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[yesterday date in shell/unix]

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jebok Posted messages 371 Registration date   Status Membre -  
 alibat -
Bonjour,

Which command (in shell-UNIX) displays yesterday's date?
PS: I clarify that it is for a ksh script

--
Thank you for your help

12 réponses

Réponse 1 / 12
jipicy Posted messages 40842 Registration date   Status Modérateur Last intervention   4 898
 
Ben if it works, the proof is:
In bash [jp@Mandrake jp]$ date --date '1 days ago' Tue Aug 2 11:38:36 CEST 2005 We switch shell (the command is an alias) [jp@Mandrake jp]$ kosh $ date --date '2 days ago' Mon Aug 1 11:39:00 CEST 2005 $ date --date '1 days ago' Tue Aug 2 11:39:22 CEST 2005
;-))
--
Z'@+...che.
JP : Zen, my Nuggets ! ;-) Knowledge is only good if it's shared.
15
jebok Posted messages 371 Registration date   Status Membre 51
 
look at
http://www.commentcamarche.net/forum/affich-1712417#1
the error message I am getting...

--
Thank you for your help
0
Réponse 2 / 12
raidcha
 
Under Solaris and HP-UX, there is a simpler solution than creating a script.
Just play with the time zone (system variable $TZ).

Example:
# echo `TZ=MET+24 date +"%D"`
gives: 11/28/05 (today is 11/29/05)

The “+24” corresponds to the number of hours to "subtract" (+) from the current time; if you want to add them, use - (see below)
%D corresponds to the date formatting (see man date): mm/dd/yy

Similarly, you could get the date for tomorrow in the format yyyymmdd, for example….
#echo `TZ=MET-24 date +"%Y%m%d"`
gives: 20051128

... or the day after tomorrow (only works under Solaris):
#echo `TZ=MET-48 date +"%Y%m%d"`
this last command doesn’t work under HP-UX because it seems you can't go around the earth more than once with HP :p
14
Lucien Hercaud
 
On Solaris, it works for a maximum of a few days' difference.
For more than that, you need to compile a file with "zic", e.g.:

-bash-3.00$ cat Delta
# Example: 100 days in the future 100*24 hours
Zone POSE/Zulu+2400 2400 - POSE
# Example: 365 days in the past 365*24 hours
Zone POSE/Zulu-8760 -8760 - POSE
# Example: 1 year and 3 months in the past: (365+90)*24 hours
Zone POSE/Zulu-10920 -10920 - POSE
# Example: 2 years and 3 months in the past: (2*365+90)*24 hours
Zone POSE/Zulu-19680 -19680 - POSE
# Example: 5 years and 3 months in the past: (5*365+90)*24 hours
Zone POSE/Zulu-46008 -46008 - POSE

-bash-3.00$ /usr/sbin/zic Delta
-bash-3.00$ TZ=POSE/Zulu+2400 date
Thu Jan 19 12:36:59 POSE 2012
-bash-3.00$ TZ=POSE/Zulu-46008
-bash-3.00$ date
Wed Jul 12 12:32:20 POSE 2006
0
ZdraL
 
Great, it works thank you so much =)
0
Réponse 3 / 12
Cyrus
 
date -d yesterday +%F
11
Réponse 4 / 12
sclamagirand Posted messages 1 Status Membre 5
 
Here is a script that should work:

#!/bin/ksh
#
set -A DAYS Sat Sun Mon Tue Wed Thu Fri Sat
set -A MONTHS Dec Jan Feb Mar Apr May Jun Jul Aug Sep Oct Nov Dec
#
# works on Linux
#
# date -d '1 days ago'
#
YESTERDAY=$((`date +%d` -1))
MONTH=`date +%m`
YEAR=`date +%Y`
NDAY=`date +%u`
WEEKDAY=${DAYS[`date +%u`]}
#
if [ $YESTERDAY -eq "0" ];
then
#
MONTH=$((MONTH-1))
#
if [ $MONTH -eq "0" ];
then
#
MONTH=12
YEAR=$((YEAR-1))
#
fi
#
set `cal $MONTH ${YEAR}`
shift $(($# - 1))
YESTERDAY=$1
#
fi
#
TMONTH=${MONTHS[MONTH]}
YEAR2=${YEAR##20}
#
# uncomment next line for debugging
#
echo ${WEEKDAY} ${YESTERDAY} ${TMONTH} ${YEAR}
#
echo ${YESTERDAY}${MONTH}${YEAR2}
#
5
skea
 
