Find, zip and remove in a single line with a condition

Question

I'm trying to simplifly a script into a single line which can find recursively files created yesterday, zip, move and remove depending of a condition.

I have this structure:

../main -> child1 (content: file01.csv, file02.xlsx, file03_20201028.xls, file04_20201028.xls)
      |____ child2 (content: file05.xls, file06.xlsx, file07_20201028)
              |___  child3 (file08.xls, file09.xlsx, file10_20201028.xls, file11_20201028.xls)

I have made this:

find $ORIGINPATH -type f -mtime -1 | zip -@ $ORIGINPATH/Backup_$YESTERDAY.zip
find $ORIGINPATH -type f -iname *$YESTERDAY* -exec rm -f {} \;
mv $ORIGINPATH/Backup_$YESTERDAY.zip $DESTINATIONPATH

I need to delete the files with date in their names and keep the rest.

Thanks in advance!

Answer 1

For a one-liner, this will work:

find $ORIGIN -type f -mtime -1 -exec zip -u ${DEST}/backup_${YESTERDAY}.zip '{}' \; -delete

You will receive an error from zip on the first file, as -u usually updates existing archives (but creates one if not present and returns a warning).

Be aware that -mtime -1 does not care about the date but about the last 24 hours! So if you run it at 23:59h you will remove today's files! If you want to search for a date string, use the following option:

Say our string is defined as YESTERDAY=20201028

find $ORIGIN -type f -name \*${YESTERDAY}\*

Note the missing hard quotes and the need for escaping the asterisks.