I have the following in a shell script. How can I subtract one hour while retaining the formatting?
DATE=`date "+%m/%d/%Y -%H:%M:%S"`
The following command works on recent versions of GNU date:
date -d '1 hour ago' "+%m/%d/%Y -%H:%M:%S"
date -v-60M "+%m/%d/%Y -%H:%M:%S"
DATE=`date -v-60M "+%m/%d/%Y -%H:%M:%S"`
If you have bash version 4.4+ you can use bash's internal date printing and arithmetics:
printf "current date: %(%m/%d/%Y -%H:%M:%S)T\n"
printf "date - 60min: %(%m/%d/%Y -%H:%M:%S)T\n" $(( $(printf "%(%s)T") - 60 * 60 ))
The $(printf "%(%s)T") prints the epoch seconds, the $(( epoch - 60*60 )) is bash-aritmetics - subtracting 1hour in seconds. Prints:
current date: 04/20/2017 -18:14:31
date - 60min: 04/20/2017 -17:14:31
if you need substract with timestamp :
timestamp=$(date +%s -d '1 hour ago');
This work on my Ubuntu 16.04 date:
date --date="#$(($(date +%s) - 3600))" "+%m/%d/%Y -%H:%M:%S"
And the date version is date (GNU coreutils) 8.25
$ date +%Y-%m-%d-%H
2019-04-09-20
$ date -v-1H +%Y-%m-%d-%H
2019-04-09-19
But in shell use as like date +%Y-%m-%d-%H, date -v-1H +%Y-%m-%d-%H
Convert to timestamp (a long integer), subtract the right number of milliseconds, reformat to the format you need.
Hard to give more details since you don't specify a programming language...
If you need change timezone before subtraction with new format too:
$(TZ=US/Eastern date -d '1 hour ago' '+%Y-%m-%d %H:%M')
Here another way to subtract 1 hour.
yesterdayDate=`date -d '2018-11-24 00:09 -1 hour' +'%Y-%m-%d %H:%M'`
echo $yesterdayDate
Output:
2018-11-23 23:09
I hope that it can help someone.
DATE=date -1H "+%m/%d/%Y -%H:%M:%S"
Related
I have the following in a shell script. How can I subtract one hour while retaining the formatting?
DATE=`date "+%m/%d/%Y -%H:%M:%S"`
The following command works on recent versions of GNU date:
date -d '1 hour ago' "+%m/%d/%Y -%H:%M:%S"
date -v-60M "+%m/%d/%Y -%H:%M:%S"
DATE=`date -v-60M "+%m/%d/%Y -%H:%M:%S"`
If you have bash version 4.4+ you can use bash's internal date printing and arithmetics:
printf "current date: %(%m/%d/%Y -%H:%M:%S)T\n"
printf "date - 60min: %(%m/%d/%Y -%H:%M:%S)T\n" $(( $(printf "%(%s)T") - 60 * 60 ))
The $(printf "%(%s)T") prints the epoch seconds, the $(( epoch - 60*60 )) is bash-aritmetics - subtracting 1hour in seconds. Prints:
current date: 04/20/2017 -18:14:31
date - 60min: 04/20/2017 -17:14:31
if you need substract with timestamp :
timestamp=$(date +%s -d '1 hour ago');
This work on my Ubuntu 16.04 date:
date --date="#$(($(date +%s) - 3600))" "+%m/%d/%Y -%H:%M:%S"
And the date version is date (GNU coreutils) 8.25
$ date +%Y-%m-%d-%H
2019-04-09-20
$ date -v-1H +%Y-%m-%d-%H
2019-04-09-19
But in shell use as like date +%Y-%m-%d-%H, date -v-1H +%Y-%m-%d-%H
Convert to timestamp (a long integer), subtract the right number of milliseconds, reformat to the format you need.
Hard to give more details since you don't specify a programming language...
If you need change timezone before subtraction with new format too:
$(TZ=US/Eastern date -d '1 hour ago' '+%Y-%m-%d %H:%M')
Here another way to subtract 1 hour.
yesterdayDate=`date -d '2018-11-24 00:09 -1 hour' +'%Y-%m-%d %H:%M'`
echo $yesterdayDate
Output:
2018-11-23 23:09
I hope that it can help someone.
DATE=date -1H "+%m/%d/%Y -%H:%M:%S"
I want to modify this command to subtract 30 days from current date automatically
$ awk -v t=$(date +%Y-%m-%d) -F "'" '$1 < t' myname.dat
When I try
$ awk -v t=$(date "--date=$(date) -30days" +%Y-%m-%d) -F "'" '$1 < t' myname.dat
I get the following error; date: illegal option
I want to do this without having to convert the dates to epoch time in the file.
#edit: The following will work with GNU date only:
You can always subtract seconds.
date --date=#$(($(date +%s) - 30 * 24 * 3600)) +%Y-%m-%d
If you are interested in subtracting 30 days form "now", just:
date --date="-30days" +%Y-%m-%d
date date formatting is so broad, it's good to specify the exact date with for example -I option, from man date:
-I[FMT], --iso-8601[=FMT]
output date/time in ISO 8601 format. FMT='date' for date only
(the default), 'hours', 'minutes', 'seconds', or 'ns' for date
and time to the indicated precision. Example:
2006-08-14T02:34:56-06:00
The following:
date --date="$(date -I) -30days" +%Y-%m-%d
works on my system as expected.