Thank you slamagirand, it's perfect!!!
0
Réponse 5 / 12
jipicy Posted messages 40842 Registration date   Status Modérateur Last intervention   4 898
 
See there: http://www.commentcamarche.net/forum/affich-1712417#1
Catch you later...che.
JP: Zen, my Nuggets! ;-) Knowledge is only good if it is shared.
1
jebok Posted messages 371 Registration date   Status Membre 51
 
No, it doesn't work...

--
Thank you for your help.
0
Réponse 6 / 12
francis
  
if [ `expr `date +%d` - 1` -le 0 ]; then cal | grep -E "28|29|30|31" | awk ........... fi


anyway, I'm going to write it anyway...
1
Réponse 7 / 12
[Dal] Posted messages 6205 Registration date   Status Contributeur Last intervention   1 108
 
Hello,

You can also convert the current date to seconds since Epoch, then subtract 86400 (24*60*60) from the result and reconvert the obtained number to the desired date format.

See man date, and man strftime for the formats.

Dal
1
Réponse 8 / 12
lionel sévérian
 
Here is the solution to your problem. It works and it's ksh AIX...

GetDate()
{ # GetDate nDays [format]
# Example of usage: export NAMEDIR=$(GetDate -1 '+%Y.%m.%d')

typeset -i nDays=$1; format=$2
eval $(echo $TZ | sed '
s!\([^-0-9]*\)\([-0-9]*\)\(.*\)!typeset -i localOffset=\2;zon1=\1;zon2=\3!')
TZ=$zon1$((localOffset-24*nDays))$zon2 date $format

}

To be used in KSH function without moderation...
Copy/Paste everything that is in bold...
1
Vag
 
Thank you for this code, it really helped me out!! (and thanks Google)

My problem was slightly different, I wanted to find the date of the last Sunday...

Thanks to your script I was able to do this:

case $(date +%a) in
Mon ) diff=-1;;
Tue ) diff=-2;;
Wed ) diff=-3;;
Thu ) diff=-4;;
Fri ) diff=-5;;
Sat ) diff=-6;;
Sun ) diff=0;;
esac
DATE1=$(GetDate $diff '+%d/%m/%Y')

It works but it's not very pretty :)

Would it be possible to make a function GetDate2? with $1 = the day to get (for last Sunday we put Sun) and $2 still the format

I'm sure it could be done recursively too :)
0
Réponse 9 / 12
francois
 
the syntax date X days ago does not exist on all Linux systems, not at all on proprietary Unix systems and not on all BSD systems either!

to be banned forever as a consequence.
one of the simple solutions: 
expr `date +%d` - 1


and of course if we want to manage the month and the year we need to create a shell script
that checks if the day is 0 then run cal of the previous month to see if it ends with 28, 29, 30, 31 and if it’s January subtract 1 from the year

nothing simpler really
0
alibat
 
Great! You're saving our lives!
0
Réponse 10 / 12
tontonserver
 
To calculate the date for J-1, don't overthink it:
DATE=$((`/bin/date +'%Y%m%d'` - 1))
There you go, it's in YYYYMMDD format, then, do it your way!
@+
0
Goldmund
 
