I want to iterate all the minutes in a month (the purpose will be to generate a CSV file).
But when I try this:
d="2016-09-01 00:00:00"
while [ "$d" != "2016-09-30 00:00:00" ]; do
echo $d
d=$(date --utc "+%Y-%m-%d %H:%M:00" -d "$d + 1 minute" )
done
Both the hour and minute are being incremented:
2016-09-01 00:00:00
2016-09-01 01:01:00
2016-09-01 02:02:00
2016-09-01 03:03:00
2016-09-01 04:04:00
2016-09-01 05:05:00
2016-09-01 06:06:00
2016-09-01 07:07:00
2016-09-01 08:08:00
What am I doing wrong and how to correctly loop minutes?
I would work with Unix timestamps instead.
d=$(date --utc +%s -d "2016-09-01 00:00:00")
end=$(date --utc +%s -d "2016-09-30 00:00:00")
while [ "$d" != "$end" ]; do
date --utc "+%Y-%m-%d %H:%M:00" -d "#$d"
d=$(( d + 60 ))
done
You can add timezone UTC after your date string variable $d to get the right output:
d="2016-09-01 00:00:00"
for ((i=0; i<=60; i++)); do
date --utc -d "$d UTC + $i minute"
done
Thu Sep 1 00:00:00 UTC 2016
Thu Sep 1 00:01:00 UTC 2016
Thu Sep 1 00:02:00 UTC 2016
Thu Sep 1 00:03:00 UTC 2016
Thu Sep 1 00:04:00 UTC 2016
Thu Sep 1 00:05:00 UTC 2016
...
...
Thu Sep 1 00:55:00 UTC 2016
Thu Sep 1 00:56:00 UTC 2016
Thu Sep 1 00:57:00 UTC 2016
Thu Sep 1 00:58:00 UTC 2016
Thu Sep 1 00:59:00 UTC 2016
Thu Sep 1 01:00:00 UTC 2016
Note use of UTC after $d.
Using + after a time component in the date string is used for ' time zone correction' not for doing what you want to do. Interestingly, inverting date and time works:
$ date "+%Y-%m-%d %H:%M:00" -d "21:31:00 2016-09-03 + 1 minute "
2016-09-03 21:32:00
while the other way around messes with timezones and offsets so the result might depend on your local configuration:
$ TZ=Europe/London date "+%Y-%m-%d %H:%M:00" -d "2016-09-03 21:32:00 + 1 minute "
2016-09-03 21:33:00
$ TZ=Europe/Brussels date "+%Y-%m-%d %H:%M:00" -d "2016-09-03 21:32:00 + 1 minute "
2016-09-03 22:33:00
$ TZ=Asia/Singapore date "+%Y-%m-%d %H:%M:00" -d "2016-09-03 21:32:00 + 1 minute "
2016-09-04 04:33:00
Related
This is an assignment. I have the following code:
#! /bin/bash
y=$1
if [ -z $1 ] # if year is not specified use the current year
then y=(`date +%Y`)
fi
for m in {1..12}; do
if [ $m -eq 12 ] # december exception
then echo $(date -d $m/1/$y +%b) - $(date -d "$(($m%12+1))/1/$y" +%A)
break
fi
echo $(date -d $m/1/$y +%b) - $(date -d "$(($m%12+1))/1/$y - 1 days" +%A) # print the last day of the week for the month
done
It lists the last day of the week for every month:
Jan - Monday
Feb - Monday
Mar - Thursday
Apr - Saturday
May - Tuesday
Jun - Thursday
Jul - Sunday
Aug - Wednesday
Sep - Friday
Oct - Monday
Nov - Wednesday
Dec - Saturday
Now I need to reverse it, so that it lists months ending on every day of the week like so:
Sunday - Jul
Monday - Jan Feb Oct
Tuesday - May
Wednesday - Aug Nov
Thursday - Mar Jun
Friday - Sep
Saturday - Apr Dec
I'm thinking of a nested loop,
for d in {1..7};
And storing months in an array?
#! /usr/bin/env bash
# if year is not specified use the current year
declare -r year="${1:-$(date +%Y)}"
# associative array (aka hash table)
declare -A months_per_day=()
for m in {01..12}; do
day_month=$(LANG=C date -d "${year}-${m}-01 +1 month -1 day" +"%A %b")
months_per_day[${day_month% *}]+=" ${day_month#* }"
done
for day in Sunday Monday Tuesday Wednesday Thursday Friday Saturday; do
echo "${day} -${months_per_day[${day}]:-}"
done
Output:
Sunday - Jul
Monday - Jan Feb Oct
Tuesday - May
Wednesday - Aug Nov
Thursday - Mar Jun
Friday - Sep
Saturday - Apr Dec
Using GNU awk for time functions, with lots of intermediate and descriptively named variables to make it easy to understand:
$ cat tst.sh
#!/usr/bin/env bash
awk -v year="$1" '
BEGIN {
OFS = " - "
year = (year == "" ? strftime("%Y") : year)
secsInDay = 24*60*60
for ( mthNr=1; mthNr<=12; mthNr++ ) {
lastDayEpochSecs = mktime(year " " (mthNr+1) " 1 12 0 0") - secsInDay
mthAbbrDayName = strftime("%b %A", lastDayEpochSecs)
split(mthAbbrDayName,m)
mthAbbr = m[1]
dayName = m[2]
mthNr2mthAbbr[mthNr] = mthAbbr
mthAbbr2dayName[mthAbbr] = dayName
dayName2mthAbbrs[dayName] = \
(dayName in dayName2mthAbbrs ? dayName2mthAbbrs[dayName] " " : "" ) mthAbbr
}
for ( mthNr=1; mthNr<=12; mthNr++ ) {
mthAbbr = mthNr2mthAbbr[mthNr]
dayName = mthAbbr2dayName[mthAbbr]
print mthAbbr, dayName
}
print "\n--------\n"
for ( dayName in dayName2mthAbbrs ) {
mthAbbrs = dayName2mthAbbrs[dayName]
print dayName, mthAbbrs
}
}
'
$ ./tst.sh
Jan - Monday
Feb - Monday
Mar - Thursday
Apr - Saturday
May - Tuesday
Jun - Thursday
Jul - Sunday
Aug - Wednesday
Sep - Friday
Oct - Monday
Nov - Wednesday
Dec - Saturday
--------
Tuesday - May
Friday - Sep
Sunday - Jul
Thursday - Mar Jun
Saturday - Apr Dec
Monday - Jan Feb Oct
Wednesday - Aug Nov
The above will be much faster than calling date multiple times in a shell loop and is trivial to modify to do anything else you need.
This answer refactors your implementation as a reusable getlastday function. Then, we go through the loop 7 times matching the getlastday to the matching months, and print it out:
#!/bin/bash
getlastday() {
if [ $m -eq 12 ]; then
echo $(date -d "$(($m%12+1))/1/$y" +%A)
return
fi
echo $(date -d "$(($m%12+1))/1/$y - 1 days" +%A) # print the last day of the week for the month
}
y=$1
if [ -z $1 ] # if year is not specified use the current year
then y=(`date +%Y`)
fi
for d in Monday Tuesday Wednesday Thursday Friday Saturday Sunday; do
months=()
for m in {1..12}; do
lastday=$(getlastday)
if [ $lastday != $d ]; then continue; fi
months+=($(date -d $m/1/$y +%b))
done
echo $d - ${months[#]}
done
[EDIT: version 2]
As per #markp-fuso's comment below the above has an inefficiency in that there are 84 iterations to build the list.
The following is an improvement where we construct an res result array with 7 entries in it. One for each day in the week. Then we use +%u to get the day number instead of the day string. This will give us the index to the res result append where to append the month to:
#!/bin/bash
y=$1
if [ -z $1 ] # if year is not specified use the current year
then y=(`date +%Y`)
fi
res=("Sunday -" "Monday -" "Tuesday -" "Wednesday -" "Thursday -" "Friday -" "Saturday -")
for m in {1..12}; do
mstr=$(date -d $m/1/$y +%b)
if [ $m -eq 12 ]; then
d=$(date -d "$(($m%12+1))/1/$y" +%u)
else
d=$(date -d "$(($m%12+1))/1/$y - 1 days" +%u)
fi
res[$d]="${res[$d]} $mstr"
done
for d in {0..6}; do
echo "${res[$d]}"
done
Using jq your task can be solved:
INPUT='
Jan - Monday
Feb - Monday
Mar - Thursday
Apr - Saturday
May - Tuesday
Jun - Thursday
Jul - Sunday
Aug - Wednesday
Sep - Friday
Oct - Monday
Nov - Wednesday
Dec - Saturday
'
jq -Rrs '
(split("\n") | map(split(" - "))) as $input | # split lines and split each line by "-"
reduce ("Sunday", "Monday", "Tuesday", "Wednesday", "Thursday", "Friday", "Saturday") as $day # iterate over all weekdays
([]; . + [ ($input | map(select(.[1] == $day)) | # select entries for $day
.[0][1] + " - " + (map(.[0]) | join(" "))) # generate output for $day
]
) | .[]
' <<< "$INPUT"
Output
Sunday - Jul
Monday - Jan Feb Oct
Tuesday - May
Wednesday - Aug Nov
Thursday - Mar Jun
Friday - Sep
Saturday - Apr Dec
Try this Shellcheck-clean code:
#! /bin/bash -p
year=${1-$(date +%Y)}
months=( Jan Feb Mar Apr May Jun Jul Aug Sep Oct Nov Dec )
day_months=( 'Sunday -' 'Monday -' 'Tuesday -' 'Wednesday -'
'Thursday -' 'Friday -' 'Saturday -' )
for m in {1..12}; do
last_day=$(date -d "$year-$m-1 +1 month -1 day" +%w)
day_months[last_day]+=" ${months[m-1]}"
done
printf '%s\n' "${day_months[#]}"
declare -A a
for i in '1/1 + 1 year' {2..12}/1; do
d=($(LC_ALL=C date '+%A %B' -d "$i - 1 second"))
a["$d"]+=${a["$d"]:+ }${d[1]}
done
for i in {Sun,Mon,Tues,Wednes,Thurs,Fri,Satur}day; do
echo "$i - ${a["$i"]}"
done
This prints the list for the current year:
Sunday - July
Monday - January February October
Tuesday - May
Wednesday - August November
Thursday - March June
Friday - September
Saturday - December April
technically the only info you'll need is what Jan ends on, and either (a) whether year is leap or not, or (b) whether Feb ends on same day of week as Jan.
The rest are easily deduced without running date 12 times
Using ksh93:
#!/bin/ksh93
# if year is not specified use the current year
year="${1:-$(date +%Y)}"
typeset -A days
for month in Jan Feb Mar Apr May Jun Jul Aug Sep Oct Nov Dec
do
lastday=$(printf "%(%A)T" "final day $month $year")
days[$lastday]+=" $month"
done
for day in Sunday Monday Tuesday Wednesday Thursday Friday Saturday
do
printf "%-9s -%s\n" "$day" "${days[$day]:-}"
done
For 2022 produces the following output:
Sunday - Jul
Monday - Jan Feb Oct
Tuesday - May
Wednesday - Aug Nov
Thursday - Mar Jun
Friday - Sep
Saturday - Apr Dec
For 2023 produces the following output:
Sunday - Apr Dec
Monday - Jul
Tuesday - Jan Feb Oct
Wednesday - May
Thursday - Aug Nov
Friday - Mar Jun
Saturday - Sep
I have a input date:
01-01-2019 08:30:00
And I would like to back it in time by:
2hours 10minutes
So the new date should be like:
01-01-2019 06:20
I have tried commands:
(1)date -d "2019-01-01 08:30:00 + 1hour"
which return: Tue Jan 1 09:30:00 STD 2019 (ok, but seems to be lucky shot)
(2)date -d "2019-01-01 08:30:00 - 1hour"
which return: Tue Jan 1 11:30:00 STD 2019 that means +3hours (wrong)
(3)date -d "2019-01-01 08:30:00 + 3hours"
which return: Tue Jan 1 07:30:00 STD 2019 that means -1hours (wrong)
(4)date -d "2019-01-01 08:30:00 + 2minutes"
which return: Tue Jan 1 07:31:00 STD 2019 that means -59minutes (wrong)
You probably need to explicitly mention your timezone:
$ TZ=STD date -d '2019-01-01 08:30:00 STD + 1 hour'
Tue Jan 1 09:30:00 STD 2019
$ TZ=STD date -d '2019-01-01 08:30:00 STD - 3 hour'
Tue Jan 1 05:30:00 STD 2019
$ TZ=STD date -d '2019-01-01 08:30:00 STD + 2 minutes'
Tue Jan 1 08:32:00 STD 2019
This seems to alleviate dates confusion.
