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"cayenne" <> wrote in message
news: m...
> Hello all,
> I'm a perl noob...and just can't quite figure out how to do something
> that should be pretty simple.
>
> Here's an example.
>
> I have $mail_address = 'fred jones <>'
>
> I want to use regular expressions to just parse out the userid here of
> fred_jones
>
> I'm trying things like this:
>
> $mail_address =~ /\w+@/;
>
> But, doesn't seem to work. I'm a little hazy on exactly how the =~
> works...through examples I've successfully used it for substitutions
> like x =~ s/tom/joe/g; but, I'm just wanting to match a regular
> expression and extract it to the variable...or even to another
> variable leaving $mail_address unchanged.
>
> I've looked in books at the substr() function, but, I don't know how
> to use regular expressions to find the offset point, etc.
>
> Can someone give me an example...or pointers to a good reference on
> this type of thing?
>
> Thanks in advance,
>
> chilecayenne

Regular expressions are not the right way to find the offset unless you want
to use $1 an $2 and $3...etc, and then use index, it still isn't an optimal
way to find the offset point. Just change up your regular expression looks
like the other code, man I'm so tired.
-Robin

Sherm Pendley wrote:

> If your pattern has subexpressions, then on a match the
> offset of each subexpression appears in the @- array. That is, the offset
> of $1 is in $-[0], $2 is in $-[1], and so forth.

Incorrect. The offset of $& is in $-[0], the offset of $1 is in $-[1], etc.
-Joe

Jürgen Exner <> wrote in comp.lang.perl.misc:
> cayenne wrote:

[...]

> > I've looked in books at the substr() function, but, I don't know how
> > to use regular expressions to find the offset point, etc.
>
> You don't.

Ah, but you do, though not in this case. The @- and @+ arrays are
there to support it.

Anno

In article <> ,
(cayenne) wrote:

> I have $mail_address = 'fred jones <>'
>
> I want to use regular expressions to just parse out the userid here of
> fred_jones
>
> I'm trying things like this:
>
> $mail_address =~ /\w+@/;

What you seem to be asking for is this:

my ($user_id) = ($mail_address =~ m/(\w+)@/);

However, please note that \w doesn't really have the complete set of
valid characters to prefix the '@' sign in an email address.

Just off the top of my head, I know that '.', '-', '?', '=', and more
are valid. Possibly any unicode character other than whitespace and '@'
are valid. It might even be valid to have '<' in an email address.

At the very least, you probably want

my ($user_id) = ($mail_address =~ m/([\w.-+=]+)@/);

HTH,
Ricky

--
Pukku

Richard Morse <> wrote:
> At the very least, you probably want
>
> my ($user_id) = ($mail_address =~ m/([\w.-+=]+)@/);


Be careful where you use '-' inside a range:
Invalid [] range ".-+" before HERE mark in regex m/([\w.-+ << HERE =]+)@/

Put the hyphen last: [\w.+=-]

--
Glenn Jackman
NCF Sysadmin

Glenn Jackman <> wrote:

> Put the hyphen last: [\w.+=-]


Or first.


--
Tad McClellan SGML consulting
Perl programming
Fort Worth, Texas

Richard Morse <> wrote in message news:<remorse->...
> In article <> ,
> (cayenne) wrote:
>
> > I have $mail_address = 'fred jones <>'
> >
> > I want to use regular expressions to just parse out the userid here of
> > fred_jones
> >
> > I'm trying things like this:
> >
> > $mail_address =~ /\w+@/;
>
> What you seem to be asking for is this:
>
> my ($user_id) = ($mail_address =~ m/(\w+)@/);
>
> However, please note that \w doesn't really have the complete set of
> valid characters to prefix the '@' sign in an email address.
>
> Just off the top of my head, I know that '.', '-', '?', '=', and more
> are valid. Possibly any unicode character other than whitespace and '@'
> are valid. It might even be valid to have '<' in an email address.
>
> At the very least, you probably want
>
> my ($user_id) = ($mail_address =~ m/([\w.-+=]+)@/);
>
> HTH,
> Ricky

Just quickly, can you explain the extensive use of parens here? I
understand the () in the regular expression, to keep those parts the
match...but, what is the function of the () around $user_id and the
entire part after the = sign?

Thanks in advance,

CC

In article <>,
(cayenne) wrote:

> Richard Morse <> wrote in message
> news:<remorse->...
>
> > my ($user_id) = ($mail_address =~ m/([\w.-+=]+)@/);
>
> Just quickly, can you explain the extensive use of parens here? I
> understand the () in the regular expression, to keep those parts the
> match...but, what is the function of the () around $user_id and the
> entire part after the = sign?

Parens around $user_id force the match to happen in a list context. A
match in a scalar context would return the number of matches, while in a
list context, it returns the various matches.

my $user_id = ($mail_address =~ m/.../)

would have $user_id be the value 1 (because there is one match, as it
isn't a /g match).

The parens around the match are there because it makes it easier for me
to read it. I've never not put them there, although a quick test I just
did seems to indicate that they aren't necessary.

HTH,
Ricky

--
Pukku

On Wed, 19 May 2004, cayenne wrote:

> Richard Morse <> wrote in message news:<remorse->...
> >
> > my ($user_id) = ($mail_address =~ m/([\w.-+=]+)@/);
> >
> Just quickly, can you explain the extensive use of parens here? I
> understand the () in the regular expression, to keep those parts the
> match...but, what is the function of the () around $user_id and the
> entire part after the = sign?
>

The parens around $user_id force the binding operation of =~ to be
evaluated in list context. This is done because a pattern match in list
context returns a list of all of the captured matches (ie, the things that
go into $1, $2, etc). This is a shorthand way of writing the two
statements:

$mail_address =~ m/([\w.-+=]+)@/
my $user_id = $1;

The parens around the whole pattern match here are actually unnecessary.
This is because the =~ operator has a higher precedence than the =
operator. They are likely used here just for clarity, to make sure the
readers of the code are aware that ($user_id) is being assigned to the
return value of the pattern match, rather than the alternate
interpretation of the assignment of $user_id to $mail_address being
pattern matched against the pattern (which would be written like so:
(my $user_id = $mail_address) =~ m/([\w.-+=]+)@/;

Please let me know if this is not clear enough.

Paul Lalli

Paul Lalli wrote:
>
> On Wed, 19 May 2004, cayenne wrote:
>
> > Richard Morse <> wrote in message news:<remorse->...
> > >
> > > my ($user_id) = ($mail_address =~ m/([\w.-+=]+)@/);
> >
> > Just quickly, can you explain the extensive use of parens here? I
> > understand the () in the regular expression, to keep those parts the
> > match...but, what is the function of the () around $user_id and the
> > entire part after the = sign?
>
> The parens around $user_id force the binding operation of =~ to be
> evaluated in list context. This is done because a pattern match in list
> context returns a list of all of the captured matches (ie, the things that
> go into $1, $2, etc). This is a shorthand way of writing the two
> statements:
>
> $mail_address =~ m/([\w.-+=]+)@/
> my $user_id = $1;

They are not the same at all. If the match fails the first will set
$user_id to undef but your version will set $user_id to the contents of
a previously successful match's capturing parentheses or ''.




John
--
use Perl;
program
fulfillment