How to find files bigger then some size, and sort them by last modification date?

Question

I need to write script which write down each files in selected catalog, which are bigger then some size. Also I need to sort them by size, name and last modification date.

So I have made the first two cases:

Sort by size

RESULTS=`find $CATALOG -size +$SIZE | sort -n -r | sed 's_.*/__'`

Sort by name

RESULTS=`find $CATALOG -size +$SIZE | sed 's_.*/__' | sort -n `

But I have no idea how to sort results by last modification date.

Any help would be appreciated.

Answer 1

One of the best approaches, provided you don't have too many files, is to use ls to do the sorting itself.

Sort by name and print one file per line:

find $CATALOG -size +$SIZE -exec ls -1 {} +

Sort by size and print one file per line:

find $CATALOG -size +$SIZE -exec ls -S1 {} +

Sort by modification time and print one file per line:

find $CATALOG -size +$SIZE -exec ls -t1 {} +

You can also play with the ls switches: Sort by modification time (small first) with long listing format, with human-readable sizes:

find $CATALOG -size +$SIZE -exec ls -hlrt {} +

Oh, you might want to only find the files (and ignore the directories):

find $CATALOG -size +$SIZE -type f -exec ls -hlrt {} +

Finally, some remarks: Avoid capitalized variable names in bash (it's considered bad practice) and avoid back ticks, use $(...) instead. E.g.,

results=$(find "$catalog" -size +$size -type f -exec ls -1rt {} +)

Also, you probably don't want to put all the results in a string like the previous line. You probably want to put the results in an array. In that case, use mapfile like this:

mapfile -t results < <(find "$catalog" -size +$size -type f -exec ls -1rt {} +)