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How to extract the value of "%" using scripting?

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Top Forums Shell Programming and Scripting How to extract the value of "%" using scripting?
# 8  
Old 06-14-2010
Hi Scrutinizer,

This time the script works perfectly.

Here is the output:
Code:
/var                            62%
/usr                            37%
/usr/misc/TEST                  38%
/tmp                            78%
/home/TEST                      91%

However, I've some questions as follow:
1) How to adjust the gap between the path and percentage? Currently the gap is quite wide.
2) Is it possible to put some remarks in the end for higher percentage? Example of remark: High usage.

Sample output:
Code:
/home/TEST   91%   High Usage

Thank you for your help and advice.



Regards,
Peter
# 9  
Old 06-14-2010
Hi, you can vary the width of the first columns and make it 24 for instance. Also you can issue a message based on the value, like so:

Code:
awk 'NR>1&&NF>1{printf "%24-s %4s  %s\n",$NF,$(NF-1),($(NF-1) > 90)?"High Usage":""}' infile

This User Gave Thanks to Scrutinizer For This Post:
superHonda123
# 10  
Old 06-15-2010
Hi Scrutinizer,

Thanks for your help.

This time the script works perfectly.

Here is the output:
Code:
/sapmnt/TEST                    91%
/oracle/stage/102_64            48%
/oracle/client                  46%
/opt                            66%
/misc                           17%
/home                           19%

However, I've 2 questions:
1) Is it possible to adjust the gap between the path and the percentage?
2) Is it possible to add a comment like: High Usage at end of line?

For example:
Code:
/sapmnt/TEST 91%  High Usage

Thank you for your help.



Regards,
Peter
# 11  
Old 06-15-2010
See post above
# 12  
Old 06-16-2010
Hi Scrutinizer,

Thanks a lot for your kind help.

The script is working perfectly.

GREAT....



Regards,
Peter

---------- Post updated 06-16-10 at 03:45 AM ---------- Previous update was 06-15-10 at 05:30 AM ----------

Hi Scrutinizer,

I've tried your script, but there is some minor issues as shown below. The condition has been set to 80, but the value that is lesser than 80 also displayed, eg: 9%.

bdf | awk 'NR>1&&NF>1{printf "%24-s %4s %s\n",$NF,$(NF-1),($(NF-1) > 80)?"High Usage":""}'

Code:
/stand                     9%   High Usage
/var                       62%
/sapmnt/TES                91%  High Usage
/oracle                    3%
/oracle/TEST/102_64        85%  High Usage

Please kindly advice.

Thanks.


Regards,
Peter
# 13  
Old 06-16-2010
Oops, try this:
Code:
bdf | awk 'NR>1&&NF>1{printf "%24-s %4s %s\n",$NF,$(NF-1),($(NF-1)+0 > 80)?"High Usage":""}'

# 14  
Old 06-16-2010
Hi Scrutinizer,

Once again, thanks a lot for your help.

The script works perfectly.



Regards,
Peter

---------- Post updated at 06:15 AM ---------- Previous update was at 06:01 AM ----------

Hi Scrutinizer,

Do yuo know how can I close this message?

Could not find any option.

Thanks.


Regards,
Peter
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