Re: Passing variables to awk

[Andreas Schwab]

TimtheEagle <tim.cheyne@safecom.co.nz> writes:

This is actually all unrelated to bash.

> + awk -v 'var1=Framed-IP-Address = 10\.6\.6\.' '/*** Received from
> 10.242.252.20 port 1645 ..../{ last_acct_status = NR; as = $0 }/Code:      

`*' and `.' are special in regexps, you need to quote them to match them
as literals.

> Access-Accept/ && last_acct_status + 1 == NR {last_user_name = NR; print
> NR-1 as; print NR $0 }/$var1/ && last_acct_status + 10 == NR { print NR $0

/.../ is the syntax for a literal regexp.  To match against a variable
use `$0 ~ var1' (awk does not use $ to refer to variables).

> }' /tmp/theout
> awk: warning: escape sequence `\.' treated as plain `.'

The value is parsed as a string.  To include a backslash in a string
write it as \\.

Andreas.

-- 
Andreas Schwab, SuSE Labs, schwab@suse.de
SuSE Linux Products GmbH, Maxfeldstraße 5, 90409 Nürnberg, Germany
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