I am sure there is a simple answer to this :
Why does the following not give me the date and time in a script? What should I use rather than awk? Awk works fine if I need to query files but I guess it is the wrong thing to use here..
Code :
$ySTD=$(date --date yesterday "+%Y-%m-%d") | awk '{print substr($1,1,10)}'
echo ${ySTD}
echo 'STD calculated from DATE - last 24 hrs data will be fetched'
${ySTM}=$(date --date yesterday "+%T")| awk '{print substr($1,1,5)}'
echo ${ySTM}
In the shell I get this when the script runs:
+ awk '{print substr($1,1,10)}'
++ date --date yesterday +%Y-%m-%d
+ =2014-07-03
logdat3: line 41: =2014-07-03: command not found
+ echo
+ echo 'STD calculated from DATE - last 24 hrs data will be fetched'
STD calculated from DATE - last 24 hrs data will be fetched
+ awk '{print substr($1,1,5)}'
++ date --date yesterday +%T
+ =11:33:34
logdat3: line 45: =11:33:34: command not found
+ echo
Many thanks in advance
You're rigth, no awk needed:
$ read ySTD ySTM < <(date --date yesterday "+%Y-%m-%d %T")
$ echo $ySTD
2014-07-03
$ echo $ySTM
12:16:37
Anyway , your original code fixed:
ySTD=$(date --date yesterday "+%Y-%m-%d")
echo ${ySTD}
echo 'STD calculated from DATE - last 24 hrs data will be fetched'
ySTM=$(date --date yesterday "+%T")
echo ${ySTM}
You don't need to get a substring.
Related
I need help in completing the below script in shell.
Requested format is to print previous, current and next business days (should exclude weekends and holidays).
text name : holiday.txt
#!/usr/bin/sh
cur_date=`date +"%A %m/%d/%Y"`
cur_day=`date +"%A"`
yesterday=`TZ=AEDT+13 date +"%A %m/%d/%Y"`;
twodd_back=`TZ=AEDT+37 date +"%A %m/%d/%Y"`;
tomorrow=`TZ=AEDT-35 date +"%A %m/%d/%Y"`;
twodd_later=`TZ=AEDT-59 date +"%A %m/%d/%Y"`;
cdate=`date +"%m/%d/%Y"`
####################
echo "Previous Business Day is $yesterday"
if echo $yesterday|awk '{print $1}'=Saturday
then echo "Previous Business Day is $twodd_back"
fi
echo "Current Business Day is $ cur_date"
echo " Next Business Day is $tomorrow"
The script in its current status does not execute successfully. It fails with below error.
awk: syntax error at source line 1
context is
{print >>> $1}= <<< Saturday
awk: bailing out at source line 1
To fix it, change your if statement.
if [[ `echo $yesterday | awk '{print $1}'` = Saturday ]];
I have tried adding -d "yesterday" but I haven't had any luck getting it to work. Here is what I have for the whole script:
#! /bin/bash
saveDir="TJ"
dd=$(date +"%m-%d-%Y")
for file in *.csv ; do
saveName="${saveDir}/TJ ${dd}.csv"
cut -d',' -f2,14 "$file" > "$saveName"
done
how do I get dd to output yesterdays date instead of the current date?
EDIT: This is what I have now
#! /bin/bash
saveDir="TJ"
dd=$(date --date='yesterday' +'%m-%d-%Y')
for file in *.csv ; do
saveName="${saveDir}/TJ ${dd}.csv"
cut -d',' -f2,14 "$file" > "$saveName"
done
the above is saving the file as TJ .csv but I'm not sure what was done incorrectly
I think you want to use -
$ cat test.sh
#!/bin/bash
dd=$(date --date='yesterday' +'%m-%d-%Y')
echo $dd
$ ./test.sh
12-31-2013
or you could use
$ date -d '1 day ago' +'%m-%d-%Y'
12/31/2013
And for tomorrow -
$ date -d '1 day' +'%m-%d-%Y'
01/02/2014
or
$ date --date='tomorrow'
Thu Jan 2 21:25:00 EST 2014
Get today's date in seconds since epoch. Subtract 86400 to get to yesterday. Then convert yesterday to the string format you want.
