How to cut the last 2 characters of a date parameter

[Suman_7]

I have a parameter which has a value
parm_date=20040401
I need to check if the day is first day of a month

how do I grep the '01' portion of this date?

Thanks

[A. Clay Stephenson]

Since your date string has no spaces, let's do this in an unconventional way. You can simply do a MOD 100.

DT=20040401
typeset -i10 DAY=$(( ${DT} % 100 ))
if [[ ${DAY} -eq 1 ]]
then
echo "First Day of Month"
fi

[Sanjay Kumar Suri · Accepted solution]

Check the following:
parm_date=20040401
dd=`expr substr $parm_date 7 2`
echo $dd

sks

[Victor Fridyev]

Try this:


if grep ......01$ >/dev/null ; then
echo this is a first day of a month
else
:
fi

[Bill Hassell]

How about something simple like cut/if rather than grep:

parm_date=20040401
DAY=$(echo $parm_date | cut -c 7-8)
if [ "$DAY" = "01" ]
then
...
fi

[Arturo Galbiati]

in_date=20040401
dd=$(echo $in_date | cut -c 7-8)
if [ "$dd" = "01" ]
then
...
fi

HTH
Art