Question

任何SQL Guru都在那里我可以使用一些帮助！我正在创建一个我认为需要Union的存储过程，以便所有结果都以1 SELECT语句返回。我已将问题简化为下表：

用户

user_id username name DOB ------------------------------------------------------

1 JohnSmith1 John Smith 01/01/1990

2 LisaGreen17 Lisa Green 03/07/1986

3 BarneyB Barney Brown 09/12/1960

user_team

user_team_id user_id team_id total_score -------------------------------------------------------------

1 1 1 29

2 2 7 37

3 3 2 15

private_league

priv_league_id league_name host_user league_password -------------------------------------------------------------------------

1 Lisa's League 2 CSUASH429d9

2 Barney's Bonanza 3 Jkap89f5I01

user_team_private_league_M2M

id priv_league_id user_team_id ----------------------------------------

1 1 1

2 1 2

3 1 3

4 2 1

5 2 3

我想运行一个带有 user_id 输入的存储过程，它会带回用户输入的所有联赛，每个人的主机那些联赛，每个联赛中有多少玩家参赛和每个联赛中用户所处的位置（按总分排序）。< / p>

目前我有：

CREATE DEFINER=`root`@`localhost` PROCEDURE `user_private_leagues`(IN v_user_id INT)
BEGIN

    DECLARE userteamid INT;

    # Retrieve user team from a user_id
    SELECT user_team_id INTO userteamid
    FROM user_team 
    WHERE user_id = v_user_id;

    # Retrieve private league name and host user (for a userteam)
    SELECT private_league.league_name, private_league.host_user
    FROM   user_team_private_league_M2M
    INNER JOIN privateleague
    ON user_team_private_league_M2M.priv_league_id=private_league.priv_league_id
    WHERE  user_team_id = userteamid;

END

此查询不包含每个联盟的玩家总数以及当前用户的位置

我创建了一个查询来恢复每个私人联盟的总用户数，没有像这样的用户过滤器：

SELECT private_league_id, COUNT(*) AS total_users
FROM   classicseasonmodel_classicseasonuserteamprivateleague
GROUP BY private_league_id;

可以使用this question的答案并使用total_score来计算用户当前位置的查询。

此刻我非常坚持这一点 - 来自SP的完美结果如下：

CALL user_private_leagues(3);（BarneyB的用户ID）

priv_league_name current_position total_users host_user -----------------------------------------------------------------------

Lisa's League 3 3 LisaGreen17

Barney's Bonanza 2 2 BarneyB

谢谢！

Answer 1

让我们一步一步......

第一个身份证明，请求理解

SELECT private_league_id, COUNT(*) AS total_users
FROM   classicseasonmodel_classicseasonuserteamprivateleague
GROUP BY private_league_id;

现在添加名称和主机用户

SELECT PL.league_name, LC.uCNT, PL.host_user  
FROM (SELECT private_league_id AS pID, COUNT(*) AS uCNT
      FROM   classicseasonmodel_classicseasonuserteamprivateleague
      GROUP BY private_league_id ) AS LC
LEFT JOIN private_league PL ON PL.priv_league_id = LC.pID

现在添加主机用户名

SELECT PL.league_name, LC.uCNT as total_users, hu.name as host_user
FROM (SELECT private_league_id AS pID, COUNT(*) AS uCNT
      FROM   classicseasonmodel_classicseasonuserteamprivateleague
      GROUP BY private_league_id ) AS LC
LEFT JOIN private_league PL ON PL.priv_league_id = LC.pID
LEFT JOIN user hu ON PL.host_user = hu.user_id

不知道目前的位置在哪里。

此查询将为每个团队提供用户和职位，加入此协议并按用户ID限制，以便为每个团队获得一个用户职位：

select UT.user_id, 
       UT.team_id, 
       ROW_NUMBER() OVER (PARTITION BY team_id ORDER BY total_score DESC) AS team_position
from private_league L
join user_team_private_league_M2M LJ ON L.priv_league_id = LJ.priv_league_id 
join user_team UT ON LJ.user_team_id = UT.user_team_id

Answer 2

很抱歉，但我没有创建数据库来测试下面的SQL。但你可以从那里开始。无需UNION。我不了解计算用户在联盟中的位置的业务规则，因为它可能来自团队或用户。

select priv_league_id, league_name, host_user_name, count(*) as total_users
from (
      select A.priv_league_id, A.league_name, D.name as host_user_name, B.user_team_id, C.user_id, D.
      from private_league A
      join user_team_private_league_M2M B
      on A.priv_league_id = B. priv_league_id
      join user_team C
      on B.user_team_id = C. user_team_id
      join user D
      on A.host_user = D.user_id
) D
group by priv_league_id

复杂的SQL联盟

2 个答案: