Operating System: Red Hat Enterprise Linux Server 7.2 (Maipo)
I want to round the time to the nearest 5 minutes, only up, not down, for example:
08:09:15 should be 08:10:00
08:11:26 should be 08:15:00
08:17:58 should be 08:20:00
I have been trying with:
(date -d #$(( (($(date +%s) + 150) / 300) * 300)) "+%H:%M:%S")
This will round the time but also down (08:11:18 will result in 08:10:00 and not 08:15:00)
Any idea how i can achieve this?
You may use this utility function for your rounding up:
roundDt() {
local n=300
local str="$1"
date -d #$(( ($(date -d "$str" '+%s') + $n)/$n * $n)) '+%H:%M:%S'
}
Then invoke this function as:
roundDt '08:09:15'
08:10:00
roundDt '08:11:26'
08:15:00
roundDt '08:17:58'
08:20:00
To trace how this function is computing use -x (trace mode) after exporting:
export -f roundDt
bash -cx "roundDt '08:11:26'"
+ roundDt 08:11:26
+ typeset n=300
+ typeset str=08:11:26
++ date -d 08:11:26 +%s
+ date -d #1535631300 +%H:%M:%S
08:15:00
GNU date can calculate already. It is explained in the manual in the chapter "Relative items in date strings". So you need just one date call.
d=$(date +%T) # get the current time
IFS=: read h m s <<< "$d" # parse it in hours, minutes and seconds
inc=$(( 300 - (m * 60 + s) % 300 )) # calculate the seconds to increment
date -d "$d $inc sec" +%T # output the new time with the offset
Btw: +%T is the same as +%H:%M:%S.
Related
I'm trying to calculate time difference stored inside of two variables inside of a shell script, I'm observing the following pattern:
hhmm -> 0950
so:
time1=1333
time2=0950
Now I need to calculate the difference in time between time1 and time2, as for now I have tried:
deltaTime=$(($time1-$time2))
but I'm facing the following error message
1333-0950: value too great for base (error token is "0950")
I'm expecting as a result: $deltaTime=0343
Unfortunately, I am strictly bound to use this time pattern. I have already researched for a solution online, some of them propose to use date -d... but I couldn't get it to work :(
Your approach has two issues.
First issue: bash recognizes numbers with leading zeroes as octal. You can force base10 by adding 10# prefix.
Second issue: it is incorrect to consider strings in hhmm format as numbers and substract them. e.g. 1333-950=383 but difference between 09:50 and 13:33 is 3 hours and 43 minutes. You should convert string values to common units, e.g. to minutes, substract them and convert back to hhmm format.
time1=1333
time2=0950
str2min()
{
printf "%u" $((10#${1%??} * 60 + 10#${1#??}))
}
min2str()
{
printf "%02u%02u" $(($1 / 60)) $(($1 % 60))
}
time1m=$(str2min $time1)
time2m=$(str2min $time2)
timediff=$(($time1m - $time2m))
deltaTime=$(min2str $timediff)
You could use this implementation maybe?
#!/usr/bin/env bash
diff_hhmm() {
local -r from=$1
local -i from_hh=10#${from:0:2} # skip 0 chars, read 2 chars (`${from:0:2}`) using base 10 (`10#`)
local -ri from_mm=10#${from:2:2} # skip 2 chars, read 2 chars (`${from:0:2}`) using base 10 (`10#`)
local -r upto=$2
local -ri upto_hh=10#${upto:0:2}
local -ri upto_mm=10#${upto:2:2}
local -i diff_hh
local -i diff_mm
# Compute difference in minutes
(( diff_mm = from_mm - upto_mm ))
