Need help in Shell scripting

[Whitehorse_1]

Hi ,

I just born new in Shell scripting., I gota to know, If Im giving "cal 7 1950", it should gimme the number of days in that calender.
Anyways please..

WH..

[Cem Tugrul]

That's correct...you can see July of 1950

[Cem Tugrul]

and if you give just "cal 1950"
and you see all month's of day of the year
1950
Good Luck,

[Whitehorse_1]

Cem, Thxs for reply, I hope I need to clarify my Qn. a bit..

I need a script/command/function(interactive), using which, If I give the "Month and Year" as input, I should just give the output as "The month/Yr which you have typed has got XY days"...I just need the number of days, if I give any month/yr..

WH..

[Cem Tugrul]

i may misunderstood..Do you need to manipulate "date" if so here are some examples;
baan01:/#date
Mon Aug 28 10:01:44 EETDST 2006

## Work out what yesterday's date was

if [ "${DAY}" = "01" ]
then
case ${MTH} in
Jan)SEARCH_MTH=Dec;((SEARCH_YEAR=YEAR-1));SEARCH_DAY=31;
;
Feb)SEARCH_MTH=Jan;SEARCH_YEAR=${YEAR};SEARCH_DAY=31;;
Mar)SEARCH_MTH=Feb;SEARCH_YEAR=${YEAR};SEARCH_DAY=28;;
Apr)SEARCH_MTH=Mar;SEARCH_YEAR=${YEAR};SEARCH_DAY=31;;
May)SEARCH_MTH=Apr;SEARCH_YEAR=${YEAR};SEARCH_DAY=30;;
Jun)SEARCH_MTH=May;SEARCH_YEAR=${YEAR};SEARCH_DAY=31;;
Jul)SEARCH_MTH=Jun;SEARCH_YEAR=${YEAR};SEARCH_DAY=30;;
Aug)SEARCH_MTH=Jul;SEARCH_YEAR=${YEAR};SEARCH_DAY=31;;
Sep)SEARCH_MTH=Aug;SEARCH_YEAR=${YEAR};SEARCH_DAY=31;;
Oct)SEARCH_MTH=Sep;SEARCH_YEAR=${YEAR};SEARCH_DAY=30;;
Nov)SEARCH_MTH=Oct;SEARCH_YEAR=${YEAR};SEARCH_DAY=31;;
Dec)SEARCH_MTH=Nov;SEARCH_YEAR=${YEAR};SEARCH_DAY=30;;
esac
else
SEARCH_MTH=${MTH}
SEARCH_YEAR=${YEAR}
((SEARCH_DAY=DAY-1))
if [ "${SEARCH_DAY}" -lt "10" ]
then
SEARCH_DAY="0${SEARCH_DAY}"
fi
fi

ONE_DAY_BACK="${SEARCH_YEAR}-${SEARCH_MTH}-${SEARCH_DAY}"
TODAY="${SEARCH_YEAR}-${SEARCH_MTH}-${DAY}"

OR;

How to schedule a job on last friday;
## get actual day
DAY=`date +"%d"`

## get the last friday of curren month
EXEC_DAY=`cal | awk 'NF >=6 { last = $6 }; END{ print last }'`

if [ $DAY -ne $EXEC_DAY ]
then
echo " - Not run because is not the last friday - "
exit;
fi

## Here must write your script
## your commands

[Whitehorse_1]

Hi Cem,

Jan)SEARCH_MTH=Dec;((SEARCH_YEAR=YEAR-1));SEARCH_DAY=31;;
Feb)SEARCH_MTH=Jan;SEARCH_YEAR=${YEAR};SEARCH_DAY=31;;

In the above sample lines of yours, from wheres the ${YEAR} taken from..?


WH

[Cem Tugrul]

oppsss....sorry..
DAY=`date +%d`
MTH=`date +%m`
YEAR=`date +%Y`

[Cem Tugrul]

baan01:/#DAY=`date +%d`
baan01:/#echo $DAY
28
baan01:/#MTH=`date +%m`
baan01:/#echo $MTH
08
baan01:/#YEAR=`date +%Y`
baan01:/#echo $YEAR
2006
baan01:/#

[Joel Girot · Accepted solution]

hi WH,

is this script appropriate ?

#!/usr/bin/sh
# count days of a month
MYNAME=${0##*/}
if [ $# != 2 ] ; then
echo "usage : $MYNAME month year"
exit 1
fi
cal $1 $2 | tail +3 | wc -w

Joel

[Arturo Galbiati]

Hi,
you can use:

#/user/bin/ksh
# day_of_month
MM=${1:-$(date +%m)}
YYYY=${2:-$(date +%Y)}
TEMP=$(cal $MM $YYYY)
DAYS=$(echo $TEMP|awk '{print $NF}')
echo "$MM $YYYY has $DAYS"
#eof

day_of_month
08 2006 has 31

day_of_month 7 1950
7 1950 has 31

it runs fine on hp hp-ux 11i. (It take care leap years of course).

HTH,
Art