I try to use find to match several alternative file patterns represented by certain numbers in the middle, but it returns an empty list. My actual pattern has a fixed beginning and variable numbers in the middle.
Reproducible example. Create a list of files
touch a10a a24b b12c a45d
Select a10a and a24b from the list using the following regex resulting in empty output
find . -regex '.*/a(10|45).*'
I expect that the issue should be easy to solve but I could not find a solution and could not figure out it. What did I miss?
system: ubuntu 16.04
The idea is right, but you need to type of the regex to use for find. Since you have alternate operator | here, you need to enable ERE (Extended Regular Expressions) support which you can do as below. The -regextype allows you to specify the regex flavor that you need for the requirement. Also the / part is optional if you have enabled a greedy match .*
find . -type f -regextype posix-extended -regex '.*/a(10|45).*'
From my version of GNU findutils, you could see from the man page
-regextype type
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.
Try specifying -regextype awk instead:
find . -regextype awk -regex '.*/a(10|45).*'
It seems you are using the wrong type of bracket. They should be square, not round.
The correct command should be:
find . -regex '.*/a[10|45].*'
Hope this helps!
Related
Am trying to do a simple file-name match with this below regex which I tested to be working from this page for a sample file-name ABC_YYYYMMDDHHMMSS.sha1
ABC_20[0-9]{2}(0[1-9]|1[0-2])([0-2][0-9]|3[0-1])([0-2][0-3])([0-5][0-9])([0-5][0-9])\.sha1
When I couple this in the -regex flag of find like
find . -type f -regex "ABC_20[0-9]{2}(0[1-9]|1[0-2])([0-2][0-9]|3[0-1])([0-2][0-3])([0-5][0-9])([0-5][0-9])\.sha1"
the command is not identifying the file present in the path (e.g ABC_20161231225950.sha1). Am aware of many existing regex-types from this page, but I realized my type is posix-extended and tried as below,
find . -type f -regextype posix-extended -regex 'ABC_20[0-9]{2}(0[1-9]|1[0-2])([0-2][0-9]|3[0-1])([0-2][0-3])([0-5][0-9])([0-5][0-9])\.sha1'
and still no result. I searched around some similar questions of this type, but they were involving giving the wrong regex leading to files not being found. In my case, though the regex is found to be proper. I need to know what am I missing here. Also possibly how to debug non matching issues when using -regex in find.
Note:- I could do some optimizations over the capturing groups in the regex, but that is not in the scope of the current question.
add .* at the start of your regex because you will always get something like ./ at start of path
find . -type f -regextype posix-extended -regex '.*ABC_20[0-9]{2}(0[1-9]|1[0-2])([0-2][0-9]|3[0-1])([0-2][0-3])([0-5][0-9])([0-5][0-9])\.sha1'
Let's say I have a file which has the path ./bar2.txt
and content of bar2.txt is
./bar2.txt
output of
grep "\bbar2\b" *
is
bar2.txt:./bar2.txt
as expected, wheres
find . -regextype posix-extended -regex "\bbar2\b"
doesn't find anything.
I know I should change the regex to
".*/bar2.*"
since find looks for full path. So, does this mean that find ignores \b which specifies the word boundary?
Thanks in advance.
grep and find will use different regular expression engines. Notably, POSIX regular expressions (which find uses) don't include "\b" as a word boundary, so it is the same as "b". On OSX, for instance, these have the same result:
find . -regex '.*bar2.txt' -print
and
find . -regex '.*\bar2.txt' -print
Double check the manpage to make sure that -regex is doing what you think. For my find the regular expression must match the entire filename, e.g. this doesn't find any files:
find . -regex 'bar' -print
but this one does:
find . -regex '.*bar.*' -print
However, what I am curious about is why it ignores \b which specifies word
boundary.
From my understanding \b is part of Perl regular expressions and would not be available to find, as find does not support the Perl regextype as grep does.
For example, I want to find a file ends with numbers+.bed, it works if used this:
find . -regex ".*/[0-9]+\.bed"
As I know, find uses emacs regex style by default, so I found this page: http://www.gnu.org/software/emacs/manual/html_node/elisp/Char-Classes.html#Char-Classes and then try to use [[:digit:]] to replace [0-9], but none of these commands works for me.
find . -regex ".*/[:digit:]\.bed"
Or
find . -regex ".*/[[:digit:]]\.bed"
Does anybody have any ideas about where I was wrong in the command?
Thanks!
You need to use a different regex type. For example, try:
find . -regextype posix-extended -regex ".*/[[:digit:]]+\.bed"
I suspect that the find documentation is out of date: the regexp syntax accepted by Emacs has changed over the years, but find's apparently hasn't changed accordingly.
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?
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
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.
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
When using find command in linux, one can add a -regex flag that uses emacs regualr expressions to match.
I want find to look for all files except .jar files and .ear files. what would be the regular expression in this case?
Thanks
You don't need a regex here. You can use find with the -name and -not options:
find . -not -name "*.jar" -not -name "*.ear"
A more concise (but less readable) version of the above is:
find . ! \( -name "*.jar" -o -name "*.ear" \)
EDIT: New approach:
Since POSIX regexes don't support lookaround, you need to negate the match result:
find . -not -regex ".*\.[je]ar"
The previously posted answer uses lookbehind and thus won't work here, but here it is for completeness' sake:
.*(?<!\.[je]ar)$
find . -regextype posix-extended -not -regex ".*\\.(jar|ear)"
This will do the job, and I personally find it a bit clearer than some of the other solutions. Unfortunately the -regextype is required (cluttering up an otherwise simple command) to make the capturing group work.
Using a regular expression in this case sounds like an overkill (you could just check if the name ends with something). I'm not sure about emacs syntax, but something like this should be generic enough to work:
\.(?!((jar$)|(ear$)))
i.e. find a dot (.) not followed by ending ($) "jar" or (|) "ear".