How to print only those numbers of a column which are greater than certain number in bash

Asked Oct 06 '21 at 08:17

Active Oct 06 '21 at 09:04

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I want to print only those numbers of a column which is greater than certain number provided by the user. My file is like this:

Name,roll_no,marks
Honey,1,40
Joe,2,34
Viky,3,56

I tried:

read -p 'enter a number= ' $num
cat class.csv | awk 'FS=",";if($3>$num){print$3}else{print "no matches"}'

if i enter 35 then output should be

40
56

And if i enter 100 then output should be

no matches

But this is giving me wrong output. Plz help me :(

edited Oct 06 '21 at 09:04

oguz ismail

asked Oct 06 '21 at 08:17

Labh deep

Try this. `awk 'FS="," -v num=$num {if($3 > num) print $3; else print "no matches" }'` – Yuri Ginsburg Oct 06 '21 at 09:20
This is not working either syntax error occurs. – Labh deep Oct 07 '21 at 06:03
My bad. Should be: `awk -F="," -v num=$num '{if ($3 > num) print $3; else print "no matches" }' – Yuri Ginsburg Oct 08 '21 at 07:48

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