today=`date +"%s"`
yesterday=`expr $today - 86400`
dd=`date --date="#${yesterday}" +"%m-%d-%Y"`
Try this
yday=$(date --date yesterday "+%d-%m-%Y")
echo $yday
And If you works in Linux
yday=$(date -d "-1 days" +"%d-%m-%Y")
echo $yday
I tried date -d "yesterday" +%m-%d-%Y on my Linux, it worked fine.
If you are on unix platform, you cannot use -d,
you can get yesterday's date using perl, this is how I do using perl
dd=$(perl -e '($a,$b,$c,$day,$mon,$year,$d,$e,$f) = localtime(time-86400);printf "%02d-%02d-%4d",$day, $mon+1, $year+1900')
echo $dd
01-01-2014
NewDate=`date +"%A %d %B %Y" --date="-1 day"`
echo $NewDate
this will give your yesterday's date (-1)
This will give you tomorrow's date (+1)
even you can check for any values like (+/-) days
I am running a shell script which accepts date in "YYYY-MM-DD" format. from this date input, how can i get year, month and day separately?
Thanks for replies in advance.
except for processing the string as text(with grep/sed/awk/cut/...), you could do with with date command:
kent$ date -d '2013-09-06' +%Y
2013
kent$ date -d '2013-09-06' +%m
09
kent$ date -d '2013-09-06' +%d
06
You could do this to store them on variables with one command:
read YEAR MONTH DAY < <(date -d '2013-09-06' '+%Y %m %d')
printf "%s\n" "$YEAR" "$MONTH" "$DAY"
Try this :
dt="2011-2-3"
arr=( $( date --date=$dt "+%Y %m %d") )
echo "Year > $arr"
echo "Month > ${arr[1]}"
echo "Month > ${arr[2]}"
I didn't want to use date command, so I got the following code. It uses awk, which I feel is more appropriate for this task. I assumed you want to store results in variables.
inputDate=2017-07-06
year=`echo $inputDate | awk -F\- '{print $1}'`
month=`echo $inputDate | awk -F\- '{print $2}'`
day=`echo $inputDate | awk -F\- '{print $3}'`
How can I print the date which is a day before current time in Bash?
if you have GNU date and i understood you correctly
$ date +%Y:%m:%d -d "yesterday"
2009:11:09
or
$ date +%Y:%m:%d -d "1 day ago"
2009:11:09
If you have BSD (OSX) date you can do it like this:
date -j -v-1d
Wed Dec 14 15:34:14 CET 2011
Or if you want to do date calculations on an arbitrary date:
date -j -v-1d -f "%Y-%m-%d" "2011-09-01" "+%Y-%m-%d"
2011-08-31
date --date='-1 day'
MAC OSX
For yesterday's date:
date -v-1d +%F
where 1d defines current day minus 1 day. Similarly,
date -v-1w +%F - for previous week date
date -v-1m +%F - for previous month date
IF YOU HAVE GNU DATE,
date --date="1 day ago"
More info: https://www.cyberciti.biz/tips/linux-unix-get-yesterdays-tomorrows-date.html
Sorry not mentioning I on Solaris system.
As such, the -date switch is not available on Solaris bash.
I find out I can get the previous date with little trick on timezone.
DATE=`TZ=MYT+16 date +%Y-%m-%d_%r`
echo $DATE
Well this is a late answer,but this seems to work!!
