How can I convert a date variable to seconds?
I have a variable called CDATE defined like this:
CDATE=$(/bin/date +"%Y-%m-%d %H:%M:%S")
CDATE example output: "2012-12-12 12:12:12"
I would like to convert this to time in seconds and save it in another variable.
x=($(date +"%Y-%m-%d %H:%M:%S %s"))
CDATE="${x[0]} ${x[1]}"
secs="${x[2]}" # seconds since 1970-01-01 00:00:00 UTC
echo $CDATE
echo $secs
Output:
2015-04-04 14:13:08
1428149588
On Linux:
CDATE=$(date "+%Y-%m-%d %H:%M:%S")
SECONDS=$(date -d "$CDATE" +%s)
On Mac:
CDATE=$(date "+%Y-%m-%d %H:%M:%S")
SECONDS=$(date -j -f "%Y-%m-%d %H:%M:%S" "$CDATE" +%s)
And the content of the two variables is (in both cases):
$ echo "$CDATE"
2015-04-04 16:24:41
$ echo "$SECONDS"
1428157497
If you are not charged for CPU time, then I would call date twice to make it more readable:
# get the timestamp:
timestamp=$( date +"%s" )
# format it:
CDATE=$( date -d #"$timestamp" +"%Y-%m-%d %H:%M:%S" )
echo "$CDATE"
echo "$timestamp"
The simplest way:
source <(date +"CDATE='%Y-%m-%d %H:%M:%S' EPOCH='%s'")
Example:
tiago#dell:/tmp$ source <(date +"CDATE='%Y-%m-%d %H:%M:%S' EPOCH='%s'")
tiago#dell:/tmp$ echo "cdate:$CDATE epoch:$EPOCH"
cdate:2015-04-05 12:39:58 epoch:1428233998
Related
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="$d" +%Y%m%d%H%M --utc`
Second one is in UTC.
I have tried adding -d "yesterday" but I haven't had any luck getting it to work. Here is what I have for the whole script:
#! /bin/bash
saveDir="TJ"
dd=$(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 $NewDate
this will give your yesterday's date (-1)
This will give you tomorrow's date (+1)
even you can check for any values like (+/-) days
In short, I want something works like this:
When I input a date like 2012-12-27 and want to expand the date by
week(start with Monday), it
outputs:2012-12-24,2012-12-25,2012-12-26,2012-12-27,2012-12-28,2012-12-29,2012-12-30
When I input a date like 2012-12-27 and want to expand the date by month, it outputs:2012-12-01,2012-12-02 ... 2012-12-31
or, how can I group a bunch of dates by week? e.g. when I input2012-12-01,2012-12-02 ... 2012-12-31. It outputs:2012-12-01,2012-12-02|2012-12-03 ... 2012-12-09|2012-12-10 ... 2012-12-16|...|2012-12-31
I have no idea how to complete this, any clue may be helpful!
DAYSECS=86400 # seconds in a day
WEEKSECS=604800
echo "Expand on week:"
g_epoch=$(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
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