List file using ls to find meet the condition - linux

Question

I am writing a batch program to delete all file in a directory with condition in filename.
In the directory there's a large number of text file (~ hundreds of thousand of files) with filename fixed as "abc" + date
abc_20180820.txt
abc_20180821.txt
abc_20180822.txt
abc_20180823.txt
abc_20180824.txt
The program try to grep all the file, compare the date to a fixed-date, delete it if filename's date < fixed date.
But the problem is it took so long to handle that large amount of file (~1 hour to delete 300k files).
My question: Is there a way to compare the date when running ls command? Not get all file in a list then compare to delete, but list only file already meet the condition then delete. I think that will have better performance.
My code is
TARGET_DATE = "5-12"
DEL_DATE = "20180823"
ls -t | grep "[0-9]\{8\}".txt\$ > ${LIST}
for EACH_FILE in `cat ${LIST}` ;
do
DATE=`echo ${EACH_FILE} | cut -c${TARGET_DATE }`
COMPARE=`expr "${DATE}" \< "${DEL_DATE}"`
if [ $COMPARE -eq 1 ] ;
then
rm -f ${EACH_FILE}
fi
done
Found some similar problem but I dont know how to get it done
List file using ls with a condition and process/grep files that only whitespaces

Answer 1

Here is a refactoring which gets rid of the pesky ls. Looping over a large directory is still going to be somewhat slow.
# Use lowercase for private variables
# to avoid clobbering a reserved system variable
# You can't have spaces around the equals sign
del_date="20180823"
# No need for ls here
# No need for a temporary file
for filename in *[0-9][0-9][0-9][0-9][0-9][0-9][0-9][0-9].txt
do
# Avoid external process; use the shell's parameter substitution
date=${filename%.txt}
# This could fail if the file name contains literal shell metacharacters!
date=${date#${date%?????????}}
# Avoid expr
if [ "$date" -lt "$del_date" ]; then
# Just print the file name, null-terminated for xargs
printf '%s\0' "$filename"
fi
done |
# For efficiency, do batch delete
xargs -r0 rm
The wildcard expansion will still take a fair amount of time because the shell will sort the list of filenames. A better solution is probably to refactor this into a find command which avoids the sorting.
find . -maxdepth 1 -type f \( \
-name '*1[89][0-9][0-9][0-9][0-9][0-9][0-9].txt' \
-o -name '*201[0-7][0-9][0-9][0-9][0-9].txt' \
-o -name '*20180[1-7][0-9][0-9].txt ' \
-o -name '*201808[01][0-9].txt' \
-o -name '*2018082[0-2].txt' \
\) -delete

Answer 2

You could do something like:
rm 201[0-7]*.txt # remove all files from 2010-2017
rm 20180[1-4]*.txt # remove all files from Jan-Apr 2018
# And so on
...
to remove a large number of files. Then your code would run faster.

Answer 3

Yes it takes a lot of time if you have so many files in one folder.
It is bad idea to keep so many files in one folder. Even simple ls or find will be killing storage. And if you have some scripts which iterate over your files, you are for sure killing storage.
So after you wait for one hour to clean it. Take time and make better folders structure. It is good idea to sort files according to years/month/days ... possibly hours
e.g.
somefolder/2018/08/24/...files here
Then you can easily delete, move compress ... whole month or year.

Answer 4

I found a solution in this thread.
https://unix.stackexchange.com/questions/199554/get-files-with-a-name-containing-a-date-value-less-than-or-equal-to-a-given-inpu
The awk command is so powerful, only take me ~1 minute to deal with hundreds of thousand of files (1/10 compare to the loop).
ls | awk -v date="$DEL_DATE" '$0 <= date' | xargs rm -vrf
I can even count, copy, move with that command with the fastest answer I've ever seen.
COUNT="$(ls | awk -v date="${DEL_DATE}" '$0 <= target' | xargs rm -vrf | wc -l)"