将多个查询合并为一个时，如何将列相加答案

【问题标题】：When combining multiple queries into one how do you sum columns together将多个查询合并为一个时，如何将列相加
【发布时间】：2014-06-20 13:17:56
【问题描述】：

我正在尝试根据比赛表计算联赛排名。

+----------------------------------+
|              Matches             |
+----------------------------------+
| id                               |
| league_id (FK League)            |   
| season_id (FK Season)            |
| home_team_id (FK Team)           |
| away_team_id (FK Team)           | 
| home_score                       |
| away_score                       |
| confirmed                        |
+----------------------------------+

我可以使用此查询正确计算主场联赛排名：

SELECT team.name, home_team_id AS team_id,
    COUNT(*) AS played,
    SUM((CASE WHEN home_score > away_score THEN 1 ELSE 0 END)) AS won,
    SUM((CASE WHEN away_score > home_score THEN 1 ELSE 0 END)) AS lost,
    SUM((CASE WHEN home_score = away_score THEN 1 ELSE 0 END)) AS drawn,
    SUM(home_score) AS goalsFor,
    SUM(away_score) AS goalsAgainst,
    SUM(home_score - away_score) AS goalDifference,
    SUM((CASE WHEN home_score > away_score THEN 3 WHEN home_score = away_score THEN 1 ELSE 0 END)) AS points
FROM matches
INNER JOIN team ON matches.home_team_id = team.id
WHERE league_id = 94
    AND season_id = 82
    AND confirmed IS NOT NULL
GROUP BY home_team_id
ORDER BY POINTS DESC;

以及使用此查询的客场联赛标准：

SELECT team.name, away_team_id AS team_id,
    COUNT(*) AS played,
    SUM((CASE WHEN away_score > home_score THEN 1 ELSE 0 END)) AS won,
    SUM((CASE WHEN home_score > away_score THEN 1 ELSE 0 END)) AS lost,
    SUM((CASE WHEN home_score = away_score THEN 1 ELSE 0 END)) as drawn,
    SUM(away_score) AS goalsFor,
    SUM(home_score) AS goalsAgainst,
    SUM(away_score - home_score) AS goalDifference,
    SUM((CASE WHEN away_score > home_score THEN 3 WHEN away_score = home_score THEN 1 ELSE 0 END)) AS points
FROM matches
INNER JOIN team ON matches.away_team_id = team.id
WHERE league_id = 94
    AND season_id = 82
    AND confirmed IS NOT NULL
GROUP BY away_team_id
ORDER BY points DESC;

但是使用 UNION ALL 结合这两个查询我没有得到正确的结果

SELECT * FROM 
(
    SELECT team.name, home_team_id AS team_id,
        COUNT(*) AS played,
        SUM((CASE WHEN home_score > away_score THEN 1 ELSE 0 END)) AS won,
        SUM((CASE WHEN away_score > home_score THEN 1 ELSE 0 END)) AS lost,
        SUM((CASE WHEN home_score = away_score THEN 1 ELSE 0 END)) AS drawn,
        SUM(home_score) AS goalsFor,
        SUM(away_score) AS goalsAgainst,
        SUM(home_score - away_score) AS goalDifference,
        SUM((CASE WHEN home_score > away_score THEN 3 WHEN home_score = away_score THEN 1 ELSE 0 END)) AS points
    FROM matches
    INNER JOIN team ON matches.home_team_id = team.id
    WHERE league_id = 94
        AND season_id = 82
        AND confirmed IS NOT NULL
    GROUP BY home_team_id
UNION
    SELECT team.name, away_team_id AS team_id,
        COUNT(*) AS played,
        SUM((CASE WHEN away_score > home_score THEN 1 ELSE 0 END)) AS won,
        SUM((CASE WHEN home_score > away_score THEN 1 ELSE 0 END)) AS lost,
        SUM((CASE WHEN home_score = away_score THEN 1 ELSE 0 END)) as drawn,
        SUM(away_score) AS goalsFor,
        SUM(home_score) AS goalsAgainst,
        SUM(away_score - home_score) AS goalDifference,
        SUM((CASE WHEN away_score > home_score THEN 3 WHEN away_score = home_score THEN 1 ELSE 0 END)) AS points
    FROM matches
    INNER JOIN team ON matches.away_team_id = team.id
    WHERE league_id = 94
        AND season_id = 82
        AND confirmed IS NOT NULL
    GROUP BY away_team_id
) x 
GROUP BY team_id
ORDER BY points DESC;

