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Full Discussion: perl reg expression
Top Forums Shell Programming and Scripting perl reg expression Post 302191441 by talashil on Friday 2nd of May 2008 03:04:25 PM
Old 05-02-2008
perl reg expression

I have regular expression like this ( replace + with \+)

($mod_server) = ($server =~ /\+/\\+/g);


the above is failing with error . what's wrong with it .

Thanks
 

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regex(3)						     Library Functions Manual							  regex(3)

Name
       re_comp, re_exec - regular expression handler

Syntax
       char *re_comp(s)
       char *s;

       re_exec(s)
       char *s;

Description
       The  subroutine	compiles  a string into an internal form suitable for pattern matching.  The subroutine checks the argument string against
       the last string passed to

       The subroutine returns 0 if the string s was compiled successfully; otherwise a string containing an  error  message  is  returned.  If	is
       passed 0 or a null string, it returns without changing the currently compiled regular expression.

       The  subroutine returns 1 if the string s matches the last compiled regular expression, 0 if the string s failed to match the last compiled
       regular expression, and -1 if the compiled regular expression was invalid (indicating an internal error).

       The strings passed to both and may have trailing or embedded newline characters; they are terminated by	nulls.	 The  regular  expressions
       recognized are described in the manual entry for given the above difference.

Diagnostics
       The subroutine returns -1 for an internal error.

       The subroutine returns one of the following strings if an error occurs:

       No previous regular expression
       Regular expression too long
       unmatched (
       missing ]
       too many () pairs
       unmatched )

See Also
       ed(1), ex(1), egrep(1), fgrep(1), grep(1)

																	  regex(3)
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