I have a file with date '2015-06-01-12', how can I get it to increment the hour in shell script? The result I want is '2015-06-01-13'. If its the 23rd hour it has to move forward a date and get 00 as hour.
I was able to do it to date but have so far had not any luck with incrementing hours.
currDate=2015-06-02
nextDate=`date '+%Y-%m-%d' -d "$currDate+1 days"`
echo $nextDate
It is reasonably easy if you keep your date as an epoch (number of seconds since January 1, 1970):
$ currDate=$( date +%s -d "2015-06-02 23:00:00" )
$ echo $currDate
1433300400
$ date +%Y-%m-%d-%H -d #$currDate
2015-06-02-23
$ nextDate=$(( $currDate + 3600 )) #adding an hour's worth of seconds
$ date +%Y-%m-%d-%H -d #$nextDate
2015-06-03-00
I would like to convert the current date and time into a hex time stamp, something like:
Tue Feb 2 10:27:46 GMT 2010 converted into 0x6d054a874449e
I would like to do this from a bash script, any idea how I might do that?
Thanks
J
printf '0x%x' $(date +%s)
Without knowing the unit or epoch for your hex timestamp, it's hard to say for sure (and I was slightly confused by your example of "Feb 2" which is not even close to the current date!).
date +%s will convert the current date into a time_t, the number of seconds since the usual Unix epoch (which is midnight on 1st Jan 1970).
printf "0x%x" some_number will convert a value from decimal to hex.
If you need to convert to a different epoch / unit, you will need to do some calculation. You can do arithmetic in bash using $(( expression )):
$ time_t=$(date +%s)
$ echo $(($time_t * 1000))
1284505668000
If you want to convert an arbitrary date (like your "Feb 2 ..." example), rather than the current one, and are happy to assume that you have the GNU version of date, then you can use the -d option along with the +%s output format to do the conversion:
$ date -d 'Tue Feb 2 10:27:46 GMT 2010' +%s
1265106466
An example of putting this all together:
$ time_t=$(date -d 'Tue Feb 2 10:27:46 GMT 2010' +%s)
$ time_t_ms=$(($time_t * 1000))
$ hexstamp=$(printf "0x%x" $time_t_ms)
$ echo $hexstamp
0x1268e38b4d0
Seconds since unix epoch, in hex:
echo "$(date +%s)"|xargs printf "0x%x"
0x59a8de5b
Milliseconds since the epoch:
echo "$(date +%s%N)/1000000"|bc|xargs printf "0x%x"
0x15e3ba702bb
Microseconds:
echo "$(date +%s%N)/1000"|bc|xargs printf "0x%x"
0x55818f6eea775
Nanoseconds:
echo "$(date +%s%N)"|xargs printf "0x%x"
0x14e0219022e3745c
In a bash script, if I have a number that represents a time, in the form hhmmss (or hmmss), what is the best way of subtracting 10 minutes?
ie, 90000 -> 85000
This is a bit tricky. Date can do general manipulations, i.e. you can do:
date --date '-10 min'
Specifying hour-min-seconds (using UTC because otherwise it seems to assume PM):
date --date '11:45:30 UTC -10 min'
To split your date string, the only way I can think of is substring expansion:
a=114530
date --date "${a:0:2}:${a:2:2}:${a:4:2} UTC -10 min"
And if you want to just get back hhmmss:
date +%H%M%S --date "${a:0:2}:${a:2:2}:${a:4:2} UTC -10 min"
why not just use epoch time and then take 600 off of it?
$ echo "`date +%s` - 600"| bc; date
1284050588
Thu Sep 9 11:53:08 CDT 2010
$ date -d '1970-01-01 UTC 1284050588 seconds' +"%Y-%m-%d %T %z"
2010-09-09 11:43:08 -0500
Since you have a 5 or 6 digit number, you have to pad it before doing string manipulation:
$ t=90100
$ while [ ${#t} -lt 6 ]; do t=0$t; done
$ echo $t
090100
$ date +%H%M%S --utc -d"today ${t:0:2}:${t:2:2}:${t:4:2} UTC - 10 minutes"
085100
Note both --utc and UTC are required to make sure the system's timezone doesn't affect the results.
For math within bash (i.e. $(( and ((), leading zeros will cause the number to be interpreted as octal. However, your data is more string-like (with a special format) than number-like, anyway. I've used a while loop above because it sounds like you're treating it as a number and thus might get 100 for 12:01 am.
My version of bash doesn't support -d or --date as used above. However, assuming a correctly 0-padded input, this does work
$ input_time=130503 # meaning "1:05:03 PM"
# next line calculates epoch seconds for today's date at stated time
$ epoch_seconds=$(date -jf '%H%M%S' $input_time '+%s')
# the 600 matches the OP's "subtract 10 minutes" spec. Note: Still relative to "today"
$ calculated_seconds=$(( epoch_seconds - 600 )) # bc would work here but $((...)) is builtin
# +%H%M%S formats the result same as input, but you can do what you like here
$ echo $(date -r $calculated_seconds '+%H%M%S')
# output is 125503: Note that the hour rolled back as expected.
For MacOS users you can do the following:
$(date -v -10M +"%H:%M:%S")
Date time without a specific format:
$(date -v -10M)
For non-macOS users:
Date time without a specific format:
date --date '-10 min'