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 .

In reply to Re: How to return $ref from: $ref = $sth->fetchall_arrayref; by ioannis
in thread How to return $ref from: $ref = $sth->fetchall_arrayref; by hipnoodle

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