How to sort 'find' results in bash by size

问题

I am wondering if there's an "easy" way (through a pipe or something) to order (by file size) the results of a "find" command in bash such as:

find /location/of/directory/ -type f -size +2G

回答1:

You can use %k for example to print the size in kilobytes:

find . -type f -size +2G -printf "%kKB %p\n" | sort -n

• By saying -printf "%kKB %p\n" you are printing the file in kilobytes and then the name.
• sort -n gets this input and sorts it accordingly.

See an example:

$ find . -type f -size +1M -printf "%p %kKB\n" | sort -n -k2
./arrr.txt.gz 1664KB
./brrr.gz 32388KB

回答2:

Try this :

find /location/of/directory/ -type f -size +2G -exec du -s {} + |sort -n

-exec executes command du -s on each search result and and sort -n sorts the result numerically.

来源：https://stackoverflow.com/questions/27189663/how-to-sort-find-results-in-bash-by-size