Unix Script: Loop over all days of a month

Assisted Solution

by:Hanno P.S.

Hanno P.S. earned 75 total points

ID: 418588822016-10-25

which date do we have today?

date '+%Y-%m-%d'

The last month would be

expr `date '+%Y%m' - 1` | sed -e 's/\(....\)/\1-'

Now you can start with you counter variable:

day=`expr `date '+%Y%m' - 1` | sed -e 's/\(....\)/\1-'`-01

Expert Comment

by:Hanno P.S.

ID: 418588832016-10-25

This variable wil vae to get incremented from day to day

LVL 5

Accepted Solution

by:Abhimanyu Suri

Abhimanyu Suri earned 225 total points

ID: 418589682016-10-25

Here is an example, please format the output as per your requirement and update ref_days with all months

/home>cat ref_days.txt ### month|days
10|31

/home>cat loop_date.sh
#!/usr/bin/sh

var_currdate=`date "+%Y-%d-%m"`
var_currmonth=`echo ${var_currdate}|awk -F'-' '{print $3}'`
var_date=`echo ${var_currdate}|awk -F'-' '{print $2}'`

echo $var_currdate $var_currmonth ${var_date}

if [[ `grep $var_currmonth ref_days.txt|wc -l` -eq 1 ]]
then
var_loopdays=`cat ref_days.txt|awk -F'|' '{print $2}'`
var_loop1=`expr $((${var_date}-1))`

counter=1

while [[ $counter -le $var_loop1 ]]
do
var_monthstartdate_calc=`expr $((${var_date}-$counter))`
echo "day $counter `date -d "-${var_monthstartdate_calc}days"`"
counter=`expr $(($counter+1))`
done

counter=${var_date}

while [[ ${counter} -le $var_loopdays ]]
do
var_monthstartdate_calc=`expr $(($counter-${var_date}))`
echo "day ${counter} `date -d "+${var_monthstartdate_calc}days"`"
counter=`expr $(($counter+1))`
done

else
echo "invalid month"
fi

/home>. ./loop_date.sh
2016-25-10 10 25
day 1 Sat Oct 1 12:45:56 EDT 2016
day 2 Sun Oct 2 12:45:56 EDT 2016
day 3 Mon Oct 3 12:45:56 EDT 2016
day 4 Tue Oct 4 12:45:56 EDT 2016
day 5 Wed Oct 5 12:45:56 EDT 2016
day 6 Thu Oct 6 12:45:56 EDT 2016
day 7 Fri Oct 7 12:45:56 EDT 2016
day 8 Sat Oct 8 12:45:56 EDT 2016
day 9 Sun Oct 9 12:45:56 EDT 2016
day 10 Mon Oct 10 12:45:56 EDT 2016
day 11 Tue Oct 11 12:45:56 EDT 2016
day 12 Wed Oct 12 12:45:56 EDT 2016
day 13 Thu Oct 13 12:45:56 EDT 2016
day 14 Fri Oct 14 12:45:56 EDT 2016
day 15 Sat Oct 15 12:45:56 EDT 2016
day 16 Sun Oct 16 12:45:56 EDT 2016
day 17 Mon Oct 17 12:45:56 EDT 2016
day 18 Tue Oct 18 12:45:56 EDT 2016
day 19 Wed Oct 19 12:45:56 EDT 2016
day 20 Thu Oct 20 12:45:56 EDT 2016
day 21 Fri Oct 21 12:45:56 EDT 2016
day 22 Sat Oct 22 12:45:56 EDT 2016
day 23 Sun Oct 23 12:45:56 EDT 2016
day 24 Mon Oct 24 12:45:56 EDT 2016
day 25 Tue Oct 25 12:45:56 EDT 2016
day 26 Wed Oct 26 12:45:56 EDT 2016
day 27 Thu Oct 27 12:45:56 EDT 2016
day 28 Fri Oct 28 12:45:56 EDT 2016
day 29 Sat Oct 29 12:45:56 EDT 2016
day 30 Sun Oct 30 12:45:56 EDT 2016
day 31 Mon Oct 31 12:45:56 EDT 2016

Author Comment

by:mruff

ID: 418598112016-10-25

Hi Abhimanu,
thx I am running this script under AIX and getting errors: date: illegal option --d usage: date [-] [+Field Descriptors]

Author Comment

by:mruff

ID: 418598152016-10-25

Dear Hanno
I am running the script at AIX
the command
day=`expr `date '+%Y%m' - 1` | sed -e 's/$....$/\1-'`-01
gives an error
syntax error at line 1 : '|' unexpecgted

Expert Comment

by:Hanno P.S.

