Script to copy files for last month

Question

I am trying to copy files using robocopy. The files I want to copy should be of last month(not one month old) i.e. on if I am running the script in the march month(any day) then it should copy all the files with timestamp of february. I am using below script:

robocopy source/*.prev destination maxage:20180201 minage:20180228

How to use the date as a variable here so I don't have to manually change it every month?

Welcome to StackOverflow! The question you've asked is hard to answer using tools within Bash alone. Can you clarify what operating system you're running, so that if a solution requires additional tools, we can be more specific with the options they'll use? Also, have you tried reviewing the man pages for [`date`](https://man.freebsd.org/date) and [`find`](https://man.freebsd.org/find)? Might also want to consider [`rsync`](https://www.samba.org/ftp/rsync/rsync.html)? — ghoti, Mar 31 '18 at 18:42
Also, can you tell us what "robocopy" you're using? The tag you've included is for a Windows utility. — ghoti, Mar 31 '18 at 21:08
I am making a batch script to copy files with 2different extensions from one windows server to other. Thanks — Yogi, Apr 01 '18 at 19:08
So .. if you need a Windows solution, why is your question tagged `bash`, and `unix`? — ghoti, Apr 01 '18 at 19:32
@ghoti really sorry about that, it mistakenly selected in place of script.,, — Yogi, Apr 01 '18 at 19:49
If you enter `RoboCopy /?` at the Command prompt, you'll note that its syntax, `RoboCopy [[ ...]] []`, seems contradictory to yours, `RoboCopy ` — Compo, Apr 01 '18 at 21:57

Jonathan · Answer 1 · 2018-03-31T18:51:13.907

0

The last date you can calculate like this: date -d "-$(date +%d) days -0 month" +%Y-%m-%d

which today outputs: 2018-02-28

The first date is a simple fixed start day: date -d "-$(date +%d) days -0 month" +%Y-%m-01

Which gaves when executed today: 2018-02-01

Your command could thus be:

robocopy source/*.prev destination \
maxage:$(date -d "-$(date +%d) days -0 month" +%Y%m01) \
minage:$(date -d "-$(date +%d) days -0 month" +%Y%m%d)

Edit: Since this is using GNU date syntax, this might not work on unix as stated by @ghoti

edited Mar 31 '18 at 18:51

answered Mar 31 '18 at 18:33

Jonathan

748
3
20

2

That usage of the `date` command is specific to GNU date; other versions of the command work differently (BSD, macOS, SysV, etc). Since the OP specified "unix", it seems reasonable to me that a portable answer would make sense. Can you add options, or at least qualify your answer as being platform-specific? – ghoti Mar 31 '18 at 18:38

Heiner Westphal · Answer 2 · 2018-03-31T18:45:18.587

0

Having GNU date around you can do it this way:

start=$(date -d "$(date +%Y-%m-1) -1 month" +%Y%m%d)
end=(date -d "$(date +%Y-%m-1) -1 day" +%Y%m%d)
robocopy source/*.prev destination maxage:${start} minage:${end}

See linux - Using date command to get previous, current and next month to see why we need -d. On 31st of a month the -1 month breaks otherwise.

edited Mar 31 '18 at 18:45

answered Mar 31 '18 at 18:41

Heiner Westphal

348
2
4

1

This answer also is not a bash answer. It depends on `date`, which is *not* part of bash, and whose usage varies depending on operating system. The question specifies **unix**, not **Linux**. Please either provide something portable, or qualify your answer as being platform specific. – ghoti Mar 31 '18 at 18:44
Thanks ghoti, you are right. It's GNU date-, not linux-specific. I edited the answer accordingly. – Heiner Westphal Mar 31 '18 at 18:49

score 0 · Answer 3 · answered Apr 01 '18 at 05:56

Let me assume POSIX compliant date is available.
Then how about:

#!/bin/bash

days[1]=31; days[2]=28; days[3]=31; days[4]=30
days[5]=31; days[6]=30; days[7]=31; days[8]=31
days[9]=30; days[10]=31; days[11]=30; days[12]=31

# current year and month
year=$(date +%Y)
month=$(( $(date +%m) + 0 ))

# previous month
if [[ "$month" = "1" ]]; then
    year=$(( $year - 1 ))
    month="12"
else
    month=$(( $month - 1 ))
fi
days=${days[$month]}

# a leap year correction
if [[ "$month" = "2" ]]; then
    days=$(( $days + ($year % 400 == 0 || ($year % 100 != 0 && $year % 4 == 0)) ))
fi

maxage=$( printf "%04d%02d%02d" $year $month 1 )
minage=$( printf "%04d%02d%02d" $year $month $days )

score 0 · Answer 4 · answered Apr 02 '18 at 18:30

@echo off
    setlocal enableextensions disabledelayedexpansion

    rem Initialize variables to hold the previous month information
    set "yyyy=" 
    set "mm="
    set "dd="

    rem The robocopy in the for command will generate an error with the current date
    for /f "tokens=1,2 delims=/" %%a in ('
        robocopy "|" . /njh
    ') do if not defined yyyy (
        rem Calculate previous month from current date
        set /a "mm=(1%%b-90) %% 12 + 1", "yyyy=%%a - !(mm %% 12)" 
        rem Calculate the last day of the previous month
        set /a "dd=30+((mm+mm/8) %% 2)+(-2+!(yyyy%%4)-!(yyyy%%100)+!(yyyy%%400))*!(%%b-3)" 
        rem Increase month and day for proper padding
        set /a "dd+=100", "mm+=100"
    )

    rem Leave only the two right digits in month and day
    set "mm=%mm:~-2%"
    set "dd=%dd:~-2%"

    rem Do the copy operation, selecting files from first to last days in previous month
    robocopy "x:\source" "y:\target" *.prev /maxage:%yyyy%%mm%01 /minage:%yyyy%%mm%%dd%

Script to copy files for last month

4 Answers4