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Hi group -

Can someone tell me why this command is adding a question mark to the file name:

#!/usr/bin/perl
$month=`/bin/date | awk '{print \$2\$6}'`;

if (blah blah blah) {
copy("total.txt","total.$month");
}



[root]# ls -l total.*
-rw-rw-r-- 1 root root 13 Oct 8 12:31 total.Oct2003?
-rw-rw-rw- 1 nobody nobody 13 Oct 8 07:35 total.txt
[root@alert StandardsAlert]# more total.Oct2003?
1|0|0|0|0|1|1
[root@alert StandardsAlert]# more total.Oct2003
total.Oct2003: No such file or directory


THANK YOU!

Syl.

Malcolm Dew-Jones <> wrote:
> Sylvie Stone () wrote:
>
> The date command may be able to output the format you want without using
> awk.
>
> $ date '+%b%Y'
> Oct2003

Or don't use date at all:
use POSIX qw(strftime);
$month = strftime "%b%Y", localtime;

> : #!/usr/bin/perl
> : $month=`/bin/date | awk '{print \$2\$6}'`;
[...]
> Perhaps it isn't a `?', I will guess the ? is a place holder to indicate a
> control character in the name, such as a new-line from the end of the awk
> output.

Indeed:
$month=`/bin/date | awk '{print \$2\$6}'`;
$len = length $month;
print "'$month' is $len characters long\n";


--
Glenn Jackman
NCF Sysadmin

Sylvie Stone () wrote:
: Hi group -

: Can someone tell me why this command is adding a question mark to the
file name:

The date command may be able to output the format you want without using
awk.

$ date '+%b%Y'
Oct2003


: #!/usr/bin/perl
: $month=`/bin/date | awk '{print \$2\$6}'`;

: if (blah blah blah) {
: copy("total.txt","total.$month");
: }


: [root]# ls -l total.*
: -rw-rw-r-- 1 root root 13 Oct 8 12:31 total.Oct2003?


Perhaps it isn't a `?', I will guess the ? is a place holder to indicate a
control character in the name, such as a new-line from the end of the awk
output.

Try chomp($month)

: [root@alert StandardsAlert]# more total.Oct2003?
: 1|0|0|0|0|1|1

That may work because ? is a wild card, so it matches any character (i.e.
a control character) in the file name, not because the filename has a
literal ? .

"Sylvie Stone" <> wrote in message ...
> Hi group -
>
> Can someone tell me why this command is adding a question mark to the file
name:
>
> #!/usr/bin/perl
> $month=`/bin/date | awk '{print \$2\$6}'`;

**** chomp($month); # maybe the ? is a newline.
>
> if (blah blah blah) {
> copy("total.txt","total.$month");
> }
>
>
>
> [root]# ls -l total.*
> -rw-rw-r-- 1 root root 13 Oct 8 12:31 total.Oct2003?
> -rw-rw-rw- 1 nobody nobody 13 Oct 8 07:35 total.txt
> [root@alert StandardsAlert]# more total.Oct2003?
> 1|0|0|0|0|1|1
> [root@alert StandardsAlert]# more total.Oct2003
> total.Oct2003: No such file or directory
>
>
> THANK YOU!
>
> Syl.

Glenn Jackman <> wrote in message news:<>...
> Malcolm Dew-Jones <> wrote:
> > Sylvie Stone () wrote:
> >
> > The date command may be able to output the format you want without using
> > awk.
> >
> > $ date '+%b%Y'
> > Oct2003
>
> Or don't use date at all:
> use POSIX qw(strftime);
> $month = strftime "%b%Y", localtime;
>
> > : #!/usr/bin/perl
> > : $month=`/bin/date | awk '{print \$2\$6}'`;
> [...]
> > Perhaps it isn't a `?', I will guess the ? is a place holder to indicate a
> > control character in the name, such as a new-line from the end of the awk
> > output.
>
> Indeed:
> $month=`/bin/date | awk '{print \$2\$6}'`;
> $len = length $month;
> print "'$month' is $len characters long\n";


Thank you all for the reply's. I ended up using the POSIX commend and
it's works great. Thanks -

Syl.