[ruby-talk:23834] Re: Probably a simple one

Hello --

(Disclaimer: I am aware of the fact that HTML cannot be parsed with a
single regular expression.  I just want to help the guy grasp the
syntax :-)

On Tue, 30 Oct 2001, Ralph Mason wrote:

> Thanks Joe,
>
> I can see how your expression works however I still mystified by the
> behavior of my one.
>
> If  .* will just keep matching to the end of the line then my version should
> have returned nil as the whole expression would never have been matched.
> But it didn't
>
> I have just been playing with this
>
> z="abc"
> z=/(.*b)/    => 0
> $1  => "ab"
> z=~/(.*)b/ => 0
> $1  => "a"
>
> So it seems that .* is matching in this case until it finds the character
> following. But in the example below it doesn't.  It seems to randomly stop.

No, it stops at exactly the point you've asked it to :-)

> > RM> s="<A  href=\"l.htm\">xxx <IMG  src=\"images/la.gif\" width=4></A>"
> > RM> s =~ /<.*A.*href *= *"(.*)".*>/   => 0
> > $1   =>>  "l.htm\">xxx <IMG  src=\"images/la.gif"

This part of your pattern:

    "(.*)"

means:

    a double quote, followed by zero or more of any character (except newline),
    followed by another double quote.

and it saves the "zero or more..." part to $1, because that's what the
parens are around.

Note that .* matches are "greedy".  That is, the match goes as far
along in the string as it can and still succeed in matching.  "Zero or
more" is really "as many as possible".  It has to stop consuming
characters when it gets to the last double quote, because if it
consumes that, the match will fail (because the match calls for a
double quote *after* the .* part).  But it can consume all the double
quotes prior to the last one.

You can do a "non-greedy" match using /.*?/ instead of /.*/.  For example:

s="<A  href=\"l.htm\">xxx <IMG  src=\"images/la.gif\" width=4></A>"
s =~ /<.*A.*href *= *"(.*?)".*>/

  # $1 =>  l.htm


David

-- 
David Alan Black
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