Using a variable as a operand in a expression - awk

Question

I tried doing something like this in awk?

limit=10000  
ls -ltr | awk '$5 >= $limit { print $5 }'

But it doesn't seem to work, it prints all size less than the limit too.
Thanks in advance!

possible duplicate of [How to use shell variables in awk script](http://stackoverflow.com/questions/19075671/how-to-use-shell-variables-in-awk-script) — fedorqui, Nov 19 '14 at 10:12

score 0 · Accepted Answer · edited Nov 19 '14 at 07:04

0

Try to set variable in AWK's paramaters using -v option:

limit=10000
ls -ltr | awk -v mylimit=$limit '$5 >= mylimit {print $5}'

edited Nov 19 '14 at 07:04

Avinash Raj

answered Nov 19 '14 at 06:51

Troublemaker-DV

You should always quote variable, so `-v mylimit="$limit"` (It will work without in this simple case, but its better to always quote it) – Jotne Nov 19 '14 at 09:58
Jotne, agree. My only excuse is that code just an example to show an idea. – Troublemaker-DV Nov 19 '14 at 21:50

score 0 · Answer 2 · answered Nov 19 '14 at 06:51

0

Change your awk command like below,

ls -ltr | awk -v limit=10000 '$5 >= limit { print $5 }'

you could declare an variable in awk itself using -v switch.

answered Nov 19 '14 at 06:51

Avinash Raj

Thanks Avinash! But I'll accept Troublemaker-DV answer since he's got it first. Please don't mind. Cheers :) – Malick Nov 19 '14 at 07:03
your wish, but why you go for two lines instead of one? – Avinash Raj Nov 19 '14 at 07:05
Oh, hadn't considered that! I'll keep in mind next time! – Malick Nov 19 '14 at 07:10

2 Answers2