Thanks, Cameron. It's a cool idea, but I can't get the leading and trailing
spaces to stick within a variable. The following script and it's output
will show you what I mean:
list="one two three four"
list=`echo $list | sed "s/^/ /"`
echo $list
list=`echo $list | sed "s/$/ /"`
echo $list
list=`echo $list | sed "s/ one / /"`
echo $list
list=`echo $list | sed "s/ four / /"`
echo $list
list=`echo $list | sed "s/^ //"`
echo $list
list=`echo $list | sed "s/ $//"`
echo $list
Output:
one two three four
one two three four
one two three four
one two three four
one two three four
one two three four
However, if I make list="x one two three four x", I get the desired results,
and I can remove the 'x ' and ' x' when done.
Is there a way to get leading and trailing spaces (and multiple spaces
between words) to stick in variables, or this a "feature" of bash?
bd
-----Original Message-----
From: [EMAIL PROTECTED]
[mailto:[EMAIL PROTECTED]]On Behalf Of Cameron Simpson
Sent: Friday, January 05, 2001 3:22 PM
To: [EMAIL PROTECTED]
Cc: Brad Doster
Subject: Re: Whole word substitution with sed
| On Wed, 03 Jan 2001 00:55:23 [EMAIL PROTECTED] wrote:
| >I've tried a slew of different things, none of which have worked.
| >Basically, if I could tell sed to only replace whole words, I'd be set.
| In perl, the \b construct denotes a word border. So /\bWORD\b/ should find
| WORD by itself (if the same works with sed).
It doesn't. However, one trick I've used a lot for this kind of thing is to
first add a space at the start and end of the line, then to go
s/ theword / somethingelse /
(i.e. _require_ spaces each side, which I've handily arranged at the
start and end as well), then to rip the spaces off again afterwards.
So:
s/^/ /
s/$/ /
s/ theword / somethingelse /
s/^ //
s/ $//
Cheers,
--
Cameron Simpson, DoD#743 [EMAIL PROTECTED] http://www.zip.com.au/~cs/
Ignorance can be cured; stupidity, on the other hand, is hereditary.
- [EMAIL PROTECTED] (Wallace Roberts)
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