YESTERDAY=`TZ=GMT+24 date +%d-%m-%Y`;
echo $YESTERDAY;
Advanced Bash-scripting Guide
date +%Y:%m:%d -d "yesterday"
For details about the date format see the man page for date
date --date='-1 day'
date -d "yesterday" '+%Y-%m-%d'
or
date=$(date -d "yesterday" '+%Y-%m-%d')
echo $date
Use Perl instead perhaps?
perl -e 'print scalar localtime( time - 86400 ) . "\n";'
Or, use nawk and (ab)use /usr/bin/adb:
nawk 'BEGIN{printf "0t%d=Y\n", srand()-86400}' | adb
Came across this too ... insane!
/usr/bin/truss /usr/bin/date 2>&1 | nawk -F= '/^time\(\)/ {gsub(/ /,"",$2);printf "0t%d=Y\n", $2-86400}' | adb
date --date='-1 day'
Not very sexy but might do the job:
perl -e 'my ($sec,$min,$hour,$mday,$mon,$year,$wday,$yday,$isdst) = localtime(time - 86400);$year += 1900; $mon+= 1; printf ("YESTERDAY: %04d%02d%02d \n", $year, $mon, $mday)'
Formated from "martin clayton" answer.
You could do a simple calculation, pimped with an regex, if the chosen date format is 'YYYYMM':
echo $(($(date +"%Y%m") - 1)) | sed -e 's/99$/12/'
In January of 2020 it will return 201912 ;-)
But, it's only a workaround, when date does not have calculation options and other dateinterpreter options (e.g. using perl) not available ;-)
short answer (GNU format):
date +%Y-%m-%d -d "-2 day"
if you are using OSX, but you need create for GNU compatible, install coreutils first
brew install coreutils
then edit your profile with:
#gnu coreutils first
export PATH="/usr/local/opt/coreutils/libexec/gnubin:$PATH"
re-start your terminal, and now you able to use GNU format!
yesterday=`date -d "-1 day" %F`
Puts yesterday's date in YYYY-MM-DD format into variable $yesterday.
#!/bin/bash
OFFSET=1;
eval `date "+day=%d; month=%m; year=%Y"`
# Subtract offset from day, if it goes below one use 'cal'
# to determine the number of days in the previous month.
day=`expr $day - $OFFSET`
if [ $day -le 0 ] ;then
month=`expr $month - 1`
if [ $month -eq 0 ] ;then
year=`expr $year - 1`
month=12
fi
set `cal $month $year`
xday=${$#}
day=`expr $xday + $day`
fi
echo $year-$month-$day
DST aware solution:
Manipulating the Timezone is possible for changing the clock some hours. Due to the daylight saving time, 24 hours ago can be today or the day before yesterday.
You are sure that yesterday is 20 or 30 hours ago. Which one? Well, the most recent one that is not today.
echo -e "$(TZ=GMT+30 date +%Y-%m-%d)\n$(TZ=GMT+20 date +%Y-%m-%d)" | grep -v $(date +%Y-%m-%d) | tail -1
The -e parameter used in the echo command is needed with bash, but will not work with ksh. In ksh you can use the same command without the -e flag.
When your script will be used in different environments, you can start the script with #!/bin/ksh or #!/bin/bash. You could also replace the \n by a newline:
echo "$(TZ=GMT+30 date +%Y-%m-%d)
$(TZ=GMT+20 date +%Y-%m-%d)" | grep -v $(date +%Y-%m-%d) | tail -1
Try the below code , which takes care of the DST part as well.
if [ $(date +%w) -eq $(date -u +%w) ]; then
tz=$(( 10#$gmthour - 10#$localhour ))
else
tz=$(( 24 - 10#$gmthour + 10#$localhour ))
fi
echo $tz
myTime=`TZ=GMT+$tz date +'%Y%m%d'`
Courtsey Ansgar Wiechers
date +%Y:%m:%d|awk -vFS=":" -vOFS=":" '{$3=$3-1;print}'
2009:11:9
I tried using $(date) in my bash shell script, however, I want the date in YYYY-MM-DD format.
How do I get this?
