
fi
else
day=`expr $day + 28`
fi
;;
esac
done

# Pad the integers before sending
[ $day -lt 10 ] && day="0$day"
[ $month -lt 10 ] && month="0$month"

# Print the month day and year according to the format
case "$3" in
1) echo $day'/'$month'/'$year;;
2) echo $month'/'$day'/'$year;;
3) echo $day'/'$month'/'`echo $year | cut -c3-4`;;
4) echo $month'/'$day'/'`echo $year | cut -c3-4`;;
esac
exit 0

}
*********** END LIBRARY SCRIPT **********

 

[Castelior]

A Perderabo's script take on a other forum
It is really good



#! /usr/bin/ksh

# datecalc -- Perderabo's date calculator
#

USAGE="datecalc -a year month day - year month day
datecalc -a year month day [-|+] n
datecalc -d year month day
datecalc -D year month day
datecalc -j year month day
datecalc -j n
datecalc -l year month
use \"datecalc -help\" use for more documentation"

DOCUMENTATION=" datecalc Version 1.1

datecalc does many manipulations with dates.
datecalc -a is for date arithmetic
datecalc -d or -D converts a date to the day of week
datecalc -j converts to date to or from julian day
datecalc -l outputs the last day of a month

All dates must be between the years 1860 and 3999.

datecalc -a followed by 7 parameters will calculate the
number of days between two dates. Parameters 2-4 and 6-8
must be dates in ymd form, and parameter 5 must be a minus
sign. The output is an integer. Example:

> datecalc -a 1960 12 31 - 1922 2 2
14212


datecalc -a followed by 5 parameters will calculate the
a new date offset from a given date, Parameters 2-4 must
be a date in ymd form, paramter 5 must be + or -, and
paramter 6 must be an integer. Output is a new date.
Example:

> datecalc -a 1960 12 31 + 7
1961 1 7


datecalc -d followed by 3 parameters will convert a date
to a day-of-week. Parameters 2-4 must be a date in ymd
form. Example:

> datecalc -d 1960 12 31
6


datecalc -D is like -d except it displays the name of
the day. Example:

> datecalc -D 1960 12 31
Saturday


datecalc -j followed by 3 parameters will convert a date
to Modified Julian Day number. Example:
> datecalc -j 1960 12 31
37299


datecalc -j followed by a single parameter will convert
a Modified Julian Day number to a date. Example:
> datecalc -j 37299
1960 12 31


datecalc -l followed by year and month will output the last
day of that month. Note that by checking the last day of
February you can test for leap year. Example:
> datecalc -l 2002 2
28"


lastday() {
integer year month leap
# ja fe ma ap ma jn jl ag se oc no de
set -A mlength xx 31 28 31 30 31 30 31 31 30 31 30 31

year=$1
if ((year<1860 || year> 3999)) ; then
print -u2 year out of range
return 1
fi
month=$2
if ((month<1 || month> 12)) ; then
print -u2 month out of range
return 1
fi

if ((month != 2)) ; then
print ${mlength[month]}
return 0
fi

leap=0
if ((!(year%100))); then
((!(year%400))) && leap=1
else
((!(year%4))) && leap=1
fi

feblength=28
((leap)) && feblength=29
print $feblength
return 0
}


date2jd() {
integer ijd day month year mnjd jd lday

year=$1
month=$2
day=$3
lday=$(lastday $year $month) || exit $?

if ((day<1 || day> lday)) ; then
print -u2 day out of range
return 1
fi

((standard_jd = day - 32075
+ 1461 * (year + 4800 - (14 - month)/12)/4
+ 367 * (month - 2 + (14 - month)/12*12)/12
- 3 * ((year + 4900 - (14 - month)/12)/100)/4))
((jd = standard_jd-2400001))


print $jd
return 0
}


jd2dow()
{
integer jd dow numeric_mode
set +A days Sunday Monday Tuesday Wednesday Thursday Friday Saturday

numeric_mode=0
if [[ $1 = -n ]] ; then
numeric_mode=1
shift
fi


jd=$1
if ((jd<1 || jd>782028)) ; then
print -u2 julian day out of range
return 1
fi

((dow=(jd+3)%7))

if ((numeric_mode)) ; then
print $dow
else
print ${days[dow]}
fi
return
}

jd2date()
{
integer standard_jd temp1 temp2 jd year month day

jd=$1
if ((jd<1 || jd>782028)) ; then
print julian day out of range
return 1
fi
((standard_jd=jd+2400001))
((temp1 = standard_jd + 68569))
((temp2 = 4*temp1/146097))
((temp1 = temp1 - (146097 * temp2 + 3) / 4))
((year = 4000 * (temp1 + 1) / 1461001))
((temp1 = temp1 - 1461 * year/4 + 31))
((month = 80 * temp1 / 2447))
((day = temp1 - 2447 * month / 80))
((temp1 = month / 11))
((month = month + 2 - 12 * temp1))
((year = 100 * (temp2 - 49) + year + temp1))
print $year $month $day
return 0
}


#
# Parse parameters and get to work.
case $1 in
-a) if (($# == 8)) ; then
if [[ $5 != - ]] ; then
print -u2 - "$USAGE"
exit 1
fi
jd1=$(date2jd $2 $3 $4) || exit $?
jd2=$(date2jd $6 $7 $8) || exit $?
((jd3=jd1-jd2))
print $jd3
exit 0
elif (($# == 6)) ; then
jd1=$(date2jd $2 $3 $4) || exit $?
case $5 in
-|+) eval '(('jd2=${jd1}${5}${6}'))'
jd2date $jd2
exit $?
;;
*)
print -u2 - "$USAGE"
exit 1
;;
esac

fi
;;

-d|-D) if (($# != 4)) ; then
print -u2 - "$USAGE"
exit 1
fi
jd1=$(date2jd $2 $3 $4) || exit $?
numeric=-n
[[ $1 = -D ]] && numeric=""
eval jd2dow $numeric $jd1
exit $?
;;

-j) if (($# == 4)) ; then
date2jd $2 $3 $4
exit $?
elif (($# == 2)) ; then
jd2date $2 $3 $4
exit $?
else
print -u2 - "$USAGE"
exit 1
fi
;;

-l) if (($# == 3)) ; then
lastday $2 $3
exit $?
else
print -u2 - "$USAGE"
exit 1
fi
;;

-help) print - "$USAGE"
print ""
print - "$DOCUMENTATION"
exit 0
;;

*) print -u2 - "$USAGE"
exit 0
;;


esac

#not reached
exit 7