# If it's negative, we've "breached" into the previous hour, so adjust
# the `diff_mm` value to be modulo 60 and compensate the `from_hh` var
# to reflect that we've already subtracted some of the minutes there.
if (( diff_mm < 0 )); then
(( diff_mm += 60 ))
(( from_hh -= 1 ))
fi
# Compute difference in hours
(( diff_hh = from_hh - upto_hh ))
# Ensure the result is modulo 24, the number of hours in a day.
if (( diff_hh < 0 )); then
(( diff_hh += 24 ))
fi
# Print the values with 0-padding if necessary.
printf '%02d%02d\n' "$diff_hh" "$diff_mm"
}
$ diff_hhmm 1333 0950
0343
$ diff_hhmm 0733 0950
2143
$ diff_hhmm 0733 0930
2203
Or an even shorter implementation using a big arithmetic compound command ((( ... )) ) and inlining some variables:
diff_hhmm_terse() {
local -i diff_hh diff_mm
((
diff_mm = 10#${1:2:2} - 10#${2:2:2},
diff_hh = 10#${1:0:2} - 10#${2:0:2},
diff_hh -= diff_mm < 0 ? 1 : 0,
diff_mm += diff_mm < 0 ? 60 : 0,
diff_hh += diff_hh < 0 ? 24 : 0
))
printf '%02d%02d\n' "$diff_hh" "$diff_mm"
}
Do you have the possibility to drop the leading zero?
As you can see from my prompt:
Prompt> echo $((1333-0950))
-bash: 1333-0950: value too great for base (error token is "0950")
Prompt> echo $((1333-950))
383
Other proposal:
date '+%s'
Let me give you some examples:
date '+%s'
1662357975
... (after some time)
date '+%s'
1662458180
=>
echo $((1662458180-1662357975))
100205 (amount of seconds)
=>
echo $(((1662458180-1662357975)/3600))
27 (amount of hours)
This bash one-liner may be used if time difference is not negative (that is, time1 >= time2):
printf '%04d\n' $(( 10#$time1 - 10#$time2 - (10#${time1: -2} < 10#${time2: -2} ? 40 : 0) ))
Well I'm trying to get the difference between two dates in Seconds.MilliSeconds The dates are in Zulu format
I have tried these two approaches doesn't work out for me
ms1=$(date -d '2022-04-22T03:47:56.744551446Z' +%N)
ms2=$(date -d '2022-04-22T03:47:57.095419744Z' +%N)
msdiff=$((ms1 - ms2))
echo "$msdiff"
$ dateutils.ddiff 2022-04-22T03:47:56.744551446Z 2022-04-22T03:47:57.095419744Z -f '%N'
0s
Is there any better way to get the difference in Seconds.MilliSeconds In linux for Z format time zones
Suggesting
ms1=$(date -d '2022-04-22T03:47:56.744551446Z' +%s.%N)
ms2=$(date -d '2022-04-22T03:47:57.095419744Z' +%s.%N)
msdiff=$(awk "BEGIN{print $ms2 - $ms1}")
echo "msdiff=$msdiff"
Output:
msdiff=0.350868
Is there an option of date function?
How do I get LOAD_TEST_START to be 2 minutes forward?
How do I get LOAD_TEST_END to be 2 minutes back?
LOAD_TEST_START=$(date -u +%FT%TZ)
LOAD_TEST_END=$(date -u +%FT%TZ)
This answer helped me and works on the linux-version of date.
LOAD_TEST_START=$(date +%FT%TZ -d "2 minutes ago")
LOAD_TEST_END=$(date +%FT%TZ -d "2 minutes")
On macos/FreeBSD one can use
LOAD_TEST_START=$(date -v+2M +%FT%TZ)
LOAD_TEST_END=$(date -v-2M +%FT%TZ)
Not sure if this is what you want, but you can get the current epoch value and then add/subtract 120 seconds (2 minutes) from it and convert the results back to date.
