Find all the files more than 10MB, sort in descending order of size and record the output of filenames and size in a text file.

find . -size +10240k -exec ls -l {} \; | awk '{ print $5,"",$9 }'|sort -rn > message.out
This command specifies the size in Kilobytes using 'k' in the -size +(N)k option. The plus sign says greater than. -exec [cmd] {} \; invokes ls -l command on each file and awk strips off the values of the 5th (size) and the 9th (filename) column from the ls -l output to display. Sort is done in reversed order (descending) numerically using sort -rn options. A cron job could be run to execute a script like this and alert the users if a dir has files exceeding certain size, and provide file details as well.
Sample Output
18932792  ./scrip5.sql
16632595  ./scrip1.sql

5
2009-02-17 19:39:56

What Others Think

A more efficient way to do this: find . -size +10240k -print0 | xargs -0 du -sm | sort -nr
milmazz · 641 weeks and 5 days ago
Hey, I am a noob at shell scripting and both the commands didn't work when I used them in my lab machine... I researched and came up with this: find . -size +10240k -exec stat -c%s" "%n {} \; | sort -rn Short, sweet and works. :D
varunagrawal · 487 weeks and 3 days ago

What do you think?

Any thoughts on this command? Does it work on your machine? Can you do the same thing with only 14 characters?

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