bash script to list the contents of a directory, separated by commas

I need a bash script that lists files recursively in a directory in the following way:

filename,size (in kb),numeric value (in filename)

e.g.:

/directory/1/file-100.txt,50,100
/directory/1/file-200.txt,45,200
/directory/2/file-100.txt,20,100
/directory/2/file-500.txt,100,500

Al开发者_运维知识库so, I need it to ignore directories that start with ".svn"

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Give this a try:

find -name ".svn" -prune -o -type f -printf "%p,%k,\n" | sed 's/\([^-]\+-\([0-9]\+\)\..*\)/\1\2/'

Only two utilities used (GNU versions probably required).

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maybe not the best way but should work:

find /path -type f | grep -v '.svn' | xargs du -k |\
  awk '{print $2","$1}' | sed 's/-\([0-9]*\)\(\..*\)/-\1\2,\1/'

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Features:

• Avoid possible expensive traverse down .svn-dirs.
• Have find report size instead of 'du'
• Make awk do some more magic, no need for sed.

Suggestion on solution:

$ find . -path '*/.svn' -prune -o -type f -printf '%p %k\n' |\
  awk -v OFS=, '{ print $1, $2, gensub(/[^0-9]*([0-9]*)/, "\\1", "g", $1) }'

find-options:

• -path ... Match .svn directories
• -prune If match (-path) is a directory, don't descend.
• -o ... Or ...
• -type f Match files only.
• -printf ... Print a custom line containing path (%p) and size in 1K-blocks (%k).

awk:

• OFS= What separator to use for fields when printing, instead of whitespace.
• $1 Filepath
• $2 Filesize
• gensub(...) Sorry, see http://www.gnu.org/manual/gawk/html_node/String-Functions.html

Edit: Updated per Dennis suggestion. Can't believe I missed %k actually, I blame 3AM.