Hi Experts,
I am using the below code to get the previous day based on the date given as a input.
#!/usr/bin/ksh
datestamp=`date '+%Y%m%d'`
yest=$((datestamp -1))
echo $yest
Output: 20130714
How can i display the same output in 14/07/2013, i tired '+%d/%m/%y'` but i am getting error.
also how can i get the first day of the month based on the date which is fetched $yest.
Any help.
Thanks
Do you have GNU date? just check man page of date if u have GNU date then you have option to get yesterday's date..
If not you have to modify TZ to get the desired date.. date - 1 wont work 
If your date command supports the -d option (gnu date)
you can do:
Yesterday:
date -d "-1 day" +%d/%m/%Y
1st of month:
YM=`date +%Y%m`
date -d ${YM}01 +%d/%m/%Y
Hi Experts,
when i run the command in unix prompt: i got the below error:
date -d "-1 day" +%d/%m/%Y
date: illegal option -- d
Usage: date [-u] [+format]
date [-u] [mmddhhmm[[cc]yy]]
date [-a [-]sss.fff]
can you help me to sort this.
regards,
OK you don't have gnu date you might have to use perl
yesterday.pl
#!/bin/perl
use POSIX; print strftime('%d/%m/%Y', localtime(time() - 24*60*60));
first.pl
#!/bin/perl
use POSIX;
my $year, $month;
($year, $month) = split(/\s/, strftime('%Y %m', localtime(time())));
printf strftime('%d/%m/%Y', localtime(mktime(0,0,0,1,$month-1,$year-1900,0,0)));
Hi Experts,
I want to use in unix script.
any help/suggestions
regards,
For your first requirement, this should work
$ date -j -f "%s" $(($(date "+%s") - 86400)) "+%d/%m/%Y"
15/07/2013
Hi Experts,
GNU date is not supporting in my UNIX prompt, so based on today's date how can i get the first day' of the month. i want the first date of the month.
for example:
if current date is 16/07/2013, i want the 01/07/2013. i mean first of the month.
Thanks,
I am not sure if this is what you need..just check out..
$ date "+%d/%m/%Y" | awk -F"/" '{print "01"FS$2FS$3}'
01/07/2013
Hi Experts,
Thanks, this is working for current month, if we pass a different date like 12/06/2013, will this work?
how can we modify the code to take any date and get the first day of that month.
for example:
if we have input as 12/06/2013,the first date of the month is 01/06/2013.
how can we do this?
thanks in davance.
regards,
$ echo "12/06/2013" | awk -F"/" '{print "01"FS$2FS$3}'
01/06/2013
$ echo "31/12/2013" | awk -F"/" '{print "01"FS$2FS$3}'
01/12/2013
Hi Experts,
can i store this value in variable like.
inputdate=12/06/2013
date= echo "inputdate" | awk -F"/" '{print "01"FS$2FS$3}'
echo $date
output: Expected
01/06/2013
regards,
Yes. I would suggest you to give an attempt and try something yourself before posting.
date=$(echo $inputdate | awk -F"/" '{print "01"FS$2FS$3}')
Hi Experts,
I tired this.
#!/bin/ksh
inputdate=15/07/2013
year=`expr substr $inputdate 7 4`
month=`expr substr $inputdate 4 2`
FROM_DATE=01/$month/$year
echo $month
echo $year
echo $FROM_DATE
let me know if correct?
---------- Post updated at 02:01 AM ---------- Previous update was at 01:26 AM ----------
Hi Experts,
i am using the below code get the date of previous day.
#!/usr/bin/ksh
datestamp=`date '+%Y%m%d'`
yest=$((datestamp -1))
echo $yest
When i execute the code i am getting output as:
20130715
What i am trying here is, based on the date passed i am fetching previus day's date.
for example:
If date passed is date = 16/07/2013, i need date of previous day
15/07/2013 but now i am getting output as 20130715, how can i convert this to format 15/07/2013
Thanks in advance.
seems, here is the issue
datestamp=`date '+%Y%m%d'
( mentioned as Year,Month and date)
please change and try.
$ dt=20130715
$ mmdd=${dt#????}
$ echo ${mmdd#??}"/"${mmdd%??}"/"${dt%????}
15/07/2013
Hi Experts,
#!/usr/bin/ksh
date = 16/07/2013
datestamp=$date '+%d/%m/%Y'
yest=$((datestamp -1))
echo $previous_day
i am getting error, how can i get the pervious days date based on the input date, any suggestions please.
My input date is
#!/usr/bin/ksh
date = 18/06/2013
i want previous date 17/06/2013.
thanks