It doesn't work on the 1st of the month!!
0
Réponse 11 / 12
uu
 
```bash #!/bin/ksh
#
set -A DAYS Sat Sun Mon Tue Wed Thu Fri Sat
set -A MONTHS Dec Jan Feb Mar Apr May Jun Jul Aug Sep Oct Nov Dec
set -A MONTH_ 12 01 02 03 04 05 06 07 08 09 10 11 12
set -A DAYS_ 09 01 02 03 04 05 06 07 08 09
#
# works on Linux
#
# date -d '1 days ago'
#
#YESTERDAY=$((`date +%d` -1))
date1=20070101
YESTERDAY=$((`date -d $date1 +%d` -1))
MONTH=`date -d $date1 '+%m' `
YEAR=`date -d $date1 '+%Y' `
NDAY=`date -d $date1 '+%d' `

WEEKDAY=${DAYS[`date +%u`]}
#
if [ $YESTERDAY -eq "0" ];
then
#
MONTH=$((MONTH-1))
#
if [ $MONTH -eq "0" ];
then
#
MONTH=12
YEAR=$((YEAR-1))
#
fi
#
set `cal $MONTH ${YEAR}`
shift $(($# - 1))
YESTERDAY=$1
#
fi
#
YESTERDAY_=$YESTERDAY
if [ $YESTERDAY -ne "0" ];
then
if [ $YESTERDAY -le 9 ];
then
#
YESTERDAY_=${DAYS_[YESTERDAY]}
#
fi
fi
#
TMONTH=${MONTHS[MONTH]}
T_MONTH=${MONTH_[MONTH]}
YEAR2=${YEAR##20}
#
# uncomment next line for debugging
#
echo ${WEEKDAY} ${YESTERDAY} ${TMONTH} ${YEAR}
#
#echo ${YESTERDAY}${MONTH}${YEAR2}
#
date=${YEAR}${T_MONTH}${YESTERDAY_}
echo $date
# ```
0
Réponse 12 / 12
ddark2002
 
Hello,

date=$(date)
set $date

date=$(date --date '1 days ago')
set $datedate=$(date)
set $date

I would like to create a file with 2 time variables to make a comparison.
diff -c /home/file_.$2$3 /home/file_.$2$3-1j > /home/test.log
0
dallas
 

DTE=`date --date '24 hours ago' '+%Y%m%d'`
0
BBH
 
```html Use the following functions:

function TimeToSecond {
# ${1} = Optional date and time (YYYYMMDDHHMMSS); it defaults to current time
( typeset -r awk_date="\"y=\" substr(\$1,01,4) \"\\nm=\" substr(\$1,05,2) \"\\nd=\" substr(\$1,07,2)"
typeset -r awk_time="\"h=\" substr(\$1,09,2) \"\\ni=\" substr(\$1,11,2) \"\\ns=\" substr(\$1,13,2)"
if [ -z "${1}" ]
then date +%Y%m%d%H%M%S
else echo "${1}"
fi |\
awk "{ print ${awk_date} \"\\n\" ${awk_time} }"
echo
echo "m += 9"
echo "if (m <= 11) {"
print "\ty -= 1 }"
echo "if (m > 11) {"
print "\tm -= 12 }"
print "(((((y-1)*1461+1)/4 - y/100 + y/400 + (m*153+2)/5 + d + 59)*24 + h)*60 + i)*60 + s")\
| bc
}

function SecondsToTime {
# ${1} = number of seconds since January 1st of year 0001
( echo "s = ${1}"
echo "i = s/60"
echo "s -= i*60"
echo "h = i/60"
echo "i -= h*60"
echo "d = h/24"
echo "h -= d*24"
echo "d += 305"
echo "b = d/146097"
echo "d -= b*146097"
echo "y = b*400"
echo "b = d/36524"
echo "if (b > 3) {"
print "\tb = 3 }"
echo "d -= b*36524"
echo "y += b*100"
echo "b = d/1461"
echo "d -= b*1461"
echo "y += b*4"
echo "b = d/365"
echo "if (b > 3) {"
print "\tb = 3 }"
echo "d -= b*365"
echo "y += b"
echo "m = (d*5+2)/153"
echo "d -= (m*153+2)/5-1"
echo "m += 3"
echo "if (m > 12) {"
print "\tm -= 12"
print "\ty += 1 }"
echo "y"
echo "m"
echo "d"
echo "h"
echo "i"
echo "s"
)\
| bc | paste -s - | awk '{ printf "%04d%02d%02d%02d%02d%02d\n",$1,$2,$3,$4,$5,$6 }'
} ```
0