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Move most /contrib README files into SGML. Some still need conversion

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Bruce Momjian
2007-11-10 23:30:46 +00:00
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contrib/intagg/README.int_aggregate
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Integer aggregator/enumerator.
Many database systems have the notion of a one to many table.
A one to many table usually sits between two indexed tables,
as:
create table one_to_many(left int, right int) ;
And it is used like this:
SELECT right.* from right JOIN one_to_many ON (right.id = one_to_many.right)
WHERE one_to_many.left = item;
This will return all the items in the right hand table for an entry
in the left hand table. This is a very common construct in SQL.
Now, this methodology can be cumbersome with a very large number of
entries in the one_to_many table. Depending on the order in which
data was entered, a join like this could result in an index scan
and a fetch for each right hand entry in the table for a particular
left hand entry.
If you have a very dynamic system, there is not much you can do.
However, if you have some data which is fairly static, you can
create a summary table with the aggregator.
CREATE TABLE summary as SELECT left, int_array_aggregate(right)
AS right FROM one_to_many GROUP BY left;
This will create a table with one row per left item, and an array
of right items. Now this is pretty useless without some way of using
the array, thats why there is an array enumerator.
SELECT left, int_array_enum(right) FROM summary WHERE left = item;
The above query using int_array_enum, produces the same results as:
SELECT left, right FROM one_to_many WHERE left = item;
The difference is that the query against the summary table has to get
only one row from the table, where as the query against "one_to_many"
must index scan and fetch a row for each entry.
On our system, an EXPLAIN shows a query with a cost of 8488 gets reduced
to a cost of 329. The query is a join between the one_to_many table,
select right, count(right) from
(
select left, int_array_enum(right) as right from summary join
(select left from left_table where left = item) as lefts
ON (summary.left = lefts.left )
) as list group by right order by count desc ;