I am trying to create a script in which 4 days ago date should be equal to to current date if it is not then add 1 more day and check. Below is the one i have created but still not clear about answer.
#!/bin/bash
batchdate=`date --date "4 day ago" '+%Y%m%d'`
matchdate=`date --date "today" '+%Y%m%d'`
for i in {0..4}
do
if [ $batchdate != $matchdate && $NEXT_DATE != $matchdate ]; then
NEXT_DATE=$(date +%Y%m%d -d "$batchdate + $i day")
echo "$NEXT_DATE"
break
fi
done
First, define a little helper function to avoid doing the same thing in slightly different ways:
get_date () {
date +%Y-%m-%d --date "$1"
}
Now, you have two variables: the current date, which will never change, and the starting date, which you will increment one day at a time until it matches the current date.
then=$(get_date "4 days ago")
now=$(get_date "today")
while [[ $then != $now ]]; do
then=$(get_date "$then + 1 day")
echo "$then"
done
Related
I'm trying to make a script that is working from 12/24/2020 through 06/01/2021. For now I olny have it working only for 2 dates with this:
if [ "$(date +'%m%d')" != "1224" ] && [ "$(date +'%m%d')" != "1226" ];
So this script should be working from 12/24 - 01/06.
Any suggestions? Thanks
You can use the +%s option in date to convert the start/finish dates to epoch time. You supply the dates using the -d option.
Then, you retrieve the current date using +%s again and compare using standard integer comparison, like this:
start=$(date +%s -d '12/24/2020')
finish=$(date +%s -d '06/02/2021')
now=$(date +%s)
if [ $now -ge $start ] && [ $now -lt $finish ]
then
echo "Do something"
else
echo "Skip"
fi
Edit: For the finish date, its necessary to use the day after your intended finishing date and compare using -lt. This is because date will return the timestamp for the start of the specified date. So doing this, you end up having the comparison succeed until midnight on 06/01/2021. Thanks to #gordon-davisson for pointing this out.
beg='20201224'
end='20210601'
now=$(date +'%Y%m%d')
if (( beg <= now )) && (( now <= end)); then
I need a scipt that gets the prior month and writes it to a variable. After that I need to check if the month is a quarter month (Mar,Jun,Sep,Dec) and if it is call another script and pass the month year as an argument in the format "Sep-2020". If its not a quarter month call another script.
monthyear="Sep-2020"
if [[ quartermonth ]]
then runscript1 $monthyear
elif [[ not a quartermonth ]]
then runscript2 $monthyear
fi
bash should be available on oracle Linux; here's a bash version:
#!/bin/bash
lastmoyr=$(date '+%b-%Y' --date='1 month ago')
lm=$(date '+%m' --date='1 month ago')
lm=${lm#0}
echo last month-year "$lastmoyr"
echo last month number "$lm"
# Mar Jun Sep Dec ==> 3 6 9 12
if [[ $((lm % 3)) == 0 ]]; then
runscript1 "$lastmoyr"
else
runscript2 "$lastmoyr"
fi
You can learn how to find previous month here
$monthyear="Sep-2020" //This wont work. Refer the above link
if [[ $quartermonth % 3 -eq 0 ]] //quartermonth is the value of month taken from monthyear
then runscript1 $monthyear
elif [[ $quartermonth % 3 -ne 0]]
then runscript2 $monthyear
fi
I can't comment on the accepted answer, so'll write it as a new one.
Attention, the accepted answer will produce an unexpected result (the wrong month) in some cases, like if you run the script on 31th.
To illustrate, run this two examples on your bash shell (if your date utility doesn't support the needed options, let me know which one you're using and i'll try to adapt it).
:: From accepted answer (for simulated day 2020-October-31)
echo "Prior month? : $(date '+%b-%Y' -d "$(date -d "2020-10-31") 1 month ago")"
Prior month? : Oct-2020 - "October", probably not what you want.
versus (a more reliable solution)
echo "Prior month? : $(date '+%b-%Y' -d "$(date +%Y-%m-1 -d "2020-10-31") 1 month ago")"
Prior month? : Sep-2020 - "September" the correct prior month.
This happens because the date utility, for 'relative' calculations, for month usually uses 30 days and not "a month".
To work around this, you'll can calculate the prior month based on day 1 (or 15) instead of using the current day.
Adapting #Milag script, it would look something like.
