Date Parsing

[moonoo]

Hi I have a shell program which takes date as a argument in the format YYYYMMDD . So i have $1 as 20031217 inside my shell Program . How do i parse this date to Year , Month and Day variables ..Someone suggested to use awk ,,but not sure how to use it..
Regards
dev

 

[vgersh99]

something like that with 'awk':

eval echo "20031221" | awk '{ print "myY=" substr($0,1,4), "myM=" substr($0,5,2), "myD=" substr($0,7,2)}'

set | grep my


vlad
+----------------------------+
| #include<disclaimer.h> |
+----------------------------+

 

[theotyflos]

You can also use cut:

Y=`echo $1|cut -c-4`
M=`echo $1|cut -c5-6`
D=`echo $1|cut -c7-`

 

[aigles]

Another solution with sed :

echo &quot;$1&quot; | sed -e 's/\(....\)\(..\)\(..\)/\1 \2 \3/' | read Y M D


Jean Pierre.

 

[PHV]

Yet another solution, with sed:
eval echo $1 | sed 's!^\(....\)\(..\)\(..\)$!Y=\1;M=\2;D=\3!'
echo &quot;Date=$1, Year=$Y, Month=$M, Day=$D&quot;

Hope This Help
PH.

 

[Ygor]

Or with expr...

Year=`expr substr $1 1 4`
Month=`expr substr $1 5 2`
Day=`expr substr $1 7 2`

 

[Chapter11]

another solution, assuming use of gnu's date:

date -d ${1} +&quot;your_format_specification&quot;