ID: 418598372016-10-26

Let's do it step by step:
1. Get today's date as "a number" consisting of current year and month

date '+%Y%m'

Now try to have the result stored in a variable (and display the variable's content)

day=`date '+%Y%m' `
echo $day

Now, decrement this by one (to get one month back):

day=`expr $day - 1'
echo $day

Expert Comment

by:Hanno P.S.

ID: 418598402016-10-26

The srcipt Abhimanu provided looks pretty good to me!
Have you checked if this works?

date "+%Y-%d-%m"

Maybe, you want to try using single instead of double quotes:

var_currdate=`date '+%Y-%d-%m'`
echo $var_currdate

Author Comment

by:mruff

ID: 418598722016-10-26

@Hanno
Yes I've checked the script
got an error (have to man date)
errors: date: illegal option --d usage: date [-] [+Field Descriptors]
And I did not fully understand yet, how a correct number of days for a month is picked if the ref_days.txt contains all the monts|days

Expert Comment

by:Hanno P.S.

ID: 418598882016-10-26

Does this work?

date '+%Y-%d-%m'

If not, check you man page for date's format string options

Expert Comment

by:tel2

ID: 418599092016-10-26

Hi mruff,

Good to say up front what flavour of UNIX you're using, especially if it's something like AIX.

I think AIX probably doesn't ship with GNU date.

But to confirm, what output does this give you?:

date -d 'next month'

And this?:

date -v -1m +%Y%m%d

And will you be running this as root, or as another user? (It could make a difference to my approach. I can explain later if required.)

Assisted Solution

by:tel2

tel2 earned 200 total points

ID: 418609642016-10-26

Hi again nruff,

I decided on a different strategy.

Firstly, where you wrote this:
"for all days in previous month (relative to currentDate)"
I assume you mean this:
"for all days in current month (relative to currentDate)"
based on your example. Right?

#!/bin/bash

currentDate=`date +'%Y-%d-%m'`
echo "currentDate = $currentDate"  # Remove this if you don't need it
yyyy=`echo $currentDate | cut -c-4`
mm=`echo $currentDate | cut -c9-`
daysInMonth=`cal | tr '\n' ' ' | awk '{print $NF}'`
day=1
while [ $day -le $daysInMonth ]
do
        currentLoopDate=`printf %d-%02d-%02d $yyyy $day $mm`
        echo $currentLoopDate    # Remove this if you don't need it
        let day=$day+1
done

Here's the output:
currentDate = 2016-27-10
2016-01-10
2016-02-10
2016-03-10
2016-04-10
2016-05-10
2016-06-10
2016-07-10
2016-08-10
2016-09-10
2016-10-10
2016-11-10
2016-12-10
2016-13-10
2016-14-10
2016-15-10
2016-16-10
2016-17-10
2016-18-10
2016-19-10
2016-20-10
2016-21-10
2016-22-10
2016-23-10
2016-24-10
2016-25-10
2016-26-10
2016-27-10
2016-28-10
2016-29-10
2016-30-10
2016-31-10

Is that what you want?

Author Comment

by:mruff

ID: 418621212016-10-27

thank you both!

Author Comment

by:mruff

ID: 418621352016-10-27

@tel2 sorry did not notice your answer until now
I've requested from moderator to select yours as best solution and provide 300 point to you

Expert Comment

by:tel2

ID: 418632342016-10-27

Thanks for that, mruff.

tel2

Author Comment

by:mruff

ID: 418633722016-10-27

@all, I reassigned the points as Abhimuanyu posted first and most likely put most effort in his solution I've assigned him a little more point then to you tel2.
I hope the points assignment is considered fair by all of you
thank you all for your assistance, with all your help I've finished my script