Bash – Defining the start and end point of a loop in bash

bash

I have a directory with folders sequentially numbered 1-1000, however I want to perform an action for some of the folders in this directory. i.e folders 1-50. How do I define the start and end point of the loop?

So far I have a script along these lines:

a=1
b=1

for i in ~/PATH/*/ ;do

(cd $i/ && action)

echo "completed"
a=`expr $a + $b`

echo "Next Folder"
done

I can define the start of the loop by changing a= but I am not sure how to define the end to the loop? As it stands, it will keep going through all the folders in that directory. Thanks 🙂

Best Answer

#!/bin/bash
a=1
b=1

cd ~/PATH/
for i in $(ls -v) ;do
  (cd $i/ && action) && echo "completed" || echo "error"
  a=$((a+b)); [ "$a" -gt 50 ] && break 
  echo "Next Folder"
done

Related Solutions

How to Execute Command on a Range of Directories in Bash

There's no reason to do either of the following mistakes for this situation:

Use non-portable GNU extensions (such as xargs -0)
Parse the filenames as a stream of text only, pretending that they can't contain newlines.

You can handle this portably without much difficulty at all:

set -- */
while [ "$#" -gt 0 ]; do
  case "$#" in
    1)
      my-command "$1"
      shift 1
      ;;
    2)
      my-command "$1" "$2"
      shift 2
      ;;
    *)
      my-command "$1" "$2" "$3"
      shift 3
      ;;
  esac
done

Really this is more verbose than necessary, but it is readable this way.

For a different type of partitioning, where you want to split all directories into exactly three different methods of handling (but don't need to handle them at the same time), you could do it like so:

x=1
for d in *; do
  [ ! -d "$d" ] && continue
  case "$x" in
    1)
      # do stuff with "$d"
      ;;
    2)
      # do stuff with "$d"
      ;;
    3)
      # do stuff with "$d"
      ;;
  esac
  x=$(($x+1))
  [ "$x" -eq 4 ] && x=1
done

Bash – a faster way to extract the year from file names to move them to year based directories than the current approach of using `cut` and `rev`

To get the YEAR portion of the filename MY_AMAZING_FILE_THAT_IMADEIN_YEAR_TY.pdf without using external utilities:

name='MY_AMAZING_FILE_THAT_IMADEIN_YEAR_TY.pdf'

year=${name%_*}    # remove everything after the last '_'
year=${year##*_}   # remove everything up to the last (remaining) '_'

After update to the question:

Moving PDF files from under topdir to a directory /some/path/destination/<year>/<parent> where <year> is the year found in the filename of the file, and <parent> is the basename of the original directory that the file was found in:

find topdir -type f -name '*.pdf' -exec bash ./movefiles.sh {} +

movefiles.sh is a shell script in the current directory:

#!/bin/bash

destdir='/some/path/destination'

for name; do
    # get basename of directory
    parent=${name%/*}
    parent=${parent##*/}

    # get the year from the filename:
    #  - Pattern:  _YYYY_         (in the middle somewhere)
    #  - Pattern:  _YYYYMMDD.pdf  (at end)
    if [[ "$name" =~ _([0-9]{4})_ ]] ||
       [[ "$name" =~ _([0-9]{4})[0-9]{4}\.pdf$ ]]; then
        year="${BASH_REMATCH[1]}"
    else
        printf 'No year in filename "%s"\n' "$name" >&2
        continue
    fi

    # make destination directory if needed
    # (remove echo when you have tested this at least once)
    if [ ! -d "$destdir/$year/$parent" ]; then
        echo mkdir -p "$destdir/$year/$parent"
    fi

    # move file
    # (remove echo when you have tested this at least once)
    echo mv "$name" "$destdir/$year/$parent"
done

Best Answer

Related Solutions

How to Execute Command on a Range of Directories in Bash

Bash – a faster way to extract the year from file names to move them to year based directories than the current approach of using `cut` and `rev`

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