On Tue, Sep 04, 2001 at 10:10:39AM -0700, [EMAIL PROTECTED] wrote:
> root Wed Aug 22 04:44:59 2001 DLs
[snip]
> for(@procs){
> /^(.{7})\s+?(.{24})\s+?(\S+?)$/;
>
>
> I want $1 to be the user (e.g. 'root' 'postfix').
> I want $2 to be something like 'Wed Aug 23 05:30:01 2001'.
> I want $3 to be the state, which was a lot of forms, but which is, for
> example 'DLs' above.
>
> Shouldn't this get the first 7, then 24 characters, then the glop that's
> left over?
Yes. Don't you? I do. I think the problem might lie in the difference
between what you say you want, what you expect, and what you get. You say
you want $1 eq 'root' from "root Wed Aug 22 04:44:59 2001 DLs", but
$1 eq 'root '. This is correct with what you expect, because you expect
the first seven characters. Space is a character.
To correct this you'll have to change your current regex from:
^(.{7})\s+?(.{24})\s+?(\S+?)$
to something like
^(\S+)\s+(\S.*\S)\s+(\S+)$
This matches on whitespace boundaries, rather than column widths. You'll
notice I removed your non-greedy specifiers (\s+?, \S+?). They aren't doing
any good; quite the opposite, they're slowing your pattern match down for no
reason.
Given that this is columnar data, a better solution may be to use unpack:
for (@procs) {
my($user, $date, $state) = unpack("A8 A29 A*", $_);
...
}
Michael
--
Administrator www.shoebox.net
Programmer, System Administrator www.gallanttech.com
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