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
– Jongware, Commented Aug 20, 2014 at 18:23

anubhava · Accepted Answer · 2014-08-20 18:25:06Z

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

dganesh2002 · Accepted Answer · 2014-08-20 18:30:07Z

1

You can use ->

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

answered Aug 20, 2014 at 18:30

dganesh2002

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