Thank you so much - that has done the trick!!
I guess this is a common mistake that people who don't take the time to read the
manual on references make!!
Thanks again,
Simon.
Sudarshan
Raghavan To: Perl beginners
<[EMAIL PROTECTED]>
<[EMAIL PROTECTED] cc:
.com> Subject: RE: Array of Hashes
19/09/2002 23:35
On Wed, 18 Sep 2002, Simon Tomlinson wrote:
>
>
> Apologies, that was bad typing in my email. I rearrange the algorithm to make it
>easier to read. Here is my exact source and the exact output. Even without the
>space there, it still doesn't work!!
>
> Any ideas?
> Simon.
>
>
> sub getEvents
> {
> my @rows = &getDBTable($tbl_events);
> my @colDefs = ("");
> @colDefs = split(",",substr($rows[0], index($rows[0],":")+1));
> my $rowCount = -1;
> my @rowCols;
>
> foreach $row (@rows)
> {
> if ($rowCount > -1)
> {
> my %tempCols = &getCols($row, @colDefs);
>
> foreach $key (keys %tempCols)
> {
> print "WORKS: $key $tempCols{$key}<br>\n";
> }
> $rowCols[$rowCount] = %tempCols;
You are assigning a hash to a scalar here, you will have to assign the
reference of the hash instead.
$rowCols[$rowCount] = \%tempCols;
> }
>
> my %event = $rowCols[$rowCount];
And change this too
my %event = %{$rowCols[$rowCount]};
or
remove the my line and change the foreach as
foreach $key (keys %{rowCols[$rowCount]})
> foreach $key (keys %event)
> {
> print "DOESN'T WORK: $key $event{$key}<br>\n";
> }
>
> $rowCount = $rowCount + 1;
> }
> return @rowCols;
> }
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