为了在我的游戏服务器上实现排名系统,我将每个玩家的一些信息存储在 MySQL 数据库中。 我有以下数据库表。
带有示例数据的表玩家
:
id | steamId | deaths
---+---------+-------
1 | asdja | 2
2 | kfjsl | 5
带有示例数据的表武器
:
playerId | weaponId | kills
---------+----------+------
1 | 5 | 8
1 | 9 | 7
2 | 3 | 3
2 | 6 | 10
2 | 7 | 2
您会看到,我没有将每个玩家的击杀数存储在 players
表中,因为我可以简单地从 weapons
表中计算出来。
我对 SQL 查询不太熟悉,但我完成了以下查询来选择具有以下字段的简单数据集:
kills,
deaths,
kill-death-rate (kdrate),
count (maximum number of players),
rank (current ranking position, sorted by kills)
查询:
SELECT
SUM(weapons.kills) AS `kills`,
`deaths`,
(SUM(kills) / IF(deaths, deaths, 1)) AS `kdrate`,
(SELECT COUNT(*) FROM `players`) AS `count`,
(
SELECT
COUNT(*)
FROM
(
SELECT
p.id AS id2,
SUM(w.kills) AS kills2,
p.deaths AS deaths2,
p.steamId AS steamId2
FROM
weapons AS w,
players AS p
WHERE
p.id = w.playerId
GROUP BY
p.id
) AS temp
WHERE
temp.kills2 >= (
SELECT
SUM(weapons.kills) AS `kills`
FROM
`players`,
`weapons`
WHERE
players.id = weapons.playerId AND
`id` = 1
GROUP BY
`id`
)
ORDER BY
temp.kills2 DESC,
temp.deaths2 ASC,
temp.steamId2 ASC
) AS `rank`
FROM
`players`
INNER JOIN
`weapons`
ON
players.id = weapons.playerId
WHERE
`id` = 1
GROUP BY
`id`
有两个问题:
1.) 查询很糟糕。
2.) 使用给定的示例数据执行此查询会产生“相同的排名”。 我的意思是,两名玩家获得相同的排名,因为两名玩家的击杀数相同。 但相反,玩家 1 应该处于排名 1,因为他的死亡次数比玩家 2 少。 我不知道如何才能实现这一点。
提前致谢!
编辑: @Manueru_mx 给了我如何做到这一点的基本想法: 我根据他的回答得到了以下代码:
SELECT
id,
kills,
deaths,
kdrate,
COUNT(*) AS rank,
(SELECT COUNT(*) FROM players) AS `count`
FROM
(
SELECT
pys.id AS id,
SUM(wps.kills) AS kills,
pys.deaths AS deaths,
(SUM(wps.kills) / pys.deaths) as kdrate
FROM
players pys
INNER JOIN
weapons wps
ON
pys.id = wps.playerid
GROUP BY
pys.id
ORDER BY
2 DESC,
3,
4 ASC
) AS tmp
WHERE
id = 1
剩下的唯一问题是,两种情况下的排名
都是1。
最佳答案
这是我的“解决方案”。这个查询更简单,但得到你正在寻找的结果
select
pys.steamID, SUM(wps.kills) as Kills,
SUM(pys.deaths) as Deaths,
(sum(wps.kills)/sum(pys.deaths)) as kdratio
--,COUNT(pys.steamID) as PlayersC
from players pys
inner join weapons wps on pys.id = wps.playerid
group by pys.steamID
order by 2 DESC, 3, 4 ASC
添加排名。
SELECT
id,
kills,
deaths,
kdrate,
ranking_usr as rank,
(SELECT COUNT(*) FROM players) AS `count`
FROM
(
SELECT
@row := @row + 1 AS ranking_usr
pys.id AS id,
SUM(wps.kills) AS kills,
pys.deaths AS deaths,
(SUM(wps.kills) / pys.deaths) as kdrate
FROM
players pys
INNER JOIN
weapons wps
ON
pys.id = wps.playerid
GROUP BY
pys.id
ORDER BY
2 DESC,
3,
4 ASC
) AS tmp
WHERE
id = 1
关于mysql - 使用复杂查询的排名系统: more precise filters,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/12976162/