Bash – find files without extension

bashfindshellwildcards

Let say I have:

foo.txt
bar
baz.ooo

If I use ls -1 !(*.*) then I'll get only bar on the output. Great, now I wish to have same results with find – some find -regex that will do the job.

NOTE:

find -name !(*.*) is not the answer as !(*.*) in it is still Bash's glob which I can NOT use.

Best Answer

you could use: find . -type f ! -name "*.*" the ! negates the following expression, here a filename that contains a '.'

you can also use the -maxdepth option to reduce the search depth.

Related Solutions

Exclude directory in find

find . -name 'foo-exclude-me' -prune -o -name 'foo*' -print

With no -print, the implicit default action applies to every match, even pruned ones. The explicit -print applies only under the specified conditions, which are -name 'foo*' only in the else branch of -name 'foo-exclude-me'.

Generally speaking, use an explicit -print whenever you're doing something more complex than a conjunction of predicates.

Your second attempt with ! -path './foo-exclude-me/*' didn't work because ./foo-exclude-me doesn't match ./foo-exclude-me/* (no trailing /). Adding ! -path ./foo-exclude-me would work.

Shell – Why file glob expansion is not suppressed

With the first invocation of your script

$ bash my.sh *

the file name globbing is performed by the shell before calling the script.

With the second invocation of your script

$ bash my.sh "*"

the file name globbing is performed in the script when running echo $@.

To avoid this, double quote $@ on that line:

echo "$@"

Further reading: Security implications of forgetting to quote a variable in bash/POSIX shells

Best Answer

Related Solutions

Exclude directory in find

Shell – Why file glob expansion is not suppressed

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