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expr, regular expressions and '^' ...
From: |
aekalman |
Subject: |
expr, regular expressions and '^' ... |
Date: |
17 Nov 2004 20:41:39 -0800 |
Hi All.
expr has the following behavior with string expressions:
`STRING : REGEX'
Perform pattern matching. The arguments are coerced to strings
and the second is considered to be a (basic, a la GNU `grep')
regular expression, with a `^' implicitly prepended. The first
argument is then matched against this regular expression.
Is there any way for me to suppress the implicitly prepended '^' in
the regex? E.g. I have
expr 'sfiaravrt-7s-sf' : '\(-7s\)'
and it fails because -7s is not at the start of the string. How can I
use expr (or something else entirely) to return what you would get if
a '^' were not explicitly prepended, i.e. '-7s'?
Thanks,
--Andrew Kalman
aek at pumpkin inc dot com
- expr, regular expressions and '^' ...,
aekalman <=