In bash (>=4.2) it is preferable to use printf's built-in date formatter (part of bash) rather than the external date (usually GNU date).
As such:
# put current date as yyyy-mm-dd in $date
# -1 -> explicit current date, bash >=4.3 defaults to current time if not provided
# -2 -> start time for shell
printf -v date '%(%Y-%m-%d)T\n' -1
# put current date as yyyy-mm-dd HH:MM:SS in $date
printf -v date '%(%Y-%m-%d %H:%M:%S)T\n' -1
# to print directly remove -v flag, as such:
printf '%(%Y-%m-%d)T\n' -1
# -> current date printed to terminal
In bash (<4.2):
# put current date as yyyy-mm-dd in $date
date=$(date '+%Y-%m-%d')
# put current date as yyyy-mm-dd HH:MM:SS in $date
date=$(date '+%Y-%m-%d %H:%M:%S')
# print current date directly
echo $(date '+%Y-%m-%d')
Other available date formats can be viewed from the date man pages (for external non-bash specific command):
man date
Try: $(date +%F)
The %F option is an alias for %Y-%m-%d
You can do something like this:
$ date +'%Y-%m-%d'
$(date +%F)
output
2018-06-20
Or if you also want time:
$(date +%F_%H-%M-%S)
can be used to remove colons (:) in between
output
2018-06-20_09-55-58
You're looking for ISO 8601 standard date format, so if you have GNU date (or any date command more modern than 1988) just do: $(date -I)
date -d '1 hour ago' '+%Y-%m-%d'
The output would be 2015-06-14.
With recent Bash (version ≥ 4.2), you can use the builtin printf with the format modifier %(strftime_format)T:
$ printf '%(%Y-%m-%d)T\n' -1 # Get YYYY-MM-DD (-1 stands for "current time")
2017-11-10
$ printf '%(%F)T\n' -1 # Synonym of the above
2017-11-10
$ printf -v date '%(%F)T' -1 # Capture as var $date
printf is much faster than date since it's a Bash builtin while date is an external command.
As well, printf -v date ... is faster than date=$(printf ...) since it doesn't require forking a subshell.
I use the following formulation:
TODAY=`date -I`
echo $TODAY
Checkout the man page for date, there is a number of other useful options:
man date
if you want the year in a two number format such as 17 rather than 2017, do the following:
DATE=`date +%d-%m-%y`
I use $(date +"%Y-%m-%d") or $(date +"%Y-%m-%d %T") with time and hours.
I used below method. Thanks for all methods/answers
ubuntu#apj:/tmp$ datevar=$(date +'%Y-%m-%d : %H-%M')
ubuntu#apj:/tmp$ echo $datevar
2022-03-31 : 10-48
Whenever I have a task like this I end up falling back to
$ man strftime
to remind myself of all the possibilities for time formatting options.
Try to use this command :
date | cut -d " " -f2-4 | tr " " "-"
The output would be like: 21-Feb-2021
#!/bin/bash -e
x='2018-01-18 10:00:00'
a=$(date -d "$x")
b=$(date -d "$a 10 min" "+%Y-%m-%d %H:%M:%S")
c=$(date -d "$b 10 min" "+%Y-%m-%d %H:%M:%S")
#date -d "$a 30 min" "+%Y-%m-%d %H:%M:%S"
echo Entered Date is $x
echo Second Date is $b
echo Third Date is $c
Here x is sample date used & then example displays both formatting of data as well as getting dates 10 mins more then current date.
You can set date as environment variable and later u can use it
setenv DATE `date "+%Y-%m-%d"`
echo "----------- ${DATE} -------------"
or
DATE =`date "+%Y-%m-%d"`
echo "----------- ${DATE} -------------"
echo "`date "+%F"`"
Will print YYYY-MM-DD
Try this code for a simple human readable timestamp:
dt=$(date)
echo $dt
Output:
Tue May 3 08:48:47 IST 2022