LinuxQuestions.org - sorting columns in bash

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sorting columns in bash

I need to show the freedisk space in any directory, for example /tmp

Code:

df /tmp | tail -1

But i need to only show the remaining free disk space. can anyone help me to to remove the other columns shown.

Quote:

Originally Posted by twistadias (Post 3256482)

I need to show the freedisk space in any directory, for example /tmp

Code:

df /tmp | tail -1

But i need to only show the remaining free disk space. can anyone help me to to remove the other columns shown.

Code:

df /tmp | tail -1 |awk '{print $3}'

Quote:

I need to show the freedisk space in any directory, for example /tmp

Ummm ... df only operates on mount-points, not
directories. While you can invoke it like that it
doesn't make too much sense unless the directories
you're iterating over are indeed on separate file-
systems.

And you don't really need tail if you're using awk:

Code:

df /tmp | awk '$4 ~ /[0-9]/ {print $4}'

Cheers,
Tink

Thanks for your fast reply. Is there any way I can show this in kb, Mb or GB

@ Tinkster. Although your code does work, i really dont understand the stuff in awk. I am using this for a project in uni wherein I have to explain each and every line. If you could make me understand what your code does, it would be great

Quote:

Originally Posted by twistadias (Post 3256547)

Thanks for your fast reply. Is there any way I can show this in kb, Mb or GB

Code:

df -k /tmp | awk '$4 ~ /[0-9]/ {print $4}'

man df
for details ....

And df still doesn't operate on a per directory basis...

Cheers,
Tink

Quote:

Originally Posted by twistadias (Post 3256547)

@ Tinkster. Although your code does work, i really dont understand the stuff in awk. I am using this for a project in uni wherein I have to explain each and every line. If you could make me understand what your code does, it would be great

awk '$4 ~ /[0-9]/ {print $4}'

Awk operates on lines and fields within lines, by default
the field separator is whitespace other than \n.

So if you see a line like
/dev/sda7 19518312 7021552 12496760 36% /
12496760 will be field $4.

What I've done is to only create output if the field
is numeric (regular expression) ~ /[0-9]/ which means
that it won't take for the df header.

In English: "Only produce output if the 4th field in any
line of input has a numeric value, and only output that
one field"

Cheers,
Tink

P.S.: If you want a response it's wiser to reply rather than
to edit pre-existing posts. People on the boards don't usually
scroll up to check for changes - that's not what a forum is about.

There is a sort utitlity in *nix. Read man sort.

Quote:

Originally Posted by twistadias (Post 3256547)

Thanks for your fast reply. Is there any way I can show this in kb, Mb or GB

Code:

df -h

Read 'man df' to find out what that does.

'df -k' doesn't produce anything different to 'df' for me.

Try these examples:-

Code:

# du -ks /var/adm

1886422 /var/adm

Code:

# du -ks /var/adm | awk '{printf "%8.2f %s\n",$1/1024,$2}'

 1842.21 /var/adm

Code:

# du -ks /var/adm | awk '{printf "%8.2f %s\n",$1/2048,$2}'

  921.10 /var/adm

Code:

# du -ks /var/adm | awk '{printf "%8.2f %s\n",$1/1024,$2}' | awk '{print $1}'

1842.21

# To be run in a directory to find dir. size sorted

Code:

for i in `ls -lrt | awk '{print $9}'`; do du -ks $i | awk '{printf "%8.2f %s\n",$1/1024,$2}'; done | sort -kr1