awk printf for user defined variables

jabberwocky · #1 (**permalink**) 10-02-2007

I am working on a SunFire 480 - uname -a gives:

SunOS bsmdb02 5.9 Generic_112233-08 sun4u sparc SUNW,Sun-Fire-480R

I am tyring to sum up the total size of all the directories for each running database using awk:

#!/usr/bin/ksh

for Database in `ps -efl | grep "ora_pmon" | grep -v grep | awk '{printf("%s\n"
,$NF)}' | cut -f3 -d'_' | sort`
do

export Database
# echo "$Database\c"

find / -type d -name $Database -exec du -sk {} \; 2>/dev/null |
awk '{
sum += $1
}
END {
printf("%10s\t%20s\n",$Database,sum) }'

done

#EOF

I can get output in the format Database NNNNNNNN using the echo command, now commented out. If I try to use printf in the awk statement to format the output for both Database name and sum, the printf statement does not pick up the value of $Database (or Database for that matter). If I use the echo command for the Database name, the output format is difficult to control. How can I get the database name recognised in the printf statement?

Thanks,

Jabberwocky

aigles · #2 (**permalink**) 10-02-2007

One way:

Code:

find / -type d -name $Database -exec du -sk {} \; 2>/dev/null |
awk -v Db=$Database '{
sum += $1
}
END {
printf("%10s\t%20s\n",Db,sum) }'

Jean-Pierre.

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