[SOLVED] Bash script to move files from date

cbing · 04-17-2012, 04:52 AM

Hello.

I'm trying to make a bash script to move multiple files to different folders depending on the date.
The files structure is YYYY-MM-DD-name.
I want to move the files based on the year and month, but not the system, creation or modification date, I want the date of the filename. If the destination folders do not exist, the script will create it.

I've tried this script from David the H. (Thank's David

), and I've added the variable DATE=$(date +%Y) to the path of mkdir and mv.
Now I have the structure I want, but based on the current year.
How I can do to get the year of the filename?

Thanks.

KenJackson · 04-17-2012, 05:52 AM

Code:

year=${file:0:4}

cbing · 04-17-2012, 06:29 AM

Quote:

Originally Posted by KenJackson

Code:

year=${file:0:4}

Thanks KenJackson.

This is what I was making, without success.

This is my script (copied from David_the_H.).

Code:

#!/bin/bash

DIR=/home/data/
target=$DIR
cd "$DIR"
year=${file:0:4}

for file in *; do

     case "${file:5:2}" in
          01)  mkdir -vp $target/$year/01_Jan
               mv -v -t $target/$year/01_Jan $file
               ;;
          02)  mkdir -vp $target/$year/02_Feb
               mv -v -t $target/$year/02_Feb $file
               ;;
          03)  mkdir -vp $target/$year/03_Mar
               mv -v -t $target/$year/03_Mar $file
               ;;
          04)  mkdir -vp $target/$year/04_Apr
               mv -v -t $target/$year/04_Apr $file
               ;;
          05)  mkdir -vp $target/$year/05_May
               mv -v -t $target/$year/05_May $file
               ;;
          06)  mkdir -vp $target/$year/06_Jun
               mv -v -t $target/$year/06_Jun $file
               ;;
          07)  mkdir -vp $target/$year/07_Jul
               mv -v -t $target/$year/07_Jul $file
               ;;
          08)  mkdir -vp $target/$year/08_Aug
               mv -v -t $target/$year/08_Aug $file
               ;;
          09)  mkdir -vp $target/$year/09_Sep
               mv -v -t $target/$year/09_Sep $file
               ;;
          10)  mkdir -vp $target/$year/10_Oct
               mv -v -t $target/$year/10_Oct $file
               ;;
          11)  mkdir -vp $target/$year/11_Nov
               mv -v -t $target/$year/11_Nov $file
               ;;
          12)  mkdir -vp $target/$year/12_Dec
               mv -v -t $target/$year/12_Dec $file
               ;;
          
        esac
  
done

exit 0

With this script, I get the files in different folders sorted by month (01_Jan, 02_Feb...), but not under its year.

Thanks.

cbing · 04-17-2012, 06:33 AM

My fault.

This

Code:

year=${file:0:4}

should go after

Code:

for file in *; do

Thanks.

David the H. · 04-17-2012, 07:24 AM

Glad to see my old code helped out.

Incidentally, with the mad coding skillz I've developed since that last post

, I'd now recommend adding a function to reduce the complexity of it a bit.

Also, variable expansions should always be quoted, for safety and consistency.

Code:

#!/bin/bash

DIR=/home/data/
target=$DIR
cd "$DIR"

movefile(){

     # $1 is the target directory to create, $2 is the file to move into it
     mkdir -vp "$1"
     mv -v -t "$1" "$2"

}

for file in *; do

     year=${file:0:4}

     case "${file:5:2}" in
          01)  movefile "$target/$year/01_Jan" "$file" ;;
          02)  movefile "$target/$year/02_Feb" "$file" ;;
          03)  movefile "$target/$year/03_Mar" "$file" ;;
          04)  movefile "$target/$year/04_Apr" "$file" ;;
          05)  movefile "$target/$year/05_May" "$file" ;;
          06)  movefile "$target/$year/06_Jun" "$file" ;;
          07)  movefile "$target/$year/07_Jul" "$file" ;;
          08)  movefile "$target/$year/08_Aug" "$file" ;;
          09)  movefile "$target/$year/09_Sep" "$file" ;;
          10)  movefile "$target/$year/10_Oct" "$file" ;;
          11)  movefile "$target/$year/11_Nov" "$file" ;;
          12)  movefile "$target/$year/12_Dec" "$file" ;;
     esac

done

exit 0

FYI, There are quite a few techniques you can use for extracting substrings from variables like this. Check out these links:

parameter substitution
string manipulation

cbing · 04-17-2012, 08:37 AM

Thanks again David the H..

Your scripts have helped me a lot.

And of course, thanks for your links. (Give a man a fish and you feed him for a day; teach a man to fish and you feed him for a lifetime).

David the H. · 04-17-2012, 10:50 AM

I thought you might be interested in seeing an even more compact version I thought up. It uses an array and a small bit of mathematic manipulation to eliminate the need for a case statement.

Code:

#!/bin/bash

DIR=/home/data
target=$DIR
cd "$DIR"

month=( 01_Jan 02_Feb 03_Mar 04_Apr 05_May 06_Jun
	07_Jul 08_Aug 09_Sep 10_Oct 11_Nov 12_Dec )

for file in *; do

	dname="$target/${file:0:4}/${month[10#${file:5:2} -1]}"
	mkdir -vp "$dname"
	mv -vt "$dname" "$file"

done

exit 0

cbing · 04-17-2012, 11:13 AM

Quote:

Originally Posted by David the H.

I thought you might be interested in seeing an even more compact version I thought up. It uses an array and a small bit of mathematic manipulation to eliminate the need for a case statement.

Code:

#!/bin/bash

DIR=/home/data
target=$DIR
cd "$DIR"

month=( 01_Jan 02_Feb 03_Mar 04_Apr 05_May 06_Jun
	07_Jul 08_Aug 09_Sep 10_Oct 11_Nov 12_Dec )

for file in *; do

	dname="$target/${file:0:4}/${month[10#${file:5:2} -1]}"
	mkdir -vp "$dname"
	mv -vt "$dname" "$file"

done

exit 0

Great!!

Very good examples. I like to learn different ways to reach the same place.

Thanks.

David the H. · 04-19-2012, 07:31 AM

And just for fun:

Code:

#!/bin/bash

DIR=/home/data
target=$DIR
cd "$DIR"

for file in *; do

	dname="$( date -d "${file%-*}" "+$target/%Y/%m_%b" )"
	mkdir -vp "${dname%/*}"
	mv -vt "$dname" "$file"

done

This one is actually quite a bit slower than the previous versions, however, due to the dependency on the external date program. But hey, it's short!

cbing · 04-20-2012, 05:45 AM

Thanks David.

Awesome examples.