#!/usr/bin/env bash
lastmoyr=$(date '+%b-%Y' -d "$(date +%Y-%m-1) 1 month ago")
lm=$(date '+%-m' -d "$(date +%Y-%m-1) 1 month ago")
echo last month-year "$lastmoyr"
echo last month number "$lm"
# Mar Jun Sep Dec ==> 3 6 9 12
if [[ $((lm % 3)) == 0 ]]; then
runscript1 "$lastmoyr"
else
runscript2 "$lastmoyr"
fi
I want to write a bash script that determines the 10 day period (decade) has ended relative to the start date (in the format YYYY-MM-DD).
If the 10 day period is finished script has to output the 10 days period.
Im new in bash and has a lot syntax errors with code, help me pls.
#!/bin/bash
# GNU bash, version 4.3.46
CURRENT_DATE=$(date +%Y-%m-%d)
START_DATE=2019-01-01
IS_TODAY_DECADE_CALCULATION_DAY = (CURRENT_DATE - START_DATE) % 10
if [ $IS_TODAY_DECADE_CALCULATION_DAY -eq 0 ]
then
BEGIN_DATE = $("$CURRENT_DATE - 11 days" +%Y-%m-%d)"
END_DATE = $("$CURRENT_DATE - 1 day" +%Y-%m-%d)"
echo "Period is="$BEGIN_DATE":"$END_DATE"
else
echo "Decade is not finished."
fi
You should compare the unix time stamps. If the time stamp "now+10 days" is larger than the start date, the period is ended.
#! /bin/bash
DATE_OLD=$(date "+%F" -d "-11 days")
DATE_NOW=$(date "+%F")
TEST_DATE_NOW=$(date "+%s" -d ${DATE_NOW})
TEST_DATE_OLD=$(date "+%s" -d ${DATE_OLD})
DIVIDER=$(( (TEST_DATE_NOW - TEST_DATE_OLD) / (60*60*24) ))
REMAINING=$(( DIVIDER % 10 ))
echo "Days between ${DATE_OLD} and ${DATE_NOW} is $DIVIDER"
if [ ${DIVIDER} -gt 0 ]; then
echo "Date ${DATE_OLD} is in the past"
else
echo "Date ${DATE_OLD} is in the future"
fi
if [ $REMAINING -eq 0 ]; then
echo "Ten days period ended"
else
echo "Still in ten day period"
fi
exit 0;
The question implies that the code should identify each 10 day period starting on a specific START_DATE. Bash does not have date math - it can not calculate difference between dates (as expected by '(CURRENT_DATE - START_DATE)'). Two options
Convert date to seconds since Unix Epoch, and do the math on those values, OR
Use date utilities package, OR
using Python, awk, perl
Implementing #1 is simple. Notice few changes to assignments - in particular no spaces are allowed in assignments variable=expression, or let variable=expression
#! /bin/bash
CURRENT_DATE=$(date +%Y-%m-%d)
START_DATE=2019-01-01
# Instead of IS_TODAY_DECADE_CALCULATION_DAY = (CURRENT_DATE - START_DATE) % 10
SEC_IN_DAY=$((60*60*24))
let D1=$(date '+%s' -d "$CURRENT_DATE Z")/SEC_IN_DAY
let D2=$(date '+%s' -d "$START_DATE Z")/SEC_IN_DAY
let IS_TODAY_DECADE_CALCULATION_DAY=(CURRENT_DATE-START_DATE)%10
# Rest of script here
if [ $IS_TODAY_DECADE_CALCULATION_DAY -eq 0 ]
then
BEGIN_DATE=$(date -d "$CURRENT_DATE - 11 days" +%Y-%m-%d)
END_DATE=$(date -d "$CURRENT_DATE - 1 day" +%Y-%m-%d)
echo "Period is=$BEGIN_DATE:$END_DATE"
else
echo "Decade is not finished."
fi
In bash on macOS, I would like to write a small script with dates (or any other program that would do) that gives me a list of dates in the format yyyymmdd of every Saturday of a given year and saves it to a variable.
For example, if I wanted to have a list of dates for all Saturdays of the year 1850, it should somehow look like this:
var = [ 18500105, 18500112, 18500119, …, 18501228 ]
with the below code:
list=()
for month in `seq -w 1 12`; do
for day in `seq -w 1 31`; do
list=( $(gdate -d "1850$month$day" '+%A %Y%m%d' | grep 'Saturday' | egrep -o '[[:digit:]]{4}[[:digit:]]{2}[[:digit:]]{2}' | tee /dev/tty) )
done
done
However, the above command does not write anything in the array list although it gives me the right output with tee.
How can I fix these issues?
Modifying Dennis Williamson's answer slightly to suit your requirement and to add the results into the array. Works on the GNU date and not on FreeBSD's version.
