Get filesystem name and usage with `df -h`

Question

 df -h | grep test | sed -e 's/*%.^ //g;s/.*[ ^I]//'

Output:

/test
/test/drv0
/test/drv1

 df -h  | grep test  | cut -d '%' -f1 | sed -e 's/*%.^ //g;s/.*[ ^I]//'

Output:

10
20
15

I want to know how can I join those outputs like this:

/test 10
/test/drv0 20
/test/drv1 15

Can someone help me?

Stéphane Chazelas · Accepted Answer · 2013-09-26T20:52:33.757

10

df -P | sed -n '/test/s/.*[[:blank:]]\(.*\)%[[:blank:]]*\(.*\)/\2 \1/p'

(that assumes the mount point paths don't contain % or newline characters)

edited Sep 26 '13 at 20:52

answered Sep 26 '13 at 20:36

Stéphane Chazelas

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+1 for `-P` (posix) usage. – Olivier Dulac Sep 27 '13 at 08:23

score 8 · Answer 2 · edited Sep 26 '13 at 21:22

8

I think something as follow. I don't understand the whole question though

df -h | awk '/test/{print $1" "$5}' | sed -e ':a;N;$!ba;s/\n/ /g' -e 's/%//g'

Since the question has changed, here is the updated answer

df -h | awk '/test/{print $1, +$5}'

edited Sep 26 '13 at 21:22

jasonwryan

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answered Sep 26 '13 at 20:34

Valentin Bajrami

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I have no idea what you mean about +$5. Do you mean the awk part?. Can you provide an example. Feel free to edit the answer. – Valentin Bajrami Sep 26 '13 at 21:01
@asonwryan This will trim the space between test/ and the percentage. So I don't think that's what the questioner want. – Valentin Bajrami Sep 26 '13 at 21:13
1

My bad: forgot to add space. Fixed. – jasonwryan Sep 26 '13 at 21:23
Yeah, that's a nice one! Eliminates the need for sed. – Valentin Bajrami Sep 26 '13 at 21:26

score 6 · Answer 3 · answered Sep 26 '13 at 20:34

6

df | awk '/test/ {print $1 " " $5}'

will print

/test 10%
/test/drv0 20%
/test/drv1 15%

answered Sep 26 '13 at 20:34

Martin von Wittich

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Many ways to remove the percent sign, including `| tr -d '%'` – glenn jackman Sep 26 '13 at 20:36
1

That assumes mount point paths don't contain blank characters. – Stéphane Chazelas Sep 26 '13 at 20:36
@StephaneChazelas yes, it does. But I'd argue that that's pretty uncommon and not in the scope of such a simple shell solution. – Martin von Wittich Sep 26 '13 at 20:47
@glennjackman I know, but I intentionally left that part as an exercise for the reader :) – Martin von Wittich Sep 26 '13 at 21:59

terdon · Answer 4 · 2013-09-27T00:01:34.317

This is not a good solution for your particular situation (it works, but is needlessly complex), you should use the answers already provided. I just wanted to mention another tool that is very useful when you want to join the output of multiple programs, paste:

DESCRIPTION
   Write  lines  consisting  of  the sequentially corresponding lines from
   each FILE, separated by TABs, to standard output.   With  no  FILE,  or
   when FILE is -, read standard input.

Combined with bash process substitution, you can combine the output of your two commands like this:

$ paste  <(df -h | grep test | sed -e 's/*%.^ //g;s/.*[ ^I]//') \
       <(df -h  | grep test  | cut -d '%' -f1 | sed -e 's/*%.^ //g;s/.*[ ^I]//')
 /test 10
 /test/drv0 20
 /test/drv1 15

Or, to use a simpler example:

$ paste <(echo -e "a\nb\nc") <(echo -e "1\n2\n3")
a   1
b   2
c   3

I have never seen (or thought of) having multiple input redirections in a case like this. No idea it would have even worked. Thank you for pointing it out. — kurtm, Oct 06 '13 at 15:55

score 4 · Answer 5 · answered Sep 26 '13 at 22:24

4

I would use int() function in awk that will remove % and important is -P flag to df because if partition in LVM then also it's print properly.

df -hP | awk '/test/{print $1,int($5)}'

answered Sep 26 '13 at 22:24

Rahul Patil

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Get filesystem name and usage with `df -h`

5 Answers5