search . and replace with /../

[Raju Sathliya]

I have writen a small script to replace the . with /../
which is as follows.

$var = "DataAccess\Dat aAccess.";
$var =~ s/(.)/(\)..(\)/;
print "value is : $var";

But in this case when I run this script it gives me result as :
value is : ()..()ataAccess DataAccess.

which not as per my expectation.

I want the result some thing like this : value is : value is : DataAccess\Data Access\..\

So what changes should i make in my current script to get desire result.
Also why it is removing '\' character when run the script.

-Thanks,
Raju

[gpraghuram]

. is a special character and when u want to replace that you shuld use escape character for that like
[code=perl]
$val =~ s/\./\.\./g;

[/code]

Raghu

[Raju Sathliya]

. is a special character and when u want to replace that you shuld use escape character for that like
[code=perl]
$val =~ s/\./\.\./g;

[/code]

Raghu

Thanks Raghu for your usefull solution...
But One problem is still there..when i run the following script

$var = "DataAccess\Dat aAccess.";
$var =~ s/\./\\.\./;

then output is : DataAccessDataA ccess\..

but in this output why '\' is removed ???....I want this in
output : DataAccess\Data Access\.. formate

Can you please suggest me some solution to maintain this \ in my output string.

Thanks,
Raju

[KevinADC]

you changed the regexp that Raghu posted for you, it should be:

s/\./\.\./g;

should not be:

s/\./\\.\./g;

Next time look at the code more carefully.

It can really be written like this:

$foo = 'this\that.';
$foo =~ s/\./../g;
print $foo;

the '.' has no special meaning on the replacement side of the regxp, only on the search side.

[gpraghuram]

the '.' has no special meaning on the replacement side of the regxpThis is a good learning for me.....

Raghuram

[KevinADC]

the '.' has no special meaning on the replacement side of the regxpThis is a good learning for me.....

Raghuram

The replacement side of s/// is treated like a double-quoted string, so variable expansion (for example $foo and @foo) and meta character expansion (for example \t and \n) do occur, but a dot in a double-quoted string is just a dot. So anything that can be interpolated/expanded in a double-quoted string will also be expanded/interplolated on the replacement side of s///. Everything else is treated literally. There is a way to alter that behavior however, namely using the "e" modifier on the end of the regexp.

[Raju Sathliya]

Thanks to all !!!!
Finaly solution is working for me....Great Forum !!!!!