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From: Phred Phungus on 31 Mar 2010 15:56
J�rgen Exner wrote:
> Phred Phungus <Phred(a)example.invalid> wrote:
>> J�rgen Exner wrote:
>>> Phred Phungus <Phred(a)example.invalid> wrote:
>>>> PerlFAQ Server wrote:
>>>>> $string = "Placido P. Octopus";
>>>>> $regex = "P.";
>>>>>
>>>>> $string =~ s/$regex/Polyp/;
>>>>> # $string is now "Polypacido P. Octopus"
>>>>>
>>>>> Because "." is special in regular expressions, and can match any single
>>>>> character, the regex "P." here has matched the <Pl> in the original
>>>>> string.
>>>>>
>>>> Am I then correct that a period is not a character?
>>> How did you come to that conclusion?
>> Because the regex would then match the middle initial and the string
>> would be "Polypacido Polyp Octopus".
>
> No, it wouldn't because there is no /g modifier in the s///-operation.

$ perl perl1.pl
Placido P. Octopus
Polypacido P. Octopus
Placido P. Octopus
Placido P. Octopus
$ cat perl1.pl
#!/usr/bin/perl

use strict;
use warnings;

my $string = "Placido P. Octopus\n";
my $regex = "P.";
print $string;
$string =~ s/$regex/Polyp/;
print $string;

my $string2 = "Placido P. Octopus\n";
print $string2;
$string =~ s/$regex/Polyp/g;
print $string2;
$

I don't know what I'm doing wrong here. Semi-colons are there. Strings
have newlines. Syntax looks like pp. 147 & 153 in Programming Perl.

??
--
fred
From: J�rgen Exner on 31 Mar 2010 16:08
Phred Phungus <Phred(a)example.invalid> wrote:
>J�rgen Exner wrote:
>> Phred Phungus <Phred(a)example.invalid> wrote:
>>> J�rgen Exner wrote:
>>>> Phred Phungus <Phred(a)example.invalid> wrote:
>>>>> PerlFAQ Server wrote:
>>>>>> $string = "Placido P. Octopus";
>>>>>> $regex = "P.";
>>>>>>
>>>>>> $string =~ s/$regex/Polyp/;
>>>>>> # $string is now "Polypacido P. Octopus"
>>>>>>
>>>>>> Because "." is special in regular expressions, and can match any single
>>>>>> character, the regex "P." here has matched the <Pl> in the original
>>>>>> string.
>>>>>>
>>>>> Am I then correct that a period is not a character?
>>>> How did you come to that conclusion?
>>> Because the regex would then match the middle initial and the string
>>> would be "Polypacido Polyp Octopus".
>>
>> No, it wouldn't because there is no /g modifier in the s///-operation.
>
>$ perl perl1.pl
>Placido P. Octopus
>Polypacido P. Octopus
>Placido P. Octopus
>Placido P. Octopus
>$ cat perl1.pl
>#!/usr/bin/perl
>
>use strict;
>use warnings;
>
>my $string = "Placido P. Octopus\n";
>my $regex = "P.";
>print $string;
>$string =~ s/$regex/Polyp/;
>print $string;
>
>my $string2 = "Placido P. Octopus\n";
>print $string2;
>$string =~ s/$regex/Polyp/g;
>print $string2;
>$
>
>I don't know what I'm doing wrong here.

You showed us your code: good.
You showed us your output: good.
You did not tell us what output you expected nor how the actual output
is different from your expected output: very bad. An error description
should always include an "expected" section.

To me your program produces exactly the output I would have expect:
First the content of $string,
then the same string again after a single substitution,
then the content of $string2
then the identical content of $string2 again.

>Semi-colons are there. Strings have newlines.

Well, fine, but not really relevant.

>Syntax looks like pp. 147 & 153 in Programming Perl.

Well, yeah, perl didn't report a syntax error, so in all probability you
don't have one.

jue
From: Willem on 31 Mar 2010 16:09
Phred Phungus wrote:
) $ perl perl1.pl
) Placido P. Octopus
) Polypacido P. Octopus
) Placido P. Octopus
) Placido P. Octopus
) $ cat perl1.pl
) #!/usr/bin/perl
)
) use strict;
) use warnings;
)
) my $string = "Placido P. Octopus\n";
) my $regex = "P.";
) print $string;
) $string =~ s/$regex/Polyp/;
) print $string;
)
) my $string2 = "Placido P. Octopus\n";
) print $string2;
) $string =~ s/$regex/Polyp/g;
) print $string2;
) $
)
) I don't know what I'm doing wrong here. Semi-colons are there. Strings
) have newlines. Syntax looks like pp. 147 & 153 in Programming Perl.

You made a typo, that's what.
For education, I'll let you figure it out yourself.

PS: Good thing that you copy-pasted the code. That shows very nicely
why copy-pasting is the only way to go when asking usenet questions.


SaSW, Willem
--
Disclaimer: I am in no way responsible for any of the statements
made in the above text. For all I know I might be
drugged or something..
No I'm not paranoid. You all think I'm paranoid, don't you !
#EOT
From: sln on 31 Mar 2010 16:11
On Wed, 31 Mar 2010 13:56:17 -0600, Phred Phungus <Phred(a)example.invalid> wrote:

>I don't know what I'm doing wrong here.

>>>>>> Because "." is special in regular expressions, and can match any single
>>>>>> character

What is it you are doing?

$hhh = "ABCDE";
$hhh =~ s/././g;

print $hhh;
From: Phred Phungus on 31 Mar 2010 16:45
Willem wrote:
> Phred Phungus wrote:

> ) my $string2 = "Placido P. Octopus\n";
> ) print $string2;
> ) $string =~ s/$regex/Polyp/g;
> ) print $string2;
> ) $
> )
> ) I don't know what I'm doing wrong here. Semi-colons are there. Strings
> ) have newlines. Syntax looks like pp. 147 & 153 in Programming Perl.
>
> You made a typo, that's what.
> For education, I'll let you figure it out yourself.
>
> PS: Good thing that you copy-pasted the code. That shows very nicely
> why copy-pasting is the only way to go when asking usenet questions.

Uff.

$ perl perl1.pl
Placido P. Octopus
Polypacido P. Octopus
Placido P. Octopus
Polypacido Polyp Octopus
$ cat perl1.pl
#!/usr/bin/perl

use strict;
use warnings;

my $string = "Placido P. Octopus\n";
my $regex = "P.";
print $string;
$string =~ s/$regex/Polyp/;
print $string;

my $string2 = "Placido P. Octopus\n";
print $string2;
$string2 =~ s/$regex/Polyp/g;
print $string2;
$ perl1.pl
bash: perl1.pl: command not found
$

Ok the regex part of it works, now, but I've got one more little thing
that doesn't really have to do with perl properly, but since the context
is here ...

I thought the idea for the shebang line was that my OS was going to
interpret the script with perl. I do have a /usr/bin/perl, so I thought
that typing perl1.pl at the prompt would be the same as typing perl
perl1.pl, but manifestly, it is not. What gives?
--
fred
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