mysql:在同一个表上查询三次
我目前正在写一个查询。 从用户、帖子和帖子中的附加信息表中检索信息(帖子视图信息) 在上面的查询中,我们对同一个表进行三次查询,以获得post'A'、'B'、'C'的数量 为了解决这个问题,我对查询做了如下更改mysql:在同一个表上查询三次,mysql,Mysql,我目前正在写一个查询。 从用户、帖子和帖子中的附加信息表中检索信息(帖子视图信息) 在上面的查询中,我们对同一个表进行三次查询,以获得post'A'、'B'、'C'的数量 为了解决这个问题,我对查询做了如下更改 SELECT u.email, u.user_nm, p.pid, p.post_ttl, p.date, p.ref_level, p.ref_origin, psi.reportCount, psi.reco
SELECT
u.email,
u.user_nm,
p.pid,
p.post_ttl,
p.date,
p.ref_level,
p.ref_origin,
psi.reportCount,
psi.recommendCount,
psi.oppositeCount,
p.ref_step,
date(p.date) = date(now()) AS is_today,
(SELECT category_path FROM post_category WHERE category_id = p.category_id) as category_full_path
FROM
user AS u,
(
SELECT *
FROM post as p
WHERE
p.is_enable = 1
ORDER BY
p.ref_origin DESC,
p.ref_step ASC
LIMIT 0, 15
) as p left join
(
SELECT
pid,
COUNT(if(status = 'A', 1, null)) AS reportCount,
COUNT(if(status = 'B', 1, null)) AS recommendCount,
COUNT(if(status = 'C', 1, null)) AS oppositeCount
FROM post_status_info
group by pid
) AS psi
on
psi.pid = p.pid
WHERE
p.uid = u.uid
ORDER BY
ref_origin DESC,
ref_step ASC
我认为最好是查询同一个表三次。
哪种代码在性能方面更好
谢谢。我认为第二种选择在性能方面更有成效。因为这里要执行的查询数量较少 您也可以使用CASE来完成
SELECT
u.email,
u.user_nm,
p.pid,
p.post_ttl,
p.date,
p.ref_level,
p.ref_origin,
p.ref_step,
date(p.date) = date(now()) AS is_today,
(SELECT category_path FROM post_category WHERE category_id = p.category_id) as category_full_path,
(SUM(CASE WHEN sub_i.status = 'A' THEN 1 ELSE 0 END)) AS recommendCount,
(SUM(CASE WHEN sub_i.status = 'B' THEN 1 ELSE 0 END)) AS oppositeCount,
(SUM(CASE WHEN sub_i.status = 'C' THEN 1 ELSE 0 END)) AS reportCount
FROM
(
SELECT *
FROM post as p
WHERE
p.is_enable = 1
ORDER BY
p.ref_origin DESC,
p.ref_step ASC
) as p,
INNER JOIN user AS u ON u.uid = p.uid
INNER JOIN post_status_info as sub_i ON p.pid = sub_i.pid
GROUP BY p.pid
ORDER BY
ref_origin DESC,
ref_step ASC
SELECT
u.email,
u.user_nm,
p.pid,
p.post_ttl,
p.date,
p.ref_level,
p.ref_origin,
p.ref_step,
date(p.date) = date(now()) AS is_today,
(SELECT category_path FROM post_category WHERE category_id = p.category_id) as category_full_path,
(SUM(CASE WHEN sub_i.status = 'A' THEN 1 ELSE 0 END)) AS recommendCount,
(SUM(CASE WHEN sub_i.status = 'B' THEN 1 ELSE 0 END)) AS oppositeCount,
(SUM(CASE WHEN sub_i.status = 'C' THEN 1 ELSE 0 END)) AS reportCount
FROM
(
SELECT *
FROM post as p
WHERE
p.is_enable = 1
ORDER BY
p.ref_origin DESC,
p.ref_step ASC
) as p,
INNER JOIN user AS u ON u.uid = p.uid
INNER JOIN post_status_info as sub_i ON p.pid = sub_i.pid
GROUP BY p.pid
ORDER BY
ref_origin DESC,
ref_step ASC