Papo Napolitano wrote:
Gunnar Hjalmarsson wrote:
Papo Napolitano wrote:
$text =~ s/\r\n/<br>/gs;
-----------------------^
Why would the /s modifier make a difference? Please study
perldoc perlre
and figure out why it doesn't. ;-)
perl -e '$_ = "line1\r\nline2\r\n\r\nline4"; s/\r\n/<br>/gs; print'
Please test, and figure out why it DOES make a difference.
perl -e '$_ = "line1\r\nline2\r\n\r\nline4"; s/\r\n/<br>/g; print'
outputs exactly the same:
line1<br>line2<br><br>line4
(as expected).
Also study what "perldoc perlop" has to say about s///.
Look for "s/PATTERN/REPLACEMENT/egimosx" and especially the
description of the "s" modifier.
Sure, but the description in "perldoc perlre" is more complete,
reducing the risk for misunderstandings:
"Treat string as single line. That is, change ``.'' to match any
character whatsoever, even a newline, which normally it would not match."
And there is no '.' in the pattern, is it?
As you may know s/// only matches to the end of line
No, I don't know that. What makes you believe so?
(And "\r\n" happens to be a line delimiter) unless you use the /s
modifier.
You are confusing the list members, Papo. It's time for *you* to test
and revise your understanding of what /s does.
--
Gunnar Hjalmarsson
Email: http://www.gunnar.cc/cgi-bin/contact.pl
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