Extract unix compatible filepath with space from command output

Question

the command :

du -sh $HOME/* | grep '[0-9]G\>' | sort -k 1rn | head -1

is giving me :

41G     /Users/user/Big folder

How can i extract the path to reuse it in my script ?

du -sh $HOME/* | grep '[0-9]G\>' | sort -k 1rn | head -1 | awk '{ print $2 }'

Will output

/Users/user/Big

Which is unusable, where i am looking for a result like :

/Users/user/Big\ folder

Answer 1

When doing automated processing, don't make your life more complicated by using numbers in different units. Don't pass -h to du. Then you can use simple numerical sorting on the output, and your script will still work if you have more than 1 TB in one of the directories.

du -s ~/* | sort -k 1n | tail -n1

This always prints out a line, even if the largest subdirectory contains less than 1 GB. If that's not what you want, you can replace the conditional “is the output empty” by “is the number less than the threshold”.

To extract the directory name, take the output and remove the part up to the first tab.

largest_directory=$(du -s ~/* | sort -k 1n | tail -n1)
largest_directory_size_kB=${largest_directory%%$(printf '\t')*}  # if you need the size
largest_directory=${largest_directory#*$(printf '\t')}

Answer 2

POSIXLY:

du -s "$HOME"/* | sort -rnk1 | LC_ALL=C sed -e "s,[^/]*\(/.*\),'\1',;q"

Without -h, du result included plain number for size, so we can easily sort it without extra grep.

Using sed to perform deletion at the first line only then quitting, we save a head invocation.

Answer 3

Another way to extract the path you want is via read command after you're done doing the du stuff and sort-ed them and then pass on to the while/read

du.. sort.. | while read x d; do echo "$d"; done