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I tried searching on "df -k", but apparently the search utility requires a 3 character, contiguous string.
Anyway, so, I have df -k, and part of that out put is:
Code:
filesystem......................avail......capacity....mounted on
swap...........................835968.........1%....../var/whatever
swap...........................835968.........1%....../tmp
/dev/dsk/c0t2d0s0..............27972233......27%....../opt/whatever
Is that 27 Gigabytes or 27 Megabytes. I think it's giga.
And 835 megabytes for the swaps.
Is that right? This seems like a dumb question, but I just want to be sure. I need a 512MB slice for an upgrade.
This is a solaris system, but the df -k question is apropos to Linux also.
You can get this by
$ man df
The -k switch stands for block-size, but I get the same thing on Debian with df and no -k. Do
$ df -h
to see the sizes in human readable (-h). Here's an example:
Sorry the formatting's sloppy. I just barely got any system up here. Basic tools and Mozilla. Just compiled a 2.4.22 kernel and came online to try to find out how to get 4 separate modules in with it.
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