Question

Simple question:

How do I do this on one line:

my $foo = $bar->{baz}; fizz(\$foo); 

I’ve tried \$bar->{baz}, \${$bar->{baz}}, and numerous others. Is this even possible?

-fREW

Update: Ok, the hashref is coming from DBI and I am passing the scalar ref into template toolkit. I guess now that I look more closely the issue is something to do with how TT does all of this. Effectively I want to say:

$template->process(\$row->{body}, $data); 

But TT doesn’t work that way, TT takes a scalar ref and puts the data there, so I’d have to do this:

$template->process(\$row->{body}, $shopdata, \$row->{data}); 

Anyway, thanks for the help. I’ll at least only have one reference instead of two.

Answer 1

\$bar->{baz} 

should work.

E.g.:

my $foo; $foo->{bar} = 123;  my $bar = \$foo->{bar};  $$bar = 456;  print '$foo->{bar}\n';   # prints '456' 

In answer to the update in the OP, you can do:

\@$row{qw(body data)}; 

This is not the same as \@array, which would create one reference to an array. The above will distribute the reference and make a list of two references.