CODEWITHSUNDEEP
bash
capser
capser

Reputation: 2635

bash keeping the higest number as variable

I am trying to get the longest_length set to the highest number. 

#!/bin/bash
for i in $(cat list_of_aways)
do
echo $i | while IFS=, read -r area name host
do
        printf "%s\n" $name
        sleep 1
        longest_length=${#name}
        printf "%s\n" "$longest_length"
        done
done

This is the data. The 9999999 is the longest string - I want to set that variable to the arrays value because it is the longest. 

__DATA__
HOME,script_name_12345,USAhost.com
AWAY,script_name_123,USAhost.com
HOME,script_name_1,EUROhost.com
AWAY,script_name_123,USAhost.com
HOME,script_name_123456,EUROhost.com
AWAY,script_name_12345678999999,USAhost.com
HOME,script_name_1234,USAhost.com
AWAY,script_name_1234578,USAhost.com
HOME,script_name_12,EUROhost.com
AWAY,script_name_123456789,USAhost.com

Upvotes: 2

Views: 45

Answers (1)

AlexM
AlexM

Reputation: 334

You can actually do it with an awk command:

awk -F '_|,' 'NR>1 {if ( length($4) > length(max) ) { max=$4 }} END { print max }' INPUTFILE

_|, = We are using 2 delimiters _ and ,

if ( length($4) > length(max) ) { max=$4 } = var max gets is compared with current 4th element in each line, and max is chosen.

print max = The max value is printed out.

If you want to set the result into a var called longest_length:

longest_length = `awk -F '_|,' 'NR>1 {if ( length($4) > length(max) ) { max=$4 }} END { print max }' INPUTFILE`

Upvotes: 1

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