MySQL 8嵌套select with count

问题描述 投票:1回答:2
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?

mysql sql database mysql-8.0
2个回答
1
投票

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;

0
投票

你错过了outersubquery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=

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