I'm running AIX with coreutils 5.0. I need to advance an arbitrary date (or time) as given conformative to ISO-8601 format YYYY-MM-DD hh:mm:ss.
For example:
Value of D1 is: 2017-07-08 19:20:01, and I need to add 30 minutes.
In a modern UNIX-system I could probably write something like
date -d "$D1 + 30 minutes" +'%H:%M'
but, alas, I need it to work on an old AIX.
Try
$ date -d "$(date -d "$D1") + 30 minutes" +'%H:%M'
This works in bash, but not in ksh.
The inner call to date will parse D1 to a date, and present it in date's "native" format.
$ date -d "$D1"
Sat Jul 8 19:20:01 CEST 2017
This output will be used with + 30 minutes to create the date that you want, with the outer call to date.
The inner call to date will be expanded so that
$ date -d "$(date -d "$D1") + 30 minutes" +'%H:%M'
will be equivalent to
$ date -d "Sat Jul 8 19:20:01 CEST 2017 + 30 minutes" +'%H:%M'
which will be
19:50
date -d #$(($(date -d "$D1" +%s) + 30 * 60)) +%H:%M
$(date -d "$D1" +%s) echoes the epoch
$((epoch + value)) calculates the wanted time
date -d#epoch +fmt formats it
If you are running AIX from 2003 you are in dire straits, my friend, but if you only need the time, not the full date, as your question implies, I think #RamanSailopal got us half way there.
echo $D1 | awk -F "[: ]" '{
m = $3+30;
h = ($2+int(m/60)) % 24;
printf("%02i:%02i\n", h, m%60)
}'
awk splits the input in different fields, with the splitter pattern given in the -F argument. The pattern denotes : or space .
The input will be split in
$1 = 2017-07-08
$2 = 19
$3 = 20
$4 = 01
Then the script calculates a fake minute value (that can be more than or equal to 60) and stores it in m. From that value it calculates the hour, modulo 24, and the actual minutes, m modulo 60.
This could fail if you hit a leap second, so if you need second precision at all times, you should use some other method.
Awk solution:
awk -F '[-: ]' '{
ram=(mktime($1" "$2" "$3" "$4" "$5" "$6)+(30*60));
print strftime("%Y-%m-%d %T",ram)
}' <<< "$D1"
Convert the date to a date string using awk's mktime function. Add 30 minutes (30*60) and then convert back to a date string with the required format using strftime.
Related
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'm trying to convert strings, describing a time interval, to the corresponding number of seconds.
After some experimenting I figured out that I can use date like this:
soon=$(date -d '5 minutes 10 seconds' +%s); now=$(date +%s)
echo $(( $soon-$now ))
but I think there should be an easier way to convert strings like "5 minutes 10 seconds" to the corresponding number of seconds, in this example 310. Is there a way to do this in one command?
Note: although portability would be useful, it isn't my top priority.
You could start at epoch
date -d"1970-01-01 00:00:00 UTC 5 minutes 10 seconds" "+%s"
310
You could also easily sub in times
Time="1 day"
date -d"1970-01-01 00:00:00 UTC $Time" "+%s"
86400
There is one way to do it, without using date command in pure bash (for portability)
Assuming you just have an input string to convert "5 minutes 10 seconds" in a bash variable with a : de-limiter as below.
$ convertString="00:05:10"
$ IFS=: read -r hour minute second <<< "$convertString"
$ secondsValue=$(((hour * 60 + minute) * 60 + second))
$ printf "%s\n" "$secondsValue"
310
You can run the above commands directly on the command-line without the $ mark.
This will do (add the epoch 19700101):
$ date -ud '19700101 5 minutes 10 seconds' +%s
310
It is important to add a -u to avoid local time (and DST) effects.
$ TZ=America/Los_Angeles date -d '19700101 5 minutes 10 seconds' +%s
29110
Note that date could do some math:
$ date -ud '19700101 +5 minutes 10 seconds -47 seconds -1 min' +%s
203
The previous suggestions didn't work properly on alpine linux, so here's a small helper function that is POSIX compliant, is easy to use and also supports calculations (just as a side effect of the implementation).
The function always returns an integer based on the provided parameters.
