MySQL 8嵌套带计数的选择 [英] MySQL 8 nested select with count

问题描述

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;

这段代码过去在MySQL8更新之前可以正常工作，但现在由于此错误而吐出

This bit of code used to work perfectly fine pre MySQL8 update, but now spits with this error

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

我浏览了google，但找不到正确的答案.用count()或count(*)替换count(1)并没有帮助.

I went through google and I couldn't find the right answer. replacing count(1) with count() or count (*) didn't help.

对于MySQL8，此查询应该是什么样?

What should this query look like for MySQL8?

推荐答案

单词 rank 是

The word rank is a reserved word in MySql 8.

因此，请使用另一个别名，或反选别名.

So use another alias name, or backtick the alias name.

在MySql 8中，您可以使用窗口函数

And in MySql 8 you can use 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;

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