RE: Reg. string matching using reg-exp

2004-02-06 Thread Tim Johnson 

It looks like there is no '\r\n' at the end of $line.



 

-Original Message-
From: Balaji Thoguluva [mailto:[EMAIL PROTECTED] 
Sent: Friday, February 06, 2004 11:45 AM
To: [EMAIL PROTECTED]
Subject: Reg. string matching using reg-exp


 
#!/usr/bin/perl -w
my $line= 'INVITE sip:[EMAIL PROTECTED] SIP/2.0'; if($line =~
/^\s*(\w+)\s*(.*)\s+(sip\/\d+\.\d+)\s*\r\n/i)

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Re: Reg. string matching using reg-exp

2004-02-06 Thread wolf blaum 
For Quality purpouses, Balaji Thoguluva 's mail on Friday 06 February 2004 
20:44 may have been monitored or recorded as:
 Hi,
  I am a novice to perl programming. When I execute the following code,
 I get always No Match. I guess my reg-exp is correct. I also tried
 changing $line= INVITE sip:[EMAIL PROTECTED] SIP/2.0; to have double
 quotes and a backslash char before @ symbol. Even then it gives me No
 Match. I appreciate your help.

 #!/usr/bin/perl -w
 my $line= 'INVITE sip:[EMAIL PROTECTED] SIP/2.0';
 if($line =~ /^\s*(\w+)\s*(.*)\s+(sip\/\d+\.\d+)\s*\r\n/i)

Do you have the \r\n at the end of $line?
Wolf


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RE: Reg. string matching using reg-exp

2004-02-06 Thread Tim Johnson 
Use the 's' option at the end of your regex after the closing '/'.
 
$var =~ /match\nsomething\nelse/s;
 
read 'perldoc perlre' for more about regexes.
 
Also, please reply to the list next time, because I might not be at my
desk or able to reply, and someone  else probably will.


  _  

From: Balaji Thoguluva [mailto:[EMAIL PROTECTED] 
Sent: Friday, February 06, 2004 12:07 PM
To: Tim Johnson
Subject: RE: Reg. string matching using reg-exp


Thanks Tim Johnson. I removed the /r/n from the reg-ex and it works. I
have another question. How to assign a multiline string or string having
many lines(strings having \n) to a $string-variable?. In C, there is a
\ operator.
 
Thanks for your help,
Balaji

RE: Reg. string matching using reg-exp

2004-02-06 Thread Jeff 'japhy' Pinyan 
On Feb 6, Balaji Thoguluva said:

Thanks Tim Johnson. I removed the /r/n from the reg-ex and it works. I
have another question. How to assign a multiline string or string having
many lines(strings having \n) to a $string-variable?. In C, there is a
\ operator.

You don't need to do anything special in Perl.

$string = This is a
very long string
that spans
many lines;

Or you can use a 'here-doc'.
$string =  END OF STRING;
this is a very
long string that
spans many lines
END OF STRING


-- 
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RPI Acacia brother #734   http://www.perlmonks.org/   http://www.cpan.org/
stu what does y/// stand for?  tenderpuss why, yansliterate of course.
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Re: Reg. string matching using reg-exp

2004-02-06 Thread Rob Dixon 
Balaji thoguluva wrote:

  I am a novice to perl programming. When I execute the following code, I get 
 always No Match. I guess my reg-exp is correct.
I also tried changing $line= INVITE sip:[EMAIL PROTECTED] SIP/2.0; to have double 
quotes and a backslash char before @ symbol.
Even then it gives me No Match. I appreciate your help.

 #!/usr/bin/perl -w
 my $line= 'INVITE sip:[EMAIL PROTECTED] SIP/2.0';
 if($line =~ /^\s*(\w+)\s*(.*)\s+(sip\/\d+\.\d+)\s*\r\n/i)
 {
 print \nSIP Method: ,$1;
 print \nRequest URI: ,$2;
 print \nSIP Version: ,$3;
 }
 else
 {
 print No Match;
 }

The string you show, as others have pointed out, has no \r\n terminator.
Even if you're pulling your record from a text file, Perl will try hard
to change all platforms' line terminators to a simple \n in the record
you actually read. And even having said that, there's no reason to match /all/
of the string if you're just extracting sub-fields: it's unlikely to help to
be told that your record doesn't actually end in /\s*\r\n/. I think your regex
should look like this:

  my $line= 'INVITE sip:[EMAIL PROTECTED] SIP/2.0';
  $line =~ /^\s*(\w+)\s+(.*)\s+(sip\/\d+\.\d+)/i;

  print map $_\n, $1, $2, $3;

**OUTPUT

  INVITE
  sip:[EMAIL PROTECTED]
  SIP/2.0

HTH,

Rob



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Re: Reg. string matching using reg-exp

2004-02-06 Thread Rob Dixon 
Jeff 'Japhy' Pinyan wrote:

 On Feb 6, Balaji Thoguluva said:

 Thanks Tim Johnson. I removed the /r/n from the reg-ex and it works. I
 have another question. How to assign a multiline string or string having
 many lines(strings having \n) to a $string-variable?. In C, there is a
 \ operator.

 You don't need to do anything special in Perl.

 $string = This is a
 very long string
 that spans
 many lines;

 Or you can use a 'here-doc'.
 $string =  END OF STRING;
 this is a very
 long string that
 spans many lines
 END OF STRING

In C, newlines have to be introduced explicitly as \n. A literal
newline character (the end of a source record) has to be escaped to
make it 'vanish', otherwise it should throw a compilation error.

In Perl:

my $string = One
Two
Three
;

In C:

char *string = One\n\
Two\n\
Three\n\
;

or, because consecutive C string constants are implicitly concatenated:

  char *string =
One\n
Two\n
Three\n
;

HTH,

Rob




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Re: Reg. string matching using reg-exp

2004-02-06 Thread david 
Rob Dixon wrote:

 In C, newlines have to be introduced explicitly as \n. A literal
 newline character (the end of a source record) has to be escaped to
 make it 'vanish', otherwise it should throw a compilation error.
 
 In Perl:
 
 my $string = One
 Two
 Three
 ;
 
 In C:
 
 char *string = One\n\
 Two\n\
 Three\n\
 ;
 
 or, because consecutive C string constants are implicitly concatenated:
 
   char *string =
 One\n
 Two\n
 Three\n
 ;

or don't quote them is you have an ANSI C compiler:

#define TEST(s)  printf(%s\n,#s)

int main(int argc,char* argv[]){

TEST(\n
\n
\n
see you!);
}

prints:




see you!

david
-- 
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sub'__{print@_ ;* \ = * ___ ,\  \}
sub'___{print@_ ;* \ = *  ,\  \}
sub'{print@_,\n}{_+Just}(another)-(Perl)-(Hacker)

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