I'm making a bash script that crawls through a directory and outputs all files of a certain type into a text file. I've got that working, it just also writes out a bunch of output to console I don't want (the names of the files)
Here's the relevant code so far, tmpFile is the file I'm writing to:
for DIR in `find . -type d` # Find problem directories
do
for FILE in `ls "$DIR"` # Loop through problems in directory
do
if [[ `echo ${FILE} | grep -e prob[0-9]*_` ]]; then
`echo ${FILE} >> ${tmpFile}`
fi
done
done
The files I'm putting into the text file are in the format described by the regex prob[0-9]*_ (something like prob12345_01)
Where I pipe the output from echo ${FILE} into grep, it still outputs to stdout, something I want to avoid. I think it's a simple fix, but it's escaping me.
All this can be done in one single find command. Consider this:
find . -type f -name "prob[0-9]*_*" -exec echo {} >> ${tmpFile} \;
EDIT:
Even simpler: (Thanks to @GlennJackman)
find . -type f -name "prob[0-9]*_*" >> $tmpFile
To answer your specific question, you can pass -q to grep for silent output.
if echo "hello" | grep -q el; then
echo "found"
fi
But since you're already using find, this can be done with just one command:
find . -regex ".*prob[0-9]*_.*" -printf '%f\n' >> ${tmpFile}
find's regex is a match on the whole path, which is why the leading and trailing .* is needed.
The -printf '%f\n' prints the file name without directory, to match what your script is doing.
what you want to do is, read the output of the find command,
for every entry find returned, you want to get all (*) the files under that location
and then you want to check whether that filename matches the pattern you want
if it matches then add it to the tmpfile
while read -r dir; do
for file in "$dir"/*; do # will not match hidden files, unless dotglob is set
if [[ "$file" =~ prob[0-9]*_ ]]; then
echo "$file" >> "$tmpfile"
fi
done < <(find . -type d)
however find can do that alone
anubhava got me there ;)
so look his answer on how that's done
