查询胜出半数以上的球队名称（数据库包含主场和客场列的结果）答案

【问题标题】：Querying the Team names with more than a half wins (Database contains Home and Away Column for results)查询胜出半数以上的球队名称（数据库包含主场和客场列的结果）
【发布时间】：2019-09-30 10:06:18
【问题描述】：

我有一个表（会议），其中包含以下列：

主队（varchar）
AwayTeam(varchar)
首页(int)
离开（整数）

对于分数（例如：3:2）。

我的问题是我需要输出所有胜率超过一半的球队。

提供的代码运行完美，但它仅适用于主场胜利，我需要主场和客场的总和，因为例如，“利物浦”有三场主场比赛和两场客场比赛，我需要双方分数。

SELECT Team, COUNT(*) AS HomeWins
FROM Teams
JOIN Meetings ON Team = Meetings.HomeTeam AND Meetings.Home > Meetings.Away
GROUP BY Team

预期输出是（从 6 支球队中，我有两支球队有 3 胜，另一支 - 4 胜）输出超过一半胜场的球队和他们的总胜场。

例子：

Team       Wins
Liverpool  4
PSG        3

【问题讨论】：

见Why should I provide a Minimal Reproducible Example for a very simple SQL query? ...另外你需要标记一个RDMS产品，这个网站上的sql标签不是RDMS产品而是ANSI/ISO SQL标准..
用您正在使用的数据库标记您的问题。

标签： sql

【解决方案1】：

如果你只想赢，你可以使用：

select t.*,
       (select count(*)
        from meetings m
        where (m.hometeam = t.team and m.home > m.away) or
              (m.awayteam = t.team and m.away > m.away)
       ) as num_wins
from team t;

注意：这会返回所有球队，包括那些没有赢得任何比赛的球队。

【讨论】：

谢谢伙计，这很有效，但我需要过滤掉 5 场比赛中胜场数少于 3 场的球队。例如：巴塞罗那1W，拜仁3W，尤文图斯1W，利物浦4W，曼联2W，PSG 3W。我只想得到拜仁、利物浦和巴黎圣日耳曼。
@S1C4R10 。 . .我建议你问一个新的问题，这个问题更符合你的实际需求。

【解决方案2】：

您只需要查询两次结果，反转查询并使用 UNION 创建组合视图。

create table #meetings (HomeTeam varchar(20), AwayTeam varchar(20), Home int, Away int);


insert into #meetings (HomeTeam, AwayTeam, Home, Away)
values
    ('Trumpton', 'Camberwick Green', 1, 0), -- Trumpton win
    ('Chigley', 'Camberwick Green', 3, 2), -- Chigley win
    ('Trumpton', 'Chigley', 3, 1), -- Trumpton win
    ('Camberwick Green', 'Trumpton', 0, 2), -- Trumpton win
    ('Camberwick Green', 'Chigley', 1, 2) -- Chigley win

/*
Trumpton - played 3, won 3
Camberwick Green - played 4, won 0
Chigley - played 3, won 2
*/

select
    Team
    ,count(*) as Played
    ,sum(Win) as Wins
from
    (
    select
        HomeTeam as Team
        ,case when Home > Away then 1 else 0 end as Win
    from
        #meetings
    union all
    select
        AwayTeam as Team
        ,case when Away > Home then 1 else 0 end as Win
    from
        #meetings
    ) results

【讨论】：

【解决方案3】：

据我了解，您希望返回胜负多于球队的信息。如果这是正确的，这是一种方法：

-- Build cte to re-use query
WITH meetings(WinningTeam, LosingTeam) AS (
  SELECT
    CASE WHEN Home > Away THEN HomeTeam ELSE AwayTeam END AS WinningTeam,
    CASE WHEN Home > Away THEN AwayTeam ELSE HomeTeam END AS LosingTeam
  FROM Meetings
)
SELECT teams.Team, wins.win_count AS Wins
FROM teams
LEFT JOIN (
  SELECT WinningTeam, COUNT(*) AS win_count
  FROM Meetings
  GROUP BY 1
) wins ON teams.team = wins.WinningTeam -- Get wins per team
LEFT JOIN (
  SELECT LosingTeam, count(*) AS loss_count
  FROM Meetings
  GROUP BY 1
) losses ON teams.team = wins.LosingTeam -- Get wins per team
WHERE win_count > loss_count -- Only return teams with winning records

【讨论】：