Trying to incorporate variable data in system call?

[b0dhi]

I'm a newbie with Perl. I'm trying to make a system call to move some files to another directory, but the directory is based on a variable that I've created.

Here is a snippet of my code:

[code=perl]
my $newsubdir = $dir . $filedate;
mkdir($newsubdi r,0777) || die "cannot mkdir";

system("move C:\\test\\files \\*.log C:\\test\\files \\$newsubdir\\" );
[/code]

The $newsubdir doesn't seem to work inside the system call.. Can someone tell me what I'm doing wrong? Thanks in advance! :)

[numberwhun]

I'm a newbie with Perl. I'm trying to make a system call to move some files to another directory, but the directory is based on a variable that I've created.

Here is a snippet of my code:

[code=perl]
my $newsubdir = $dir . $filedate;
mkdir($newsubdi r,0777) || die "cannot mkdir";

system("move C:\\test\\files \\*.log C:\\test\\files \\$newsubdir\\" );
[/code]

The $newsubdir doesn't seem to work inside the system call.. Can someone tell me what I'm doing wrong? Thanks in advance! :)

Have you tried doing a print of the $newsubdir variable to ensure that it is getting set correctly? We don't have the rest of your code and I don't like making assumptions.

Regards,

Jeff

[new1234]

Hi, I'm actually having the same problem..
when I run:

print "making directory ";
print $directory;
print "\n";

mkdir($directory,0777) ||  print $!;

I get the output 'No such file or directory'.
I've tried this:

print "making directory ";
print $directory;
print "\n";
$directory="\"".$directory."\"";
print $directory;
print "\n";
mkdir($directory,0777) ||  print $!;

Error = No such file or directory
and this:

mkdir("$directory",0777) ||  print $!;

which creates the directory ./$directory ...
As you can see I've been checking that the variable prints out correctly. (Output for the first snippet of code is: "making directory ./testdir/testdir2/")
Any help would be really appreciated..

[new1234]

I think it's because I may or may not have subdirectories that I want to make too - I caved and used
[code=perl]
system "mkdir -p $directory" || print $!;
[/code]
which works very nicely.. I'd still be interested in any alternative solutions as this seems a bit clumsy to me.
Cheers
p.s. i'm also struggling to remove the ./$directory that i created above (oops). Command line returns "directory: Undefined variable" I've tried hashing out the $
rm \$directory
and
rm "$directory "
--> same result.. d'oh!

[numberwhun]

I think it's because I may or may not have subdirectories that I want to make too - I caved and used
[code=perl]
system "mkdir -p $directory" || print $!;
[/code]
which works very nicely.. I'd still be interested in any alternative solutions as this seems a bit clumsy to me.
Cheers
p.s. i'm also struggling to remove the ./$directory that i created above (oops). Command line returns "directory: Undefined variable" I've tried hashing out the $
rm \$directory
and
rm "$directory "
--> same result.. d'oh!

Question, are you just wanting to print the error if it doesn't work? It is more customary to use the die() function for that and not just a print statement. That way, processing can be stopped at point of failure.

Regards,

Jeff

[KevinADC]

this looks wrong:

$directory="\"" .$directory."\" ";

why do you want to add double-quotes around $directory? If $directory were: foo/bar it would equal "foo/bar" after being processed by the above line.