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Old 11-10-2016, 03:06 PM   #1
Owais.Ahmad
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Post AWK Command Help Needed


Hello All,

I have following lines and i want an output displaying the disc space greater than 80%.

df -kh
Filesystem Size Used Avail Capacity iused ifree %iused Mounted on
/dev/disk1 465Gi 229Gi 235Gi 50% 909633 4294057646 0% /
devfs 182Ki 182Ki 0Bi 100% 628 0 100% /dev
map -hosts 0Bi 0Bi 0Bi 100% 0 0 100% /net
map auto_home 0Bi 0Bi 0Bi 100% 0 0 100% /home

I'm using the AWK Command so extract the column so checked can be applied on its output but it is giving me the following lines,

Owaiss-MBP:Scripting $ awk '{print $9}' DF.rtf








Mounted
/\
/dev\
100%
100%
Owaiss-MBP:Scripting $ awk '{print $8}' DF.rtf







-kh\
%iused
0%
100%
0
0


Help me out in this regard

Thank you!
 
Old 11-10-2016, 03:30 PM   #2
rtmistler
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Member Response

Hi,

Awk is giving you exactly what you've asked for. Column 9 and then column 8. The issues are that the lines of input have different counts of columns.

To print the last field or column you use the notation:
Code:
awk '{print $NF}'
This is assuming that the lines input are already filtered to only show resources in excess of 80%.
 
1 members found this post helpful.
Old 11-10-2016, 10:41 PM   #3
grail
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Firstly, please use [code/][code] tags around your code and data to make it clearer and preserve formatting.

I am a little concerned at seeing we are using an 'rtf' type file, was it created ion a Windows machine? I only ask as it may skew the results.

As above, when columns vary you will nede a way of identifying the required column from a specific line or as shown if always the last item is required you can use the NF variable.

Maybe you could explain how you came by your data? ie. what have you already used on the original output to get the current output?
 
Old 11-11-2016, 01:11 PM   #4
MadeInGermany
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Maybe you only want to see local disks?
The all start with a / usually /dev/
Code:
df -kP | awk '($1~/^\/dev\//) {print $1,$5,$6}'
If you want to only list the disks that are over 80% used, then add another condition into the implicit if clause
Code:
df -kP | awk '($1~/^\/dev\// && $5>80) {print $1,$5,$6}'
 
  


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