我有以下查询来按 8 小时分段显示机器的性能（尽管它可以更改为 1 小时、1 周、1 个月等）。我的表有200万条记录，查询执行需要10秒。这是一个合理的速度，还是可以改进？
我有
tbl_piezacreate table tbl_pieza
(
id_nu_pieza integer not null,
id_nu_orden_fabricacion integer,
id_nu_referencia integer,
id_nu_operacion integer,
id_nu_maquina integer,
id_nu_usuario integer,
ind_paro integer,
ind_validada integer default 0,
nu_segundos integer,
dtm_inicio_at timestamp default CURRENT_TIMESTAMP not null,
dtm_fin_at timestamp,
ind_estatus integer default 1,
dtm_create_at timestamp,
dtm_update_at timestamp default CURRENT_TIMESTAMP,
ind_retrabajo integer default 0,
primary key (id_nu_pieza, dtm_inicio_at)
);
create index tbl_pieza_dtm_inicio_at_idx
on tbl_pieza (dtm_inicio_at desc);
create index idx_time_range
on tbl_pieza (dtm_inicio_at, dtm_fin_at);
WITH Rangos AS (
SELECT
generate_series(
'2023-05-22 16:23:14'::timestamp,
'2023-05-26 08:23:14'::timestamp,
'8 hour'::interval
) AS inicio,
generate_series(
'2023-05-23 00:23:14'::timestamp,
'2023-05-26 16:23:14'::timestamp,
'8 hour'::interval
) AS fin
),
PiezasPorIntervalo AS (
SELECT
r.inicio,
r.fin,
p.id_nu_operacion,
p.id_nu_maquina,
SUM(
CASE
WHEN EXTRACT(epoch FROM p.dtm_fin_at - p.dtm_inicio_at) = 0 THEN 0
ELSE GREATEST(0, EXTRACT(epoch FROM LEAST(r.fin, p.dtm_fin_at) - GREATEST(r.inicio, p.dtm_inicio_at)) / EXTRACT(epoch FROM p.dtm_fin_at - p.dtm_inicio_at))
END
) as PiezasReales
FROM Rangos r
JOIN tbl_pieza p ON p.dtm_inicio_at < r.fin AND p.dtm_fin_at > r.inicio
AND p.id_nu_usuario in (1,8,11,43,44,45,46,47,48,49)
AND p.id_nu_operacion in (84,85,86,87,88,89,90,91,92,93,118,119)
AND p.id_nu_referencia in (46,58,59,60)
AND p.id_nu_maquina in (1,2,3,8)
GROUP BY r.inicio, r.fin, p.id_nu_operacion, p.id_nu_maquina
)
SELECT
p.inicio as fecha_inicio,
p.fin as fecha_fin,
p.id_nu_maquina as id_maquina,
CASE
WHEN o.ciclo_estimado + o.tiempo_cambio_estimado = 0 THEN 0
ELSE (p.PiezasReales::decimal / (28800 / (o.ciclo_estimado + o.tiempo_cambio_estimado))) * 100
END as resultado
FROM PiezasPorIntervalo p
JOIN operacion o ON o.id_operacion = p.id_nu_operacion
ORDER BY fecha_inicio;
我正在具有上述规格的系统上运行它。任何有关如何优化此查询以获得更好性能的建议将不胜感激。谢谢！
EXPLAIN（分析、缓冲区）的输出
PiezasPorIntervalo 是耗时最长的部分，我会立即解释我所做的
假设我们有一个包含以下条目的生产表：
Piece | Production Start | Production End
----- | ------------------------| -------------------------
A | 2023-05-23 08:00:00 | 2023-05-23 10:00:00
B | 2023-05-23 09:30:00 | 2023-05-23 12:00:00
C | 2023-05-23 10:30:00 | 2023-05-23 11:30:00
D | 2023-05-23 12:00:00 | 2023-05-23 13:30:00
假设我们要计算从“2023-05-23 09:00:00”到“2023-05-23 11:00:00”的特定时间间隔内“PiezasReales”的数量。这是逐步计算：
时间范围与单件生产时间交叉的持续时间：
For Piece A: MIN(2023-05-23 11:00:00, 2023-05-23 10:00:00) - MAX(2023-05-23 09:00:00, 2023-05-23 08:00:00) = 1 hour
For Piece B: MIN(2023-05-23 11:00:00, 2023-05-23 12:00:00) - MAX(2023-05-23 09:00:00, 2023-05-23 09:30:00) = 0.5 hours
For Piece C: MIN(2023-05-23 11:00:00, 2023-05-23 11:30:00) - MAX(2023-05-23 09:00:00, 2023-05-23 10:30:00) = 0.5 hours
For Piece D: No intersection with the interval, so the duration is 0.
该作品的总时长：
For Piece A: 2023-05-23 10:00:00 - 2023-05-23 08:00:00 = 2 hours
For Piece B: 2023-05-23 12:00:00 - 2023-05-23 09:30:00 = 2.5 hours
For Piece C: 2023-05-23 11:30:00 - 2023-05-23 10:30:00 = 1 hour
For Piece D: 2023-05-23 13:30:00 - 2023-05-23 12:00:00 = 1.5 hours
间隔中的时间分数：
For Piece A: 1 hour / 2 hours = 0.5
For Piece B: 0.5 hours / 2.5 hours = 0.2
For Piece C: 0.5 hours / 1 hour = 0.5
For Piece D: 0 (since there's no intersection with the interval).
区间内的压电实数数量：
间隔内时间分数的总和：0.5 + 0.2 + 0.5 + 0 = 1.2 因此，在这个特定的时间间隔内，考虑到每件作品在该时间间隔内生产的时间比例，“RealPieces”相当于 1.2 件。
首先要做的是创建一个表时间表，以唯一列作为 PK 来替换查询中的generate series语句：
CREATE TABLE TIME_SERIES AS
SELECT *
FROM generate_series(
'2023-05-22 16:23:14'::timestamp,
'2023-05-26 08:23:14'::timestamp,
'8 hour'::interval
) AS d UNION
SELECT *
FROM generate_series(
'2023-05-23 00:23:14'::timestamp,
'2023-05-26 16:23:14'::timestamp,
'8 hour'::interval
) AS f;