Re^3: Getting only the link in a line

I'll give a quick breakdown here. Refer to perlre for details (I've indicated the appropriate sections).

You were originally missing your first line because it was https and you'd only specified http. The s? means zero or more 's's (see Quantifiers).
\S+? says match all non-whitespace non-greedily which stops it capturing the terminal period if it exists (further down under Quantifiers).
[.] stops '.' being a special (match anything) character by placing it in a character class (see Metacharacters).
[.]? just says zero or one non-special '.' (that's Quantifiers again).
\s at the end anchors the URL (and optional '.') to the whitespace that follows it (see Character Classes and other Special Escapes).

-- Ken

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Re^4: Getting only the link in a line by iphone (Beadle) on Nov 04, 2010 at 06:05 UTC
1.That's a very detailed explanation.but one question,why should we make "." a non-special character?even without it I am able to remove the "." 2.i understand what \S+? does but fail to understand how it removes the terminal period?	[reply]
Re^5: Getting only the link in a line by kcott (Archbishop) on Nov 04, 2010 at 06:40 UTC
When you want to match a '.' exactly, you need to remove the special meaning. `[.]` is one way, `\.` is another. Here's what happens if you don't remove the special meaning and the URL doesn't have a trailing '.': `$ perl -wE 'while (<>) { my ($x) = $_ =~ /(https?:\S+?).?\s/; say $x; +}' Bug found in the build. Please check check https://web.com/fluent/x/JI +OUAQ for more details. https://web.com/fluent/x/JIOUA` [download] In this instance it matches any character; there's no '.' so it matches the 'Q' which you can see has been removed in the output. The `\S+?` does not remove the '.', it captures up to the '.'. I've been saying any character throughout - it actually doesn't match a newline unless you use the 's' modifier - see Modifiers in perlre. -- Ken	[reply] [d/l] [select]
Re^6: Getting only the link in a line by Anonymous Monk on Nov 04, 2010 at 07:54 UTC
ah..got it ,so .? captures "." if it is present but doesnt print ,right? >/p>	[reply]