Escape from backslash hell

[Static]

greetings monks --

trying to convert string '11/29/2001' to string '11\/29\/2001'.

i have attempted several conversions to no avail:

$dt =~ s/\//\\\//g
$dt =~ s{/}{\/}g
$dt =~ s{/}{\\/}g

all of my attempts thus far have failed to produce the desired result, all return either the original string or 11\\/29\\/2001.

i know this is an easy one, but it's trapped behind the cobwebs of many beers in my mind.

thanks for the help.

edited by footpad, ~Fri Nov 30 21:16:31 2001 (GMT)

[Rich36]

Try using a different delimiter for the substitution other than "/". That tends to help cut down confusion when working with patterns that contain "/"'s.

my $dt = '11/29/2001';
$dt =~ s|\/|\\/|g;
print qq($dt);


Rich36

---

There's more than one way to screw it up...


---

[tilly]

If you are going to avoid "leaning toothpicks" by using a different delimiter, then you might as well drop all of the useless ones:

s:/:\\/:g;

(Note that I don't like using metacharacters which are meaningful within a substitution. Hence my avoiding the pipes.)

[patgas]

Hmm, I'm not sure... I can't reproduce your problem...

$_ = '11/29/2001';

# First I tried this...
# s,/,\\/,g;

s{/}{\\/}g;

print;

On my system, at least, both substitutions print 11\/29\/2001.

"We're experiencing some Godzilla-related turbulence..."

[Purdy]

I know several other Monks will jump on this, too, but here's my answer. :)

#!/usr/bin/perl -w

use strict;

my $date = '11/29/2001';

print "DATE (before): $date\n";

$date =~ s/\//\\\//g;

print "DATE (after): $date\n";

Output:
$ ./slash.pl
DATE (before): 11/29/2001
DATE (after): 11\/29\/2001

Jason

[lestrrat]

Eh, if you just want to convert the '/'s to '\/'s,

my $str = '11/29/2001';
$str =~ s{/}{\\/}g;
print $str, "\n";

This seems to work for me...

[Static]

this just in: i am a tool. i was using the debugger for this and after all attempts 'x $dt' was returning '11\\/29\\/2001', but if i print the variable it comes out golden. thx for the help.

[dws]

trying to convert string '11/29/2001' to string '11\/29\/2001'

$converted = join '\/', split '/', '11/29/2001';

[chipmunk]

all of my attempts thus far have failed to produce the desired result, all return either the original string or 11\\/29\\/2001.

I wonder if you're checking the results in the debugger... Be aware that if you examine the contents of a variable (with x, X, or V), the debugger will escape quotes and backslashes, in order to produce a valid quoted string. However, if you print the contents of a variable (with p or print), you get just the contents.

Here's an example, showing that your second attempt works, although the debugger's output might be a bit misleading:

  DB<1> $_ = "2001/11/30"

  DB<2> s{/}{\\/}g

  DB<3> x $_
0  '2001\\/11\\/30'
  DB<4> p $_
2001\/11\/30
  DB<5>

[crazyinsomniac]

Wow, at least 4 replies and all are off base. When you want to substitute characters with different ones you need to use tr aka y (see perlop).

#untested but should work
$dt =~ y{/}{\\/};
#or
$dt =~ tr{/}{\\/};

update: Bravo! You are soooo right! Glad to see some quality control here. I don't know why that "escaped" me, but hey, it is a valuable learning experience to realize that not all monks know what they are talking, and that you have to check out their references (like I should've done). Live and learn, and then go crazy.

 
___crazyinsomniac_______________________________________
Disclaimer: Don't blame. It came from inside the void
perl -e "$q=$_;map({chr unpack qq;H*;,$_}split(q;;,q*H*));print;$q/$q;"

[impossiblerobot]

I don't think this will work, since (AFAIK) tr can only translate a single character to another single character.

Impossible Robot