Re: How to return $ref from: $ref = $sth->fetchall

in reply to How to return $ref from: $ref = $sth->fetchall_arrayref;

I followed your code, and everything is working fine. Data are returned to $ref for outside processing.

Check if your $sqlstatement is properly quoted

I doubt there could possibly be a problem with $sth being out of scope: the statement $a = $b does not copy pointers, it copies the actual data of $b .

Comment on Re: How to return $ref from: $ref = $sth->fetchall_arrayref;

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Re^2: How to return $ref from: $ref = $sth->fetchall_arrayref; by hipnoodle (Novice) on Mar 20, 2006 at 13:43 UTC
Thanks for your help. I persisted, changed the for loop to foreach and it worked. (crazy). Anyway, thanks for letting me know I was on the right track.	[reply]

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