SELECT mapname,
(SELECT count(1)+1 FROM ck_bonus b WHERE a.mapname=b.mapname AND a.runtime > b.runtime AND a.zonegroup = b.zonegroup AND b.style = %i) AS rank,
(SELECT count(1) FROM ck_bonus b WHERE a.mapname = b.mapname AND a.zonegroup = b.zonegroup AND b.style = %i) as total
FROM ck_bonus a WHERE steamid = '%s' AND style = %i;
这段代码用于在MySQL8更新之前完美地工作,但现在吐出这个错误
You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near
'rank, (SELECT count(1) FROM ck_bonus b WHERE a.mapname = b.mapname AND a.zonegro' at line 1
我经历了谷歌,我找不到正确的答案。用count()或count(*)替换count(1)没有帮助。
MySQL8的查询应该是什么样的?
rank
这个词在MySql 8中是一个reserved word。
因此,请使用其他别名,或者反复添加别名。
在MySql 8中,您可以使用window functions
SELECT
mapname,
DENSE_RANK() OVER (PARTITION BY mapname, zonegroup, steamid, style ORDER BY runtime DESC) AS `rank`,
COUNT(*) OVER (PARTITION BY mapname, zonegroup, steamid, style) AS total
FROM ck_bonus
WHERE steamid = '%s' AND style = %i;
你错过了outer
和subquery
的单引号:
SELECT mapname,
(SELECT count(1) FROM ck_bonus b WHERE a.mapname = b.mapname AND a.runtime > b.runtime AND a.zonegroup = b.zonegroup AND b.style = '%i') + 1 AS rnk,
(SELECT count(1) FROM ck_bonus b WHERE a.mapname = b.mapname AND a.zonegroup = b.zonegroup AND b.style = '%i') as total
FROM ck_bonus a
WHERE steamid = '%s' AND style = '%i';
我怀疑你需要LIKE
谓词而不是=
。