Bash - my script doesnt remember variables

Question

My script doesn't work. It doesn't remember changed values of variables. I made following script:

#!/bin/bash

kon_p=100
kon_m="nothing"

cat /etc/list | grep -v '#' | while read machine priority ;
do  
    if [ $priority -le $kon_p ]
    then
        kon_p=$priority 
        kon_m=$machine

    fi  

done

 echo priority machine is $kon_m with priority $kon_p

and the result is:

  priority machine is nothing with priority 100

why no changes?

file "list" is following

         Debian_pokus   1
         debian_2       2
         debian_3       3

anyone Can you help me please?

The variables are set in the subshell opened by the while, so the main process does not have access to them. — fedorqui
– fedorqui, Commented Dec 10, 2013 at 12:59
See mywiki.wooledge.org/BashFAQ/024 for a thorough explanation, including workarounds. — Guntram Blohm
– Guntram Blohm, Commented Dec 10, 2013 at 13:04

Édouard Lopez · Accepted Answer · 2013-12-10 13:28:33Z

2

There is a FAQ on this exact question on Greycat wiki (a bash reference).

What's wrong

your shebang is bash so use the adequate test: [[ … ]] or (( … )) ;
you don't need the cat, grep can't take a file as input (e.g.: grep <pattern> <file>).
your while loop is wrong, use a redirect or process substitution: while …; do …; done < <(cmd)

Solution

#!/bin/bash

kon_p=100
kon_m="nothing"

while read machine priority ; do  
    if (( $priority < $kon_p )); then
        kon_p=$priority 
        kon_m=$machine
    fi  
done < <(grep -v '#' /etc/list)

printf "priority machine is %s with priority %s\n" "$kon_m" $kon_p

edited Dec 10, 2013 at 13:28

answered Dec 10, 2013 at 13:15

Édouard Lopez

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Comments

cforbish · Accepted Answer · 2013-12-10 13:18:11Z

0

The subshell was doing you in, this should work:

#!/bin/bash

kon_p=100
kon_m="nothing"

IFS=$'\n'
for line in $(cat /etc/list | grep -v '#')
do  
    IFS=' ' read -ra values <<< "${line}"
    if [ ${values[1]} -le $kon_p ]
    then
        echo "doing it."
        kon_p=${values[1]} 
        kon_m=${values[0]}
    fi  
done

echo priority machine is $kon_m with priority $kon_p

answered Dec 10, 2013 at 13:18

cforbish

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3 Comments

Édouard Lopez Over a year ago

don't do this can of loop for … in $(), it behaves inconsistently depending on the source. Also read about word splitting

Édouard Lopez Over a year ago

also see my response there is plenty of links to support it.

Édouard Lopez Over a year ago

BTW, you can affect kon_p and kon_m directly in the read. Something like: read -r kon_m kon_p leftover <<< "$line"

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