Problem in sustitution of a string

Anonymous Monk has asked for the wisdom of the Perl Monks concerning the following question:

hi monks,
I have one string "/morse_value" that i want to convert it into "\morse_value"
for this i am using s/// operator and my code is as below.

$a="/morse_value";
$b=$a =~ s///\/g;
print "$a\n";
[download]

after running this i am getting error as below
Search pattern not terminated at x.pl line 3. solution for this will be a greate help for me.. thx in advace ....for reply.

Comment on Problem in sustitution of a string Download Code

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Re: Problem in sustitution of a string by GrandFather (Saint) on Aug 14, 2007 at 03:48 UTC
Actually, that is not the message I see (nor quite what I would expect). I see: `Backslash found where operator expected at noname.pl line 5, near "s// +/\" syntax error at noname.pl line 5, near "s///\" Search pattern not terminated at noname.pl line 5.` [download] The rule is: resolve the first error first. Perl is expecting to see either flag characters or a termination character such as ; - Perl doesn't expect to see a \, it makes no sense there. What were you trying to achieve with the regex substitution? Replace the / with \? Instead try: `$a =~ s!/!\\!g;` [download] Note the use if ! to delimit the regex so that / can be used unquoted, and also that \ needs to be quoted. BTW: avoid the use of $a and $b except in the special context of sort. DWIM is Perl's answer to Gödel	[reply] [d/l] [select]
Re: Problem in sustitution of a string by dynamo (Chaplain) on Aug 14, 2007 at 06:38 UTC
Your only real issue was that you have to backslash (verb) the slash (noun) and the backslash (noun) in that expression. Also, you weren't using $b, so I'd drop it: `$a = "/morse_value"; $a =~ s/\//\\/g; print "$a\n";` [download] Good luck.	[reply] [d/l]