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I have a web application that access a remote storage running Linux to get some files, the problem is that the remote storage have currently 3 million files , so accessing the normal way is a bit tricky.

So I needed to work on a script that is going to make it a little bit more easy to use , this script is going to reorganize the files into multiple folders depending on their creation date and specially their names,i made the script and it worked just fine, it intended to do what it meant to do, but it was too slow, 12 hours to perform the work completely (12:13:48 to be precise).

I think that the slowness is coming from the multiple cut and rev calls I make.

example :

I get the file names with an ls command that I loop into with for, and for each file I get the parent directory and, depending on the parent directory, I can get the correct year:

 case "$parent" in ( "Type1" ) year=$(echo "$fichier" | rev | cut -d '_' -f 2 | rev );; ( "Type2" ) year=$(echo "$fichier" | rev | cut -d '_' -f 2 | rev);; ( "Type3" ) year=$(echo "$fichier" | rev | cut -d '_' -f 1 | rev | cut -c 1-4);; ( "Type4" ) year=$(echo "$fichier" | rev | cut -d '_' -f 1 | rev | cut -c 1-4);; ( "Type5" ) year=$(echo "$fichier" | rev | cut -d '_' -f 1 | rev | cut -c 1-4);; esac

for type1 of files :

the file==>MY_AMAZING_FILE_THAT_IMADEIN_YEAR_TY.pdf

I need to get the year so I perform a reverse cut:

year=$(echo "$file" | rev | cut -d '_' -f 2 | rev );;

for type2 of files :

the file==>MY_AMAZING_FILE_THAT_IMADE_IN_YEAR_WITH_TY.pdf

etc...

and then I can mv the file freely : mv $file /some/path/destination/$year/$parent

and yet this is the simplest example, there are some files that are much more complex, so to get 1 information I need to do 4 operations, 1 echo , 2rev and 1echo.

While the script is running I am getting speeds of 50 files/sec to 100 files\s , I got this info by doing a wc-l output.txt of the script.

Is there anything I can do to make it faster? or another way to cut the files name? I know that I can use sed or awk or string operations but I did not really understand how.

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    Is there anything i can do to make it faster ? - Sure, but you don't give enough information for a more useful answer. Commented Oct 11, 2017 at 13:24
  • @SatōKatsura what information i need to add ? i'll give them gladly Commented Oct 11, 2017 at 13:27
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    What do the "much more complex" filenames look like? What does the directory structure look like? How does $file get its value? What does your current code look like? Please edit your question. Commented Oct 11, 2017 at 13:41
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    You're doing way too many operations per file, and also getting the list of files with ls is not a good idea. You can probably do everything with a single find and Perl rename. But again, you don't give enough information for a full answer. Good luck. Commented Oct 11, 2017 at 14:13
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    @Kingofkech You can do it with find, no problem. Are the years the only four-digit number in the filenames? Commented Oct 11, 2017 at 14:19

2 Answers 2

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To get the YEAR portion of the filename MY_AMAZING_FILE_THAT_IMADEIN_YEAR_TY.pdf without using external utilities:

name='MY_AMAZING_FILE_THAT_IMADEIN_YEAR_TY.pdf' year=${name%_*} # remove everything after the last '_' year=${year##*_} # remove everything up to the last (remaining) '_'

After update to the question:

Moving PDF files from under topdir to a directory /some/path/destination/<year>/<parent> where <year> is the year found in the filename of the file, and <parent> is the basename of the original directory that the file was found in:

find topdir -type f -name '*.pdf' -exec bash ./movefiles.sh {} +

movefiles.sh is a shell script in the current directory:

#!/bin/bash destdir='/some/path/destination' for name; do # get basename of directory parent=${name%/*} parent=${parent##*/} # get the year from the filename: # - Pattern: _YYYY_ (in the middle somewhere) # - Pattern: _YYYYMMDD.pdf (at end) if [[ "$name" =~ _([0-9]{4})_ ]] || [[ "$name" =~ _([0-9]{4})[0-9]{4}\.pdf$ ]]; then year="${BASH_REMATCH[1]}" else printf 'No year in filename "%s"\n' "$name" >&2 continue fi # make destination directory if needed # (remove echo when you have tested this at least once) if [ ! -d "$destdir/$year/$parent" ]; then echo mkdir -p "$destdir/$year/$parent" fi # move file # (remove echo when you have tested this at least once) echo mv "$name" "$destdir/$year/$parent" done
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  • Thank you sir for your answer but are those instructions going to make my code faster then the state he is in right now ? (compared to the cut and double rev) Commented Oct 11, 2017 at 13:33
  • @Kingofkech Yes, but it also depends on the context of your code. How does $file get its value, for example? Update your question to give a bit more context to what it is you're doing. Commented Oct 11, 2017 at 13:34
  • i've edited the post now Commented Oct 11, 2017 at 14:05
  • @Kingofkech See updated answer. Commented Oct 11, 2017 at 14:35
  • Thank you so much i think that this is much more optimal , i did not really thought of using find to call a script , that will be more optimal. Commented Oct 11, 2017 at 14:46
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You may apply sed approach to extract year value:

year=$(sed -E 's/.*_([0-9]{4})_TY\.pdf/\1/' <<<"$file")
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