Re: SQL Questions

this should do it:

for (keys %$ref) {
  $data{$_} = $ref->{$_};
}
[download]

---- kurt

Comment on Re: SQL Questions Download Code

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Re^2: SQL Questions by tadman (Prior) on Jul 08, 2002 at 10:05 UTC
That was kind of a lacklustre answer, wasn't it? Either way, you could be doing this: `@data{keys %$ref} = values %$ref;` No `for` loop, one line, no fuss.	[reply] [d/l] [select]