Re: [Expectperl-discuss] $ in regex has duplicate meaning/ambiguity

[Jo R. <jr...@sv...>]

On Fri, Mar 10, 2006 at 02:37:36PM +0100, Roland Giersig wrote:
> Expect collects all output into a buffer (the accumulator) and matches
> against that. This is the mindset you should program in: you have a
> string with multiple lines in it.
> 
> So here are a few rules:
> 
> To match a line, use "^.*\n". Note that "." will not match a newline
> until you use single-line matching (the option or "(?s)").

My syntax reading fouled up, and I'm confused. "the option" is what?
And what did you mean by "(?s)"  My first read was using ? for smallest
match (ie "(.*?)\n"), but now I'm not so sure.
 
> Note that "\s" WILL match a newline, so it gives unexpected results with
> empty lines (as you have already found out). Use "[ \t]" instead.
> 
> To match against the end of buffer (which IMHO is a bad idea), use "\z".
> 
> Don't use "$" or "\Z", instead always anchor your patterns at the start
> of the line.

Anchoring at the start of the line is fine, but how to properly detect the
end of the line is the question.  Lines where I know the start but don't
know the end (and thus need to find and output the lines) is the issue.

/^[ \t]+Line starts with(.*?)[\r\n]+/
	..doesn't seem to work

/^[ \t]+Line starts with(.*?)$/
	.. does work, but can also match a partial line in the accumulator.
 
/^[ \t]+Line starts with(.*?)\r+$/
	.. does work

In short, I'm not convinced that \n is actually matching right now.
 
And this all makes sense, but goes explicitly against the documentation 
that says that line-matching can be done with ^ and $.  It seems likely 
that the documentation is out of date, but...

-- 
Jo Rhett
senior geek
SVcolo : Silicon Valley Colocation