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Regular expression

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# 1  
Old 05-30-2008
Regular expression
Hi

I have to extract the first field and the last %field of the following out put..

/home (/abc/def/bhd ) : 522328 total allocated Kb
319448 free allocated Kb
202880 used allocated Kb
38 % allocation used

i need to to get only /home and 38% allocation used in a array.
How to do it in a perl script using sed or awk or regular expression.

Regards...
Last edited by Harikrishna; 05-30-2008 at 11:02 AM..
# 2  
Old 05-31-2008
In seperate lines:

Code:
awk 'NR==1{print $1}END{print}'

In one line:

Code:
awk 'NR==1{printf("%s ",$1)}END{print}'

Regards
# 3  
Old 05-31-2008
Quote:
Originally Posted by Harikrishna
i need to to get only /home and 38% allocation used in a array.
Code:
awk '/^\// {a=$1;b=NR+3;}; NR==b {print a, $1, $2}' data.file

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