The is only one most reliable way of doing that:
date -d "#$(( $(date -d '2019-01-01 08:30:00' +%s) - ( 2 * 60 + 10 ) * 60 ))"
First convert to seconds since epoch.
Then subtract the time you want.
Then use date to format the output.
I am trying to create some sort of loop in bash that will iterate through a specific time range, one second at a time.
At each interval, it will perform a duty with that timestamp I.e. "Wed Mar 2 12:00:03 CDT 2018"
I am having a hard time wrapping my head around how to make a loop that will iterate every second of time, and when it hits 60 seconds, update the minute, etc.
Thoughts? This seems obvious but the right syntax escapes me.
this is one way of doing it, delegate time computations to date
$ for i in {1..10}; do date -d "+$i seconds"; done
Sun Mar 11 20:40:57 UTC 2018
Sun Mar 11 20:40:58 UTC 2018
Sun Mar 11 20:40:59 UTC 2018
Sun Mar 11 20:41:00 UTC 2018
Sun Mar 11 20:41:01 UTC 2018
Sun Mar 11 20:41:02 UTC 2018
Sun Mar 11 20:41:03 UTC 2018
Sun Mar 11 20:41:04 UTC 2018
Sun Mar 11 20:41:05 UTC 2018
Sun Mar 11 20:41:06 UTC 2018
Sun Mar 11 20:41:07 UTC 2018
if you want to iterate up to a certain time, change the for loop to a while loop and a counter.
To make a loop with time you need to format date/time in such a way that can be comparable. The most easy trick is to transform date to something that will look like an integer and you then use lt,le,eq,gt,ge operators of bash.
Consider this :
$ date
Mon Mar 12 00:16:29 EET 2018 #this format/data type is not comparable by bash
$ date +%Y%m%d%H%M%S
20180312001629 #this is an integer representing the current time
This is a sample loop from now up to a specific time point:
stop=$(date +%Y%m%d%H%M%S --date "2018-03-12 00:20:55")
while [[ $(date +%Y%m%d%H%M%S) -le $stop ]];do #comparing current date & time with stop date & time
echo "date now is $(date)";
sleep 1; #sleep 1 second
done
date now is Mon Mar 12 00:20:51 EET 2018
date now is Mon Mar 12 00:20:52 EET 2018
date now is Mon Mar 12 00:20:53 EET 2018
date now is Mon Mar 12 00:20:54 EET 2018
date now is Mon Mar 12 00:20:55 EET 2018
The trick here is that the command
stop=$(date +%Y%m%d%H%M%S --date "2018-03-12 00:20:55")
converts given date (using the --date flag) to a kind of integer format.
Then the while loop keeps comparing current date in the same integer format with the stop date.
