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comp.lang.ruby

multiline regexp and newlines

Markus Schirp

9/28/2008 10:42:00 PM

Hello Community,

Im trying to reject any newline in a string using regexp.
The string must not be empty.

Yes, this can be done without regexp.

So my Regexp is like:

"Match any character from the begin to the end of a string witch is not
newline, and ensure there is at least one character, and to this match
over multiple lines (so if there are multiple lines reject it)"

Results in /^[^\n]+$/m

My ruby 1.8.6 does:

/^[^\n]+$/m =~ "a\n" -> 0

without multiline flag the same:

/^[^\n]+$/ =~ "a\n" -> 0

But:

/^[^\n]+$/m =~ "\n" -> nil

If I replace "reject newline" with "reject character x" it worked as
expected (x is 'a' in this example):

/^[^a]+$/m =~ "ba" -> nil

Is the multiline-regexp implementation broken or my multiline behaviour
expectation?

Mfg

Markus Schirp
2 Answers

Daniel DeLorme

9/29/2008 4:49:00 AM

0

Markus Schirp wrote:
> So my Regexp is like:
>
> "Match any character from the begin to the end of a string witch is not
> newline, and ensure there is at least one character, and to this match
> over multiple lines (so if there are multiple lines reject it)"
>
> Results in /^[^\n]+$/m
>
> My ruby 1.8.6 does:
>
> /^[^\n]+$/m =~ "a\n" -> 0
>
> without multiline flag the same:
>
> /^[^\n]+$/ =~ "a\n" -> 0

The multiline flag only changes the meaning of "." to include the \n
character. It doesn't change the meaning of ^ and $ which mean
respectively "beginning of line" and "end of line". True true "beginning
of string" metacharacter is not ^ but \A, and for "end of string" it's
not $ but \z (or \Z if you want to allow a trailing \n on your string)

>> "a\nb\n"[ /^.*$/ ]
=> "a"
>> "a\nb\n"[ /\A.*\z/ ]
=> nil
>> "a\nb\n"[ /\A.*\z/m ]
=> "a\nb\n"

Yeah, it's tricky.

--
Daniel

Markus Schirp

9/29/2008 7:07:00 AM

0

On Mon, Sep 29, 2008 at 01:49:23PM +0900, Daniel DeLorme wrote:
> Markus Schirp wrote:
>> So my Regexp is like:
>> "Match any character from the begin to the end of a string witch is not
>> newline, and ensure there is at least one character, and to this match
>> over multiple lines (so if there are multiple lines reject it)"
>> Results in /^[^\n]+$/m
>> My ruby 1.8.6 does:
>> /^[^\n]+$/m =~ "a\n" -> 0
>> without multiline flag the same:
>> /^[^\n]+$/ =~ "a\n" -> 0
>
> The multiline flag only changes the meaning of "." to include the \n
> character. It doesn't change the meaning of ^ and $ which mean respectively
> "beginning of line" and "end of line". True true "beginning of string"
> metacharacter is not ^ but \A, and for "end of string" it's not $ but \z
> (or \Z if you want to allow a trailing \n on your string)
>
> >> "a\nb\n"[ /^.*$/ ]
> => "a"
> >> "a\nb\n"[ /\A.*\z/ ]
> => nil
> >> "a\nb\n"[ /\A.*\z/m ]
> => "a\nb\n"
>
> Yeah, it's tricky.
>
> --
> Daniel
>

Thx, it works now!