in reply to Data Set Combination
If so, a solution is straightforward. You'll find elements in the first hash but not the second as you iterate over it. If you delete elements from the second hash as you find them, anything left in the second hash will be items that weren't in the first hash.
If there can be a many-to-one correspondence between the two, you can pull the first array into a hash with selectall_hashref, find elements in the first but not the second while iterating over the first array, then iterate over the second array to search for items which aren't in the first.
You can also have the SQL server do most of the work for you by JOINing the tables in an appropriate way. For example, a SELECT ... FROM table1 LEFT JOIN table2 ... will return all elements of table1, where possible joining the data from table2, and where not replacing it with null. SELECT ... FROM table2 LEFT JOIN table1 WHERE ... AND table1.something IS NULL will return all items in table2 with no corresponding row in table1. Maybe a better SQL hacker than I would know one query to efficiently do the whole thing.
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Re^2: Data Set Combination
by Anonymous Monk on Aug 26, 2005 at 20:07 UTC | |
by sgifford (Prior) on Aug 26, 2005 at 20:21 UTC |