I was hoping to get some help with finding the difference in days between two dates.
Date 1: Sysdate
Date2: Mar 20 2022 (future)
What I was trying to do is convert it to EPOCH time, subtract and then divide by 86400 to get the number of days between the two dates. However, I've having issues with the systax. Here's what I've tried:
days_remaining=('date "+%s" -d "Mar 20 2022"'-'date "+%s" -d "$(date '+%b %d %Y')"')/86400
But the above isn't working. Any assistance would be appreciated.
To do calculations in bash you need an arithmetic context (( )). Also, to execute your date commands you have to put them inside $() instead of string quotes ''.
By the way: The last date command in date "+%s" -d "$(date '+%b %d %Y') isn't necessary. date -d 0:00 +%s will print the same unix time.
(( days_remaining = ($(date -d "Mar 20 2022" +%s) - $(date -d 0:00 +%s)) / 86400 ))
you can use this
days_remaining=$(($(($(date "+%s" -d "Mar 20 2022") - $(date "+%s" -d "$(date '+%b %d %Y')"))) / 86400))
Related
I am trying to write a bash script, to merge 24 files in a given day. The requirement changes during Day light saving time changes, where I get 23 or 25 files.
So, with further research I realized that day-light savings begins on the second Sunday of March(23) of every year and ends on first sunday of Novemeber(25).
I need more inputs to get second sunday in a given month to do the check of finding 23 or 25 files for March and November respectively.
Any inputs to help me with this will be really appreciated.
Thank you
Here is the sample code to find 24 files in a day-
if [ -z "$1" ];then
now=$(date -d "-1 days" +%Y-%m-%d);
else now=$1;
fi
load_date='load_date='$now
singlePath="$newPath/$load_date"
fileCount=$(hdfs dfs -ls -R $hdfsPath/$load_date/ | grep -E '^-' | wc -l)
path=$hdfsPath/$load_date
if [ $fileCount -eq 24 ]; then
echo "All files are available for "$load_date;
hadoop fs -cat $path/* | hadoop fs -put - $singlePath/messages.txt
else echo $fileCount" files are available for "$load_date"! Please note, few files are being missed";
fi
I wouldn't hardcode the dates of DST transistions. I would just count "how many hours did today have":
a "normal" day:
$ diff=$(( $(date -d now +%s) - $(date -d yesterday +%s) ))
$ echo $(( diff / 3600 ))
24
"spring forward"
$ diff=$(( $(date -d "2019-03-10 23:59:59" +%s) - $(date -d "2019-03-09 23:59:59" +%s) ))
$ echo $(( diff / 3600 ))
23
"fall back"
$ diff=$(( $(date -d "2019-11-03 23:59:59" +%s) - $(date -d "2019-11-02 23:59:59" +%s) ))
$ echo $(( diff / 3600 ))
25
One thing to note: since bash only does integer arithmetic, if the difference is not 86400 but 86399, you get:
$ echo $((86399 / 3600))
23
So, better to query yesterday's time first in the tiny-but-non-zero chance that the seconds tick over between the 2 date calls:
diff=$(( -$(date -d yesterday +%s) + $(date -d now +%s) ))
Here, $diff will be 86400 or 86401 (for non DST transition days), and dividing by 3600 will give 24 not 23.
To get the current command we simple shell script command
date
if you want to get a custom date format
NOW=$(date +"%Y-%m-%d %T")
Am not getting simple command to get the 3hrs back date with above format , any one can help me please .
i used below shell command with perl like below , but i do not want another additional dependency perl , i just need to shell command
CURRENT_DATE=`perl -e 'use POSIX qw(strftime);$d = strftime "%Y-%m-%d %H:%M:%S", localtime(time());print $d'`
CURRENT_DATE_MINUS_3_HRS_DATE=`perl -e 'use POSIX qw(strftime);$d = strftime "%Y-%m-%d %H:%M:%S", localtime(time() - 10800);print $d'`
The date command has the nice keyword ago for the date you can give with -d option:
date -d '3 hours ago' +"%Y-%m-%d %T"
Shows the current date:
date "+%Y-%m-%d %T"
Show date back to 3 hours:
date -d "-3 Hours" "+%Y-%m-%d %T"
Some of the other things you can try:
# Back to 3 days
date -d "-3 Days" "+%Y-%m-%d %T"
# Next to 3 days
date -d "3 Days" "+%Y-%m-%d %T"
# Back to 2 Years and 3 Hours
date -d "-2 Years -3 Hours" "+%Y-%m-%d %T"
Using the BSD date that ships with macOS, use the -v option.
$ date +"%Y-%m-%d %T"
2018-11-08 16:25:21
$ date -v -3H +"%Y-%m-%d %T"
2018-11-08 13:25:26
How to change this time output ?
date --date="#$(echo $(TZ=UTC date +%s) - $(date +%s)'%(5*60)-(5*60)' | bc)"
Output: za apr 12 00:25:00 CEST 2014
Should output in this layout: %Y%m%d%H%M
How to implement this in the string ?
Thanks!!
i think this should do what you want:
d="#$(echo $(TZ=UTC date +%s) - $(date +%s)'%(5*60)-(5*60)' | bc)"&&echo `date --date="$d"` `date --date="$d" +%Y%m%d%H%M`
d="#$(echo $(TZ=UTC date +%s) - $(date +%s)'%(5*60)-(5*60)' | bc)"&&echo `date --date="$d" --utc` `date --date="$d" +%Y%m%d%H%M --utc`
Second one is in UTC.
