repetition in GNU find regexp

Question

I am trying to find all the files whose name contains exactly 14 digits (I'm trying to match a timestamp in the filename). I'm not sure how to get the GNU find regexp syntax for repetitions right.

I've tried find -regex ".*[0-9]{14} and find -regex ".*[0-9]\{14\}, neither of these turns up any results. Can you help me with the syntax?

The top answer here by Susam Pal worked for me: http://stackoverflow.com/questions/6844785/how-to-use-regex-with-find-command — , Jun 09 '12 at 07:18

score 3 · Accepted Answer · answered Aug 30 '11 at 15:37

3

remember, GNU find's -regex matches a whole path. Anyway, you can use a combination of find and grep to do the task, eg to find exactly 14 digits with no other characters

find . -type f -printf "%f\n" | grep -E "\b[0-9]{14}\b"

modify to suit your needs

answered Aug 30 '11 at 15:37

ghostdog74

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score 2 · Answer 2 · answered Aug 30 '11 at 15:03

Try changing the -regextype parameter to find.

Changes the regular expression syntax understood by -regex and -iregex tests which occur later on the command line. Currently-implemented types are emacs (this is the default), posix-awk, posix-basic, posix-egrep and posix-extended.

score 0 · Answer 3 · answered Aug 30 '11 at 15:11

0

Strange, I just gave it a try and I could not get this work. Here's a workaround anyway (matching 2 consecutive numbers):

$ls
a123.txt  a1b2c3.txt  a45.txt  b123.txt
$find -regex '.*[^0-9][0-9][0-9][^0-9].*'
./a45.txt

answered Aug 30 '11 at 15:11

bpgergo

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repetition in GNU find regexp

3 Answers3