这应该是预期的结果：

知道我做错了什么吗？谢谢！

更新 1：

尝试 Dans 查询不起作用：

选择球队名称、主场积分 + 客场积分从团队加入（选择team.id，总和（当 home.home_score > home.away_score 然后 3 的情况当 home.home_score = home.away_score 然后 1 else 0 end) HomePoints, 总和（当 away.away_score > away.home_score 然后 3 else 0 end 时的情况）AwayPoints 来自团队在 team.id = home.home_team_id 上加入比赛在 team.id = away.away_team_id 上加入比赛在哪里 home.league_id = 94 和 home.season_id = 82 AND home.confirmed 不为空按 ID 分组 ) team.id 上的温度 = temp.id 按点顺序排列;

我得到这个结果：

【问题讨论】：

您的预期结果包括未被选中的字段。
@DanBracuk 在查看预期结果时，请仅考虑“P W ... PTS”部分。这张图片来自我使用的一种旧方法，我有一个 LeagueStanding 表，并在每次使用 PHP 比赛后更新值，但我计划的新功能迫使我改用数据库查询。
请稍等，我将更新查询以显示姓名。

标签： mysql sql union-all

【解决方案1】：

试试这个：

SELECT team.name, 
       team_id AS team_id,
       COUNT(*) AS played,
       SUM((CASE WHEN team_score > other_team_score THEN 1 ELSE 0 END)) AS won,
       SUM((CASE WHEN team_score < other_team_score THEN 1 ELSE 0 END)) AS lost,
       SUM((CASE WHEN team_score = other_team_score THEN 1 ELSE 0 END)) AS drawn,
       SUM(team_score) AS goalsFor,
       SUM(other_team_score) AS goalsAgainst,
       SUM(team_score - other_team_score) AS goalDifference,
       SUM((CASE WHEN team_score > other_team_score THEN 3 
                 WHEN team_score = other_team_score THEN 1 
                 ELSE 0 END)) AS points
FROM
    (
        -- LIST TEAM STATS WHEN PLAYED AS HOME_TEAM
        SELECT 
             id                               
             league_id
             season_id
             home_team_id as team_id,
             home_score   as team_score,
             away_score   as other_team_score, 
             confirmed 
        FROM    matches
        UNION ALL
        -- LIST TEAM STATS WHEN PLAYED AS AWAY_TEAM
        SELECT 
             id                               
             league_id
             season_id
             away_team_id as team_id,
             away_score   as team_score,
             home_score   as other_team_score, 
             confirmed 
        FROM matches
    ) matches
INNER JOIN team ON matches.team_id = team.id
WHERE league_id = 94
    AND season_id = 82
    AND confirmed IS NOT NULL
GROUP BY team.name, team_id
ORDER BY POINTS DESC;

【讨论】：

【解决方案2】：

您可能想使用 JOIN 而不是 UNION。在对都具有 team_id 的子查询使用 UNION 之后，您在 team_id 上使用 GROUP BY。那是行不通的……如果你只使用join，你甚至可以将group by 排除在外。

【讨论】：

【解决方案3】：

我想我知道发生了什么，但没有安装 MySQL 来测试它。

当您进行分组查询时，不是分组分组子句一部分的每一列必须是一个聚合函数（SUM、MAX 等）。如果您不这样做，大多数数据库引擎都会给您一个错误； MySQL试图提供帮助？通过选择随机行来代替。

tl;dr 您的外部选择需要一堆 SUM，而不仅仅是选择 *。

【讨论】：

它选择它首先在磁盘上遇到的任何值。

【解决方案4】：

我会使用这种方法。我只是要做点。你想要的其他东西的逻辑是一样的。

select TeamName, HomePoints + AwayPoints points
from team join (
select team_id
, sum(case when home.home_score > home.away_score then 3
when home.home_score = home.away_score then 1 else 0 end) HomePoints
, sum(case when away.away_score > away.home_score then 3
when away.home_score = away.away_score then 1 else 0 end) AwayPoints
from team join matches home on team.team_id = home.home_team_id
join matches away on team.team_id = away.away_team_id
where blah blah blah
group by team_id
) temp on team.team_id = temp.team_id
order by points desc

【讨论】：

谢谢！我明天早上试试。
尝试了这个查询，但它给了我太多的分数（例如，264 代表第一名的团队）