I got different results in Perl and Ruby of this regular expression. Can
somebody maybe give me a "Ruby Way" solution of this? The output from
Perl is what I want. But I'm currently programming this is Ruby.
In Perl: #snippet start ==========
sub transform_data
{
my $data = shift;
if ($data=~/^[\d\.]+$/) #numbers
{
print "Got here!\n";
}
else
{
print "'$data'","\n";
}
}
my $data = "patched 3 systems:
134.27.56.237
134.27.59.6
134.27.55.43";
transform_data($data); #snippet end =========
Output is:
'patched 3 systems:
134.27.56.237
134.27.59.6
134.27.55.43'
I got different results in Perl and Ruby of this regular expression. Can
somebody maybe give me a "Ruby Way" solution of this? The output from
Perl is what I want. But I'm currently programming this is Ruby.
def transform_data(data)
if (data=~/^[\d\.]+$/) #numbers
puts "Got here!"
else
puts("'" + data + "'")
end
end
data = "patched 3 systems:
134.27.56.237
134.27.59.6
134.27.55.43"
transform_data(data) #snippet end ============
Output is:
Got here!
Now why would it match in Ruby? Am I missing something here?
^ and $ match start and end of line, not string. For start and end of
string, you want \A and \z (or \Z to ignore final newline).
David
--
David A. Black
dblack@wobblini.net
"Ruby for Rails", from Manning Publications, coming April 2006!
Yes, after I replaced ^ with \A and $ with \a, it works now.
Another difference learned. I'm discovering new stuffs in Ruby every day.
Regards,
Sam
dblack@wobblini.net
01/09/2006 05:54 PM
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ruby-talk@ruby-lang.org
To
ruby-talk@ruby-lang.org (ruby-talk ML)
cc
Subject
Re: regexp problem - differences in Perl and Ruby
Classification
Hi --
···
On Tue, 10 Jan 2006, Sam Dela Cruz wrote:
I got different results in Perl and Ruby of this regular expression. Can
somebody maybe give me a "Ruby Way" solution of this? The output from
Perl is what I want. But I'm currently programming this is Ruby.
def transform_data(data)
if (data=~/^[\d\.]+$/) #numbers
puts "Got here!"
else
puts("'" + data + "'")
end
end
data = "patched 3 systems:
134.27.56.237
134.27.59.6
134.27.55.43"
transform_data(data) #snippet end ============
Output is:
Got here!
Now why would it match in Ruby? Am I missing something here?
^ and $ match start and end of line, not string. For start and end of
string, you want \A and \z (or \Z to ignore final newline).
David
--
David A. Black
dblack@wobblini.net
"Ruby for Rails", from Manning Publications, coming April 2006!
Oops sorry, typo, I mean when I replaced ^ with \A and $ with \z
Regards,
Sam
Sam Dela Cruz <sam.dela.cruz+FromInternet@philips.com>
01/09/2006 06:05 PM
Please respond to
ruby-talk@ruby-lang.org
To
ruby-talk@ruby-lang.org (ruby-talk ML)
cc
Subject
Re: regexp problem - differences in Perl and Ruby
Classification
Thanks David.
Yes, after I replaced ^ with \A and $ with \a, it works now.
Another difference learned. I'm discovering new stuffs in Ruby every day.
Regards,
Sam
dblack@wobblini.net
01/09/2006 05:54 PM
Please respond to
ruby-talk@ruby-lang.org
To
ruby-talk@ruby-lang.org (ruby-talk ML)
cc
Subject
Re: regexp problem - differences in Perl and Ruby
Classification
Hi --
···
On Tue, 10 Jan 2006, Sam Dela Cruz wrote:
I got different results in Perl and Ruby of this regular expression. Can
somebody maybe give me a "Ruby Way" solution of this? The output from
Perl is what I want. But I'm currently programming this is Ruby.
def transform_data(data)
if (data=~/^[\d\.]+$/) #numbers
puts "Got here!"
else
puts("'" + data + "'")
end
end
data = "patched 3 systems:
134.27.56.237
134.27.59.6
134.27.55.43"
transform_data(data) #snippet end ============
Output is:
Got here!
Now why would it match in Ruby? Am I missing something here?
^ and $ match start and end of line, not string. For start and end of
string, you want \A and \z (or \Z to ignore final newline).
David
--
David A. Black
dblack@wobblini.net
"Ruby for Rails", from Manning Publications, coming April 2006!