Find directory with regular expression: white spaces, command find

Question

I had to write a bash script that have to find the directories in the current directory and that directories must have a name that start with a letter of the alphabet [A-z]. For shell I wrote:

find . -maxdepth 1 -name '[[:alpha:]]*' -type d

and it's ok. But in the script I wrote:

#! /bin/bash files=$(find . -maxdepth 1 -name '[[:alpha:]]*' -type d) for FILE in $files; do echo 'you are in', $FILE; done;

But, when it finds a directory with whitespace (ex. ./New Directory) the output is

./New Directory

as it were 2 different directories. Why? How can i resolve this problem?

By the time files is set, it is too late for quoting $FILE to make a difference. mywiki.wooledge.org/BashFAQ/001 — chepner
– chepner, Commented Dec 24, 2015 at 16:58

repzero · Accepted Answer · 2015-12-24 17:24:47Z

1

find -maxdepth 1 -type d -regextype posix-awk -regex ".*/[A-Z].*" -exec echo "you are in" {} \;

answered Dec 24, 2015 at 17:24

repzero

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sjsam · Accepted Answer · 2015-12-24 18:01:12Z

1

This may work for you :

find . -maxdepth 1 -name '[[:alpha:]]*' -type d | sed 's/^/You are in /'

edited Dec 24, 2015 at 18:01

answered Dec 24, 2015 at 17:05

sjsam

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