$ durationToSeconds '<value>' '<fallback>'
$ durationToSeconds "1h 30m"
5400
$ durationToSeconds "$someemptyvar" 1h
3600
$ durationToSeconds "$someemptyvar" "1h 30m"
5400
# Calculations also work
$ durationToSeconds "1h * 3"
10800
$ durationToSeconds "1h - 1h"
0
# And also supports long forms for year, day, hour, minute, second
$ durationToSeconds "3 days 1 hour"
262800
# It's also case insensitive
$ durationToSeconds "3 Days"
259200
function durationToSeconds () {
set -f
normalize () { echo $1 | tr '[:upper:]' '[:lower:]' | tr -d "\"\\\'" | sed 's/years\{0,1\}/y/g; s/months\{0,1\}/m/g; s/days\{0,1\}/d/g; s/hours\{0,1\}/h/g; s/minutes\{0,1\}/m/g; s/min/m/g; s/seconds\{0,1\}/s/g; s/sec/s/g; s/ //g;'; }
local value=$(normalize "$1")
local fallback=$(normalize "$2")
echo $value | grep -v '^[-+*/0-9ydhms]\{0,30\}$' > /dev/null 2>&1
if [ $? -eq 0 ]
then
>&2 echo Invalid duration pattern \"$value\"
else
if [ "$value" = "" ]; then
[ "$fallback" != "" ] && durationToSeconds "$fallback"
else
sedtmpl () { echo "s/\([0-9]\+\)$1/(0\1 * $2)/g;"; }
local template="$(sedtmpl '\( \|$\)' 1) $(sedtmpl y '365 * 86400') $(sedtmpl d 86400) $(sedtmpl h 3600) $(sedtmpl m 60) $(sedtmpl s 1) s/) *(/) + (/g;"
echo $value | sed "$template" | bc
fi
fi
set +f
}
Edit : Yes. I developed for OP after comment and checked on Mac OS X, CentOS and Ubuntu. One liner, POSIX compliant command for converting "X minutes Y seconds" format to seconds. That was the question.
echo $(($(echo "5 minutes 10 seconds" | cut -c1-2)*60 + $(echo "5 minutes 10 seconds" | cut -c1-12 | awk '{print substr($0,11)}')))
OP told me via comment that he wants for "X minutes Y seconds" format not for HH:MM:SS format. The command with date and "+%s" is throwing error on (my) Mac. OP wanted to grab the numerical values from "X minutes Y seconds" format and convert it to seconds. First I extracted the minute in digit (take it as equation A) :
echo "5 minutes 10 seconds" | cut -c1-2)
then I extracted the seconds part (take it as equation B) :
echo "5 minutes 10 seconds" | cut -c1-12 | awk '{print substr($0,11)}'
Now multiply minute by 60 then add with the other :
echo $((equation A)*60) + (equation B))
OP should ask the others to check my developmental version (but working) of command before using it for automatic repeated usage like we do with cron on a production server.
If we want to run this on a log file with values in "X minutes Y seconds" format, we have to change echo "5 minutes 10 seconds" to cat file | ... like command. I kept a gist of it too if I or others ever need we can use it with cat to run on server log files with x minutes y seconds like log format.
Although off-topic (what I understood, question has not much to do with current time), this is not working for POSIX-compliant OS to get current time in seconds :
date -d "1970-01-01 00:00:00 UTC 5 minutes 10 seconds" "+%s"
It will throw error on MacOS X but work on most GNU/Linux distro. That +%s part will throw error on POSIX-compliant OS upon complicated usage. These commands are mostly suitable to get current time in seconds on POSIX compliant to any kind of unix like OS :
awk 'BEGIN{srand(); print srand()}'
perl -le 'print time'
If OP needs can extend it by generating current time in seconds and subtract. I hope it will help.
---- OLD Answer before EDIT ----
You can get the current time without that date -- echo | awk '{print systime();}' or wget -qO- http://www.timeapi.org/utc/now?\\s. Other way to convert time to second is echo "00:20:40.25" | awk -F: '{ print ($1 * 3600) + ($2 * 60) + $3 }'.
The example with printf shown in another answer is near perfect.
That thing you want is always needed by the basic utilities of GNU/Linux - gnu.org/../../../../../Setting-an-Alarm.html
Way to approach really depends how much foolproof way you need.
I have a file with date '2015-06-01-12', how can I get it to increment the hour in shell script? The result I want is '2015-06-01-13'. If its the 23rd hour it has to move forward a date and get 00 as hour.
I was able to do it to date but have so far had not any luck with incrementing hours.
currDate=2015-06-02
nextDate=`date '+%Y-%m-%d' -d "$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'