cannot use awk in perl module??

Hi,

I am trying to fetch the folder names under main_wa using a perl module. For some reason when i use opendir () is not working..
$filename = "/.iwmnt/EMEA/main/Intranet/...."
opendir ( DIR, "$filename" )...
it output cannot open the dir.. . (though the permssion are set properly..)
Then i tried another logic
@web_dir = qx( ls -l $filename | grep ^d |awk '{print \$9}' );
now it displays list of all files under $filename.. the grep and awk function is not working..

Can somebody help me out.. its very urgent

Sara...

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nipper

Check your path. Depending on the flavor of Unix awk may be in /usr/sbin which may not be picked up.

Do which awk from the command line and then hard code the path in your command

Also, IIRC, qx uses /bin/sh so be certain your syntax does not require csh/ksh/etc

HTH

Andy

Adam Stoller

Hi,

I am trying to fetch the folder names under main_wa using a perl module. For some reason when i use opendir () is not working..
$filename = "/.iwmnt/EMEA/main/Intranet/...."
opendir ( DIR, "$filename" )...
it output cannot open the dir.. . (though the permssion are set properly..)
Then i tried another logic
@web_dir = qx( ls -l $filename | grep ^d |awk '{print \$9}' );
now it displays list of all files under $filename.. the grep and awk function is not working..

Can somebody help me out.. its very urgent

Sara...

First of all - there is no reason to invoke awk within Perl - it's inefficient at best and certainly not portable.

Second - learn to name your variables correctly: "$filename" implies that you are providing an actual file name, possibly without a path. "$dir", "$directory", "$path", "$vpath" are all much better choices for a variable that contains the path to a directory.

Third - learn to use better error handling:

...
my $dir = "/.iwmnt/EMEA/main/Intranet/....";
if (! opendir(DIR, $dir)) {
    die $!; # ERROR - do something to handle it
}
my @subdirs = grep { -d "$dir/$_" && !/^\.+$/ } readdir(DIR);
closedir(DIR);
print "$dir\n\t" . join("\n\t", @subdirs) . "\n";

The above should get you a list of all subdirectories directly within $dir (no recursion performed).

There are numerous alternatives and/or ways of augmenting the above, depending on what you actually are trying to do - but the above should at least get you started in the right direction.

tsmaster

Thanks you so much Ghoti.. I have taken all your points..

I have mentioned the sample piece of code, tats y the variable names are inappropriate.

Moreover, the grep { -d "$dir/$_" && !/^\.+$ } gives error and then i modified to { -d "$dir/$_" && !/^(\.)(.*)$/ }.. it worked now..

Thanks a lot.... again

But do you know the reason, why the awk is not working ....??

Regards,
Sara...

Adam Stoller

Moreover, the grep { -d "$dir/$_" && !/^\.+$ } gives error and then i modified to { -d "$dir/$_" && !/^(\.)(.*)$/ }.. it worked now..

The error was simpler than that - I accidently left off the last slash:

 ... grep { -d "$dir/$_" && !/^\.+$/ } ...

a simple, but obviously serious, typo on my part. Sorry about that.