dateval=`date +%s`
echo "orig "`date -d #$dateval +%FT%T%Z`
start=`expr $dateval + 120`
end=`expr $dateval - 120`
LOAD_TEST_START=`date -d #$start +%FT%T%Z`
LOAD_TEST_END=`date -d #$end +%FT%T%Z`
echo "start "$LOAD_TEST_START
echo "end "$LOAD_TEST_END
I have a backup script which is written in bash/shell scripting language. I calculate the total runtime/execution time by doing this:
#!/bin/bash
# start time
res1=$(date +%s.%N)
### do some work here
# end time & calculate
res2=$(date +%s.%N)
dt=$(echo "$res2 - $res1" | bc)
dd=$(echo "$dt/86400" | bc)
dt2=$(echo "$dt-86400*$dd" | bc)
dh=$(echo "$dt2/3600" | bc)
dt3=$(echo "$dt2-3600*$dh" | bc)
dm=$(echo "$dt3/60" | bc)
ds=$(echo "$dt3-60*$dm" | bc)
# finished
printf " >>> Process Completed - Total Runtime (d:h:m:s) : %d:%02d:%02d:%02.4f\n" $dd $dh $dm $ds
echo " "
exit 0
This outputs something like this:
How do you format the result, so it looks something like this:
0 Days, 0 Hours, 0 Minutes and 0.0968 Seconds
If it can intelligently show only values > 0, like these examples - it would be abonus:
7 Minutes and 5.126 Seconds
or
2 hours, 4 Minutes and 1.106 Seconds
or
7.215 Seconds etc...
You can use your last printf like this:
printf " >>> Process Completed - Total Runtime (d:h:m:s) : %d Days, %02d Hours, %02d Minutes, %02.4f Seconds\n" $dd $dh $dm $ds
However I would suggest you to use awk and do all calculations and formatting in awk itself so that you can avoid many invocations of bc.
Suggested awk script:
awk -v res1="$res1" -v res2="$res2" 'BEGIN {dt=res2-res1; dd=dt/86400; dt2=dt-86400*dd;
dh=dt2/3600; dt3=dt2-3600*dh; dm=dt3/60; ds=dt3-60*dm;
printf " >>> Process Completed - Total Runtime (d:h:m:s) : %d Days, %02d Hours, %02d Minutes, %02.4f Seconds\n",
dt/86400, dd, dh, dm, ds}'
In a Linux script: I have a file that has two time entries for each message within the file. A 'received time' and a 'source time'. there are hundreds of messages within the file.
I want to calculate the elapsed time between the two times.
2014-07-16T18:40:48Z (received time)
2014-07-16T18:38:27Z (source time)
The source time is 3 lines after the received time, not that it matters.
info on the input data:
The input has a lines are as follows:
TimeStamp: 2014-07-16T18:40:48Z
2 lines later: a bunch of messages in one line and within each line, multiple times is:
sourceTimeStamp="2014-07-16T18:38:27Z"
If you have GNU's date (not busybox's), you can give difference in seconds with:
#!/bin/bash
A=$(date -d '2014-07-16T18:40:48Z' '+%s')
B=$(date -d '2014-07-16T18:38:27Z' '+%s')
echo "$(( A - B )) seconds"
For busybox's date and ash (modern probably / BusyBox v1.21.0):
#!/bin/ash
A=$(busybox date -d '2014-07-16 18:40:48' '+%s')
B=$(busybox date -d '2014-07-16 18:38:27' '+%s')
echo "$(( A - B )) seconds"
you should be able to use date like this (e.g.)
date +%s --date="2014-07-16T18:40:48Z"
to convert both timestamps into a unix timestamp. Getting the time difference between them is then reduced to a simple subtraction.
Does this help?
I would use awk. The following script searches for the lines of interest, converts the time value into a UNIX timestamp and saves them in the start, end variables. At the end of the script the difference will get calculated and printed:
timediff.awk:
/received time/ {
"date -d "$1" +%s" | getline end
}
/source time/ {
"date -d "$1" +%s" | getline start
exit
}
END {
printf "%s seconds in between", end - start
}
Execute it like this:
awk -f timediff.awk log.file
Output:
141 seconds in between