Problem with Perl script that creates directories with names

[vlsi_fanatic]

Help with perl

Hi,
i am new to perl. I got a problem. I am writing a script to create a
directory whose name will be generated in the script. for example one
of the directory name I need to create will be 9.1.3C I am getting
this value in a variable ($dirname) and I am using
print `mkdir C:/perl/$dirname` command to create it. Since the value
includes a Charecter perl is giving me syntax error. If the value is
just 9.1.3 the print line is working fine. But with that charecter at
the end I am getting problems. Can anyone solve my problem?

Thanks

 

[nand_gates]

Re: Help with perl

Try this one instead

system("mkdir C:/perl/$dirname");

Hope this helps!!

 

[vlsi_fanatic]

Re: Help with perl

again I am getting the same error

 

[jammed]

Re: Help with perl

make sure your directory name is enclosed in quotes.

$dirname = "9.1.3C";

hope this helps.

 

[disney]

Re: Help with perl

have it in quotes

 

[kovalja]

Re: Help with perl

Hi,

when I see your code, I'm really a little bit confused. Shouldn't you use back-slash instead of slash in Win directory path?

Let me know if windows changed the command line behaviour

BTW. If you want to create the directory structure similar you presented, try this:

system( "mkdir C:\\perl\\$dirname" );

HTH,
kovalja

 

[jarodz]

Help with perl

Hi,

Perl had already implemented "mkdir" function.
Here is a good old book can be found on the google.
https://affy.blogspot.com/p5be/ch09.htm

Sincerely,
Jarod

 

[vlsi_fanatic]

Re: Help with perl

Hi,

when I see your code, I'm really a little bit confused. Shouldn't you use back-slash instead of slash in Win directory path?

Let me know if windows changed the command line behaviour

BTW. If you want to create the directory structure similar you presented, try this:

system( "mkdir C:\\perl\\$dirname" );

HTH,
kovalja

Windows didn't change the command line behavior. In this way you don't need to escape the backslash in the path. Try using it. It works.

Added after 1 minutes:

make sure your directory name is enclosed in quotes.

$dirname = "9.1.3C";

hope this helps.

If I put it in Quotes it works.

Thanks.