hi

i need to get the size of one directory but with "du -m |awk '{print $1}' " i got it including the sub directories. In the case below i need only "23" as a output..

nazmin@darkstar:/windows/shared/working_iso$ du -m |awk '{print $1}'
2
1
1
1
16
2
2
1
2
1
1
22
22
22
22
23
23

Why not pipe it through tail -1?

not working...thanx

nazmin@darkstar:/windows/shared/working_iso$ du -m iso_dir/ | tail -l
2 iso_dir/pythontut/python.org/doc/2.4/ref
1 iso_dir/pythontut/python.org/doc/2.4/tut
2 iso_dir/pythontut/python.org/doc/2.4/dist
1 iso_dir/pythontut/python.org/doc/2.4/inst
1 iso_dir/pythontut/python.org/doc/2.4/icons
22 iso_dir/pythontut/python.org/doc/2.4
22 iso_dir/pythontut/python.org/doc
22 iso_dir/pythontut/python.org
22 iso_dir/pythontut
23 iso_dir/
nazmin@darkstar:/windows/shared/working_iso$

How about du -sh

ah...perfect...

i should "man du" first....:-P

thanx

"du...de, mannnnn, duude..." LOL

Your first post ever on LQ and this is the best you can do? A pointless comment to a 5 year old post. If your intention is to try out a test post, feel free to introduce yourself in the new member forum:

http://www.linuxquestions.org/questi...mber-intro-24/

I thought it was funny and noone got hurt.

Back on topic though: I didn't know of sort or its options but to list it will list input in reverse, numeric order. With the du prepending directories' sizes, it saves a lot of skimming, if, for example, you're looking for the first 5 directories using the most storage.