I'm using DB2 v9.1 LUW. I'm looking at "df" on Unix now and below is an example. It's lists out in 512-blocks, I need this in 4k blocks. Is there a way to do this in Unix or do I manually convert and how?

So for container 1 there is 7,340,032 in size in 512-blocks. What would the 4k block be and how is it converted? Or am I not understanding the 512-blocks and what this means?


Filesystem 512-blocks Free %Used Iused %Iused Mounted on
/dev/copy811_ctr1 7340032 1564416 79% 26 1% /db2_data/copy811/container1
/dev/copy811_ctr2 7340032 1564416 79% 26 1% /db2_data/copy811/container2
/dev/copy811_ctr3 7340032 1564408 79% 26 1% /db2_data/copy811/container3
/dev/copy811_ctr4 7340032 1564416 79% 26 1% /db2_data/copy811/container4
/dev/copy811_ctr5 7340032 1564416 79% 26 1% /db2_data/copy811/container5
/dev/copy811_logs1 2097152 1907864 10% 12 1% /db2_data/copy811/logs


Thank you.


CC

What does it have to do with DB2 though?

You're right, it has nothing to do with DB2. It Unix. My bad.

I tried to do this command and got this message:

devdb# df -B4096
df: illegal option -- B
df: illegal option -- 4
Usage: df [-P] | [-IMitv] [-gkm] [-s] [filesystem ...] [file ...]

How about taking the number, adding 7*512 (for the last non-full block), and dividing it by 8?