我有两张桌子,Cars
和Defect
。我通过将car_id
作为外键传递给缺陷表来维护关系。
我在cars表中有3000条记录,在缺陷表中有16K条记录(有open_defect
和close_defect
)。我试图找出所有的汽车(3000)计数的开放缺陷(如果没有任何开放的缺陷,对汽车应该返回0)。
我正在尝试一些疑问:
SELECT cars.cars_id_primary ,IF(COUNT(defect.defect_id_primary)>0,1,0) AS `def_count`
FROM cars
LEFT JOIN defect ON cars.cars_id_primary = defect.cars_id AND defect.defect_status_id =1
WHERE cars.stage_id !=5
GROUP BY cars.cars_id_primary
ORDER BY cars.updated_on
这个查询给出了结果,但执行起来花费了太多时间。需要优化此查询。我被困在优化。
任何帮助都欢迎,谢谢。
发布于 2019-01-18 11:59:16
不需要在GROUP BY
中使用cars表。将查询重写如下:
SELECT cars.cars_id_primary, COALESCE(agg.open_defect_count, 0) AS open_defect_count
FROM cars
LEFT JOIN (
SELECT cars_id, COUNT(*) AS open_defect_count
FROM defect
WHERE defect_status_id = 1
GROUP BY cars_id
) AS agg ON cars.cars_id_primary = agg.cars_id
WHERE cars.stage_id != 5
ORDER BY cars.updated_on
您还需要创建索引。我建议从ix_defect(defect_status_id, cars_id)
开始。
发布于 2019-01-18 11:58:14
可以通过使用索引来提高性能。您能否在cars_id_primary
of cars
表上创建索引,在cars_id
of defect
上创建索引,如下面所示。然后您可以尝试查询。
CREATE INDEX idx1 ON cars (cars_id_primary);
CREATE INDEX idx2 ON defect (cars_id);
发布于 2019-01-18 11:59:19
我不明白你的问题,我在你的问题上做了一些改动,
SELECT cars.cars_id_primary ,COUNT(CASE WHEN (defect.defect_id_primary)>0 THEN 1 ELSE 0
END) AS `def_count`
FROM cars
LEFT JOIN defect ON cars.cars_id_primary = defect.cars_id
WHERE cars.stage_id !=5 AND defect.defect_status_id =1
GROUP BY cars.cars_id_primary
ORDER BY cars.updated_on
尝尝这个
https://stackoverflow.com/questions/54253457
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