If I'm working with two hashes which are actually HoH how would that work? This
is the snippet of code I'm working with:
sub timesheet {
my ($dept, $env) = @_;
if (exists $dept->{username}) {
open TIMESHEET,
">/work_reports/user/ops_timesheet_weekof_$endDate.txt";
}else{
open TIMESHEET,
">/work_reports/user/eng_timesheet_weekof_$endDate.txt";
}
print TIMESHEET "Timesheet for $startDate to $endDate\n\n\n";
foreach my $environ (sort keys %$dept) {
#Print the header for our data
print TIMESHEET "$environ", "\n";
printf TIMESHEET "%10s%8s\n", "User", "hh:mm";
print TIMESHEET ("-" x 30);
print TIMESHEET "\n";
foreach my $name (sort keys %${ $dept->{$environ} }) {
printf TIMESHEET "%10s%8s\n", "$name",
"$dept->{$environ->{$name}}";
}
printf TIMESHEET ("-" x 30);
print TIMESHEET "\n";
printf TIMESHEET "%18s\n\n", "$env->{$environ}";
}
close TIMESHEET;
}
Does that look right?
Mathew
Keep up with me and what I'm up to: http://theillien.blogspot.com
Chas Owens wrote:
> On 5/12/07, Mathew Snyder <[EMAIL PROTECTED]> wrote:
>> When passing two hashes into a subroutine how do I use them separately
>> if they
>> are placed into one flat list?
>>
>> Mathew
>> --
>> Keep up with me and what I'm up to: http://theillien.blogspot.com
>
> Use references:
>
> func(\%h1, \%h2);
>
> sub func {
> my ($h1, $h2) = @_;
>
> print "h1\n";
> for my $key (%$h1) {
> print "$key => $h1->{$key}\n";
> }
>
> print "h2\n";
> for my $key (%$h2) {
> print "$key => $h2->{$key}\n";
> }
> }
>
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