How to get files do not match a pattern

Question

Here is my script

data_dir="/home/data"
shopt extglob
files=!($data_dir/*08142014*)
echo ${files[@]}

for file in $files[@]
do 
  #blabla
done

/home/data contains multiple files with different date info within file name, thus I should be able to get a list of files that does not contain "08142014".

But kept get syntax error. It seems files is just "!(/home/data/08202014)", while I want a list of file names.

Did I miss anything? Thanks

Not enough information. Is there only one file matching that pattern, or should there be more? Does `ls` show you the same? — Jongware, Aug 20 '14 at 18:23
http://stackoverflow.com/questions/15253702/regex-find-files-that-dont-match — laughing_man, Aug 20 '14 at 18:25

score 4 · Accepted Answer · answered Aug 20 '14 at 18:25

You can use:

data_dir="/home/data"
shopt -s extglob
files=($data_dir/!(*08142014*))

for file in "${files[@]}"
do 
  echo "$file"
done

To set extglob you need to use shopt -s extglob
To set array your syntax isn't right
Check how array is correctly iterated

score 1 · Answer 2 · answered Aug 20 '14 at 18:30

1

You can use ->

files=`ls $data_dir | grep -v 08142014`

answered Aug 20 '14 at 18:30

dganesh2002

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How to get files do not match a pattern

2 Answers2