Re^8: Why is my code assigning the last-retrieved value to all elements in my hash?

Thanks a heap for taking time to deal with this completely and in detail.

I do believe that the truly salient point is this:

There's one other misconception you seem to have that I'd like to clear up. You can't have a $b{foo}{bar} and have $b{foo} set to some unrelated value (such as 1). If there is a $b{foo}{bar}, then $b{foo} must somehow be a reference to a hash where you access ->{bar}. In your original code, that "reference" was "1", treated as a symbolic reference to %1. In the code I suggested, it's a hard reference to an anonymous hash that gets printed as 'HASH(0x8529f88)'.

This appears to be the real root of the problem. I started with a one-dimensional hash, then extended it to two dimensions. $booklist_1{$book_id} therefore becomes a reference to a hash and I can no longer use it the way I was trying to.

On this basis, I changed it to

while ($book_id = $sth->fetchrow_array()) {
    $booklist_1{$book_id}{'book_id'} = 1;
}
[download]

and that works perfectly.

Time flies like an arrow. Fruit flies like a banana.

Comment on Re^8: Why is my code assigning the last-retrieved value to all elements in my hash? Download Code

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Re^9: Why is my code assigning the last-retrieved value to all elements in my hash? by kyle (Abbot) on Jul 10, 2008 at 03:30 UTC
Unless there's some reason for wanting '1', I'd put the actual book ID in the 'book_id' slot of that hash. `while (($book_id) = $sth->fetchrow_array()) { $booklist_1{$book_id}{'book_id'} = $book_id; }` [download] Unless you have a book ID of '0', it will still be true in boolean tests, and it will be less confusing to future programmers.	[reply] [d/l]
Re^10: Why is my code assigning the last-retrieved value to all elements in my hash? by punch_card_don (Curate) on Jul 10, 2008 at 12:30 UTC
Getting too squished on my screen - see bottom of thread for my conclusion in the light of the next day.... Time flies like an arrow. Fruit flies like a banana.	[reply]


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