In short, I want something works like this:
When I input a date like 2012-12-27 and want to expand the date by
week(start with Monday), it
outputs:2012-12-24,2012-12-25,2012-12-26,2012-12-27,2012-12-28,2012-12-29,2012-12-30
When I input a date like 2012-12-27 and want to expand the date by month, it outputs:2012-12-01,2012-12-02 ... 2012-12-31
or, how can I group a bunch of dates by week? e.g. when I input2012-12-01,2012-12-02 ... 2012-12-31. It outputs:2012-12-01,2012-12-02|2012-12-03 ... 2012-12-09|2012-12-10 ... 2012-12-16|...|2012-12-31
I have no idea how to complete this, any clue may be helpful!
DAYSECS=86400 # seconds in a day
WEEKSECS=604800
echo "Expand on week:"
g_epoch=$(date +"%s" -d $1) # given date as seconds from epoch
g_dayno=$(date +"%u" -d $1) # given date as day of week
g_month=$(date +"%m" -d $1) # given month
g_year=$(date +"%Y" -d $1) # given year
s_epoch=$(($g_epoch - $DAYSECS * ($g_dayno - 1)))
e_epoch=$(($s_epoch + $WEEKSECS))
for etime in $(seq $s_epoch $DAYSECS $e_epoch); do
date +"%Y-%m-%d" -d "#$etime"
done
echo "Expand on month:"
s_epoch=$(date +"%s" -d "$g_year-$g_month-01")
e_epoch=$(($s_epoch + 4 * $WEEKSECS))
for etime in $(seq $s_epoch $DAYSECS $e_epoch); do
if [ $(date +"%m" -d "#$etime") -ne "$g_month" ]; then
break;
fi
date +"%Y-%m-%d" -d "#$etime"
done
The script below work out from #perreal, I leave it here because:
It shows the power of GNU date.
It makes #perreal 's idea more clear and more universal.
Thank you, perreal!
Here it is
dd="2012-12-27" # test date
DAYSECS=86400 # seconds in a day
echo "expand by week:"
s_epoch=$(date +%s -d "$dd -$(($(date +%u -d $dd) - 1)) day") # start date of the week
e_epoch=$(date +%s -d "1970-01-01 00:00:00 +0000 +${s_epoch} seconds +6 days") # end date of the week
for etime in $(seq $s_epoch $DAYSECS $e_epoch); do
date +"%Y-%m-%d" -d "#$etime";
done
echo "expand by month:"
s_epoch=$(date +%s -d "$dd -$(($(date +%d -d $dd) - 1)) day") # start date of the month
e_epoch=$(date +%s -d "1970-01-01 00:00:00 +0000 +${s_epoch} seconds +1 month -1 day") # end date of the month
for etime in $(seq $s_epoch $DAYSECS $e_epoch); do
date +"%Y-%m-%d" -d "#$etime";
done
If there are dates as 2010-06-01 and another as 2010-05-15
Using shell script or date command how to get the number of days between the two dates
Thanks..
Using only date and shell arithmetics:
echo $((($(date -d "2010-06-01" "+%s") - $(date -d "2010-05-15" "+%s")) / 86400))
There's a solution that almost works: use the %s date format of GNU date, which prints the number of seconds since 1970-01-01 00:00. These can be subtracted to find the time difference between two dates.
echo $(( ($(date -d 2010-06-01 +%s) - $(date -d 2010-05-15 +%s)) / 86400))
But the following displays 0 in some locations:
echo $((($(date -d 2010-03-29 +%s) - $(date -d 2010-03-28 +%s)) / 86400))
Because of daylight savings time, there are only 23 hours between those times. You need to add at least one hour (and at most 23) to be safe.
echo $((($(date -d 2010-03-29 +%s) - $(date -d 2010-03-28 +%s) + 43200) / 86400))
Or you can tell date to work in a timezone without DST.
echo $((($(date -u -d 2010-03-29 +%s) - $(date -u -d 2010-03-28 +%s)) / 86400))
(POSIX says to call the reference timezone is UTC, but it also says not to count leap seconds, so the number of seconds in a day is always exactly 86400 in a GMT+xx timezone.)
OSX date is different than GNU date. Got it working like this in OSX. This is not portable solution.
start_date=$(date -j -f "%Y-%m-%d" "2010-05-15" "+%s")
end_date=$(date -j -f "%Y-%m-%d" "2010-06-01" "+%s")
echo $(( ($end_date - $start_date) / (60 * 60 * 24) ))
Idea is still same as in the other answers. Convert dates to epoch time, subtract and convert result to days.
Got it
d1=`date +%s -d $1`
d2=`date +%s -d $2`
((diff_sec=d2-d1))
echo - | awk -v SECS=$diff_sec '{printf "Number of days : %d",SECS/(60*60*24)}'
thanks..
Gnu date knows %j to display the day in year:
echo $(($(date -d 2010-06-01 +%j) - $(date -d 2010-05-15 +%j)))
crossing year-boundaries will give wrong results, but since you gave fixed dates ...