LinuxQuestions.org - [SOLVED] summinf values in bash

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julirio

10-10-2012 08:33 AM

summinf values in bash

hi! i'm trying to get all the regular files opened by all the processes in the current session. I have this code

while read pid
do

FILES_ACTUAL=$(lsof -p $pid | grep REG | wc -l)

done < <(ps -o pid,tt -u $USER | grep $CURRENT_TERMINAL | awk '{print $1}')

echo $FILES_ACTUAL

but i dont know how to add up, inside the while, the variable $FILES_ACTUAL.. I tried to use the awk command, but i couldn´t do it.

pan64

10-10-2012 08:42 AM

Code:

SUM=0

while read pid

do



FILES_ACTUAL=$(lsof -p $pid | grep REG | wc -l)

SUM=$(( SUM + FILES_ACTUAL ))



done < <(ps -o pid,tt -u $USER | grep $CURRENT_TERMINAL | awk '{print $1}')



echo $FILES_ACTUAL

echo $SUM

something like this may help

grail

10-10-2012 09:03 AM

Some other suggestions for your code:

Code:

SUM=0

while read pid

do



FILES_ACTUAL=$(lsof -p $pid | grep -c REG)

(( SUM ++ FILES_ACTUAL ))



done < <(ps -o pid,tt -u $USER | awk -vct="$CURRENT_TERMINAL" '$2 ~ ct{print $1}')



echo $FILES_ACTUAL # not sure the point of showing the last value this was set to??

echo $SUM

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