#!/usr/bin/env bash
y=1850
for d in {0..6}
do
# Identify the first day of the year that is a Saturday and break out of
# the loop
if (( $(date -d "$y-1-1 + $d day" '+%u') == 6))
then
break
fi
done
array=()
# Loop until the last day of the year, increment 7 days at a
# time and append the results to the array
for ((w = d; w <= $(date -d "$y-12-31" '+%j'); w += 7))
do
array+=( $(date -d "$y-1-1 + $w day" '+%Y%m%d') )
done
Now you can just print the results as
printf '%s\n' "${array[#]}"
To set up the GNU date on MacOS you need to do brew install coreutils and access the command as gdate to distinguish it from the native version provided.
Argh, just realised you need it for MacOS date.
I will leave the answer for others that do not have that restriction, but it will not work for you.
This is not quite what you want, but close:
year=1850
firstsat=$(date -d $year-01-01-$(date -d $year-01-01 +%w)days+6days +%Y%m%d)
parset a 'date -d '$firstsat'+{=$_*=7=}days +%Y%m%d' ::: {0..52}
echo ${a[#]}
It has the bug, that it finds the next 53 Saturdays, and the last of those may not be in current year.
parset is part of GNU Parallel.
I didn't do much error checking, but here's another implementation.
Takes day of week and target year as arguments.
Gets the julian day of the first matching weekday requested -
gets the epoch seconds of noon on that day -
as long as the year matches what was requested, adds that date to the array and adds a week's worth of seconds to the tracking variable.
Lather, rinse, repeat until no longer in that year.
$: typeset -f alldays
alldays () { local dow=$1 year=$2 julian=1;
until [[ "$dow" == "$( date +%a -d $year-01-0$julian )" ]]; do (( julian++ )); done;
es=$( date +%s -d "12pm $year-01-0$julian" );
allhits=( $( while [[ $year == "$( date +%Y -d #$es )" ]]; do date +%Y%m%d -d #$es; (( es+=604800 )); done; ) )
}
$: time alldays Sat 1850
real 0m9.931s
user 0m1.025s
sys 0m6.695s
$: printf "%s\n" "${allhits[#]}"
18500105
18500112
18500119
18500126
18500202
18500209
18500216
18500223
18500302
18500309
18500316
18500323
18500330
18500406
18500413
18500420
18500427
18500504
18500511
18500518
18500525
18500601
18500608
18500615
18500622
18500629
18500706
18500713
18500720
18500727
18500803
18500810
18500817
18500824
18500831
18500907
18500914
18500921
18500928
18501005
18501012
18501019
18501026
18501102
18501109
18501116
18501123
18501130
18501207
18501214
18501221
18501228
I am writing a bash script which will print the number of weeks,days and hours left for user's birthday from the time it is run. The user will only fill in the his birth-date and month.
I searched and found out that we can take difference of two dates:
let DIFF=(`date +%s -d 20150108`-`date +%s -d today`)/86400
echo $DIFF
The above script will print the number of days left for the given date. We can similarly find out the number of weeks and hours.
But the problem is, if user gives only the birth-date excluding the year, (here he gives only 8 Jan) how do I figure out the least difference possible between upcoming birthday and today?
Example, if the birthday just passed , then I have to consider the birthday which comes next year.
Just check if the date happened already this year, and if it did append next year to the input.
> input="8 Feb"
> input=$(date -d "$input" '+%d %b')
> [[ $(date -d "$input" '+%s') -lt $(date -d "$(date '+%d/%m/%y')" '+%s') ]] && input="$input $(( $(date '+%Y') + 1 ))"
> let DIFF=(`date +%s -d "$input"`-`date +%s -d today`)/86400
> echo $DIFF
360
Couple of caveats. 2nd line strips year and other information (time, etc...) from date if user enters it with that. And $(date -d "$(date '+%d/%m/%y')" '+%s') in third line is needed so it compares only against the current date, rather than the date with time.
Another way to count days is to use date's %j format code, which returns the day number in the year (an three-digit integer between 001 and 366, inclusive). (Note: we prepend 1 to the output of date +%j to make it a four-digit integer not starting with a 0, to avoid octal conversion.)
DATE="Feb 7"
TODAY=1$(date +%j)
BDAY=1$(date +%j -d "$DATE")
if ((BDAY<TODAY)); then
BDAY=1$(date +%j -d "$DATE "$(($(date +%Y)+1)))
fi
DAYS_TO_GO=$(($BDAY-$TODAY))