This is a similar script to start/stop at a specific time, but it is a bit "resources hungry" since it keeps comparing current time to start time:
date; #just print the current date & time
compl=false;
start=$(date +%Y%m%d%H%M%S --date "2018-03-12 01:45:50");
stop=$(date +%Y%m%d%H%M%S --date "2018-03-12 01:45:55");
while true;do
while [[ $(date +%Y%m%d%H%M%S) -ge $start ]] && [[ $(date +%Y%m%d%H%M%S) -le $stop ]];do
echo "date now is $(date)";
sleep 1;
compl=true;
done;
($compl) && break;
done
Mon Mar 12 01:45:37 EET 2018
date now is Mon Mar 12 01:45:50 EET 2018
date now is Mon Mar 12 01:45:51 EET 2018
date now is Mon Mar 12 01:45:52 EET 2018
date now is Mon Mar 12 01:45:53 EET 2018
date now is Mon Mar 12 01:45:54 EET 2018
date now is Mon Mar 12 01:45:55 EET 2018
An alternative to start/stop a script at specific time would be to use epoch date / epoch seconds.
Epoch date is measured in seconds since the epoch (1970-01-01 UTC).
Every date can be expressed as epoch seconds, using the format date +%s.
date #just print the current date & time
start="2018-03-12 02:17:52"
stop="2018-03-12 02:17:57"
timerequired=$(( $(date +%s --date "$start") - $(date +%s) ))
sleep $(($timerequired)) #sleep till the starting time
while [[ $(date +%s) -le $(date +%s --date "$stop") ]];do
echo "date now is $(date)";
sleep 1;
done
Mon Mar 12 02:17:39 EET 2018
date now is Mon Mar 12 02:17:52 EET 2018
date now is Mon Mar 12 02:17:53 EET 2018
date now is Mon Mar 12 02:17:54 EET 2018
date now is Mon Mar 12 02:17:55 EET 2018
date now is Mon Mar 12 02:17:56 EET 2018
date now is Mon Mar 12 02:17:57 EET 2018
Finally, you can use external tools to automate things based on date & time like cron, anacron , at, etc
Echo a date for every second in a range:
from=$(date -d "Wed Mar 2 12:59:58 CDT 2018" +%s)
to=$(date -d "Wed Mar 2 13:00:04 CDT 2018" +%s)
for s in $(seq $from $to); do echo $(date -d #$s); done
The date -d takes a well formed date as input. +%s formats it into seconds since epoch. So you get two big values like from=1520013598, to=...3604. The date -d #1520013598 transforms the timestamp back to a human readable date.
Your range might be in a variable like start and stop, and you might like to foobar ($time) instead of echoing it:
start="Wed Mar 2 12:59:58 CDT 2018"
stop="Wed Mar 2 13:00:04 CDT 2018"
from=$(date -d "$start" +%s)
to=$(date -d "$stop" +%s)
for s in $(seq $from $to)
do
timestamp=$(date -d #$s)
foobar $timestamp
done
I have this data:
`date +%Y-%m-%d`" 00:00:00"
that return 2015-10-08 00:00:00
I would like cancel 5 minute:
2015-10-07 23:55:00
Many thanks
You need to subtract 5 minutes from a known point in time:
$ date -d "00:00:00 today"
Thu Oct 8 00:00:00 EDT 2015
$ date -d "00:00:00 today -5 minutes"
Wed Oct 7 23:55:00 EDT 2015
You just need to add your format string.
There's more than one way to subtract a value from the current time, although this should match the format shown in your question:
date -d "-5 min" "+%Y-%m-%d %H:%M:%S"
Result:
2015-10-08 15:26:13
How to get epoch time in shell script (for ksh)?
I am interested in getting epoch time for the start of day (so e.g. now is July 28th, 2011 ~ 14:25:00 EST, I need time at midnight).
If you have GNU date,
epoch=$( date -d 00:00 +%s )
Otherwise, if you have tclsh,
epoch=$( echo 'puts [clock scan 00:00]' | tclsh )
Otherwise,
epoch=$( perl -MTime::Local -le 'print timelocal(0,0,0,(localtime)[3..8])' )
ksh's printf '%(fmt)T' supports time calculating. For example:
$ printf '%T\n' now
Mon Mar 18 15:11:46 CST 2013
$ printf '%T\n' '2 days ago'
Sat Mar 16 15:11:55 CST 2013
$ printf '%T\n' 'midnight today'
Mon Mar 18 00:00:00 CST 2013
$