WITH q AS
(
SELECT (
SELECT MIN(start_date)
FROM mytable
) + level - 1 AS mydate
FROM dual
CONNECT BY
level <= (
SELECT MAX(end_date) - MIN(start_date)
FROM mytable
)
)
SELECT group, mydate,
(
SELECT COUNT(*)
FROM mytable mi
WHERE mi.group = mo.group
AND q BETWEEN mi.start_date AND mi.end_date
)
FROM q
CROSS JOIN
(
SELECT DISTINCT group
FROM mytable
) mo
Bijwerken:
Een betere en snellere zoekopdracht door gebruik te maken van analytische functies.
Het belangrijkste idee is dat het aantal bereiken met elke datum het verschil is vóór het aantal bereiken dat vóór die datum begon en het aantal bereiken dat ervoor eindigde.
SELECT cur_date,
grouper,
SUM(COALESCE(scnt, 0) - COALESCE(ecnt, 0)) OVER (PARTITION BY grouper ORDER BY cur_date) AS ranges
FROM (
SELECT (
SELECT MIN(start_date)
FROM t_range
) + level - 1 AS cur_date
FROM dual
CONNECT BY
level <=
(
SELECT MAX(end_date)
FROM t_range
) -
(
SELECT MIN(start_date)
FROM t_range
) + 1
) dates
CROSS JOIN
(
SELECT DISTINCT grouper AS grouper
FROM t_range
) groups
LEFT JOIN
(
SELECT grouper AS sgrp, start_date, COUNT(*) AS scnt
FROM t_range
GROUP BY
grouper, start_date
) starts
ON sgrp = grouper
AND start_date = cur_date
LEFT JOIN
(
SELECT grouper AS egrp, end_date, COUNT(*) AS ecnt
FROM t_range
GROUP BY
grouper, end_date
) ends
ON egrp = grouper
AND end_date = cur_date - 1
ORDER BY
grouper, cur_date
Deze zoekopdracht wordt voltooid in 1
tweede op 1,000,000
rijen.
Zie dit bericht in mijn blog voor meer details: