Question

我不知道我的标题是正确的，但让我解释一下我在寻找什么。

我有两张桌子。

用户评分

maxContourarea

游戏

user (primary key)
game (varchar)
score (int)

客户用户评分数据如下所示

game (varchar)

游戏表数据如下所示

user    game             score
Chris   naruto           10
Gale    dungen           9
Donna   winning eleven   8

是否可以创建如下所示的视图：

game
dungen
naruto
winning eleven

提前致谢。

Answer 1

您首先需要使用交叉联接获取用户+游戏的所有可能组合的一组不同的组合：

select distinct u.user, g.game
from game g
cross join user u
;

这将给出以下结果：

|  USER |           game |
|-------|----------------|
| Chris |         dungen |
| Chris |         naruto |
| Chris | winning eleven |
|  Gale |         dungen |
|  Gale |         naruto |
|  Gale | winning eleven |
| Donna |         dungen |
| Donna |         naruto |
| Donna | winning eleven |

然后您可以通过这种方式将此结果集（使用外部联接）加入user表：

SELECT * 
FROM (
    select distinct u.user, g.game
    from game g
    cross join user u
) x
LEFT JOIN user u ON x.user = u.user and x.game = u.game
;

以上查询为您提供以下结果集：

|  user |           game |   USER |           game |  score |
|-------|----------------|--------|----------------|--------|
| Chris |         naruto |  Chris |         naruto |     10 |
|  Gale |         dungen |   Gale |         dungen |      9 |
| Donna | winning eleven |  Donna | winning eleven |      8 |
| Chris |         dungen | (null) |         (null) | (null) |
| Chris | winning eleven | (null) |         (null) | (null) |
|  Gale |         naruto | (null) |         (null) | (null) |
|  Gale | winning eleven | (null) |         (null) | (null) |
| Donna |         dungen | (null) |         (null) | (null) |
| Donna |         naruto | (null) |         (null) | (null) |

接下来，您可以使用上述结果集使用简单的SUM和GROUB BY生成报告但首先你需要用0替换所有空值，因为空值之和给出空值这样：

SELECT x.user, x.game, 
       sum( case when score is not null then score else 0 end ) as score
FROM (
    select distinct u.user, g.game
    from game g
    cross join user u
) x
LEFT JOIN user u ON x.user = u.user and x.game = u.game
GROUP BY x.user, x.game
order by x.user, x.game
;

此查询产生以下内容：

|  user |           game | score |
|-------|----------------|-------|
| Chris |         dungen |     0 |
| Chris |         naruto |    10 |
| Chris | winning eleven |     0 |
| Donna |         dungen |     0 |
| Donna |         naruto |     0 |
| Donna | winning eleven |     8 |
|  Gale |         dungen |     9 |
|  Gale |         naruto |     0 |
|  Gale | winning eleven |     0 |

您可以在此处找到完整的可运行示例：http://sqlfiddle.com/#!9/657cdf/12

我在这些示例中使用SQL的MySql方言，因为它在SqlFiddle上可用但我相信将其转换为MS-Access应该不是问题。

===编辑 - 如何将此查询转换为MS-Access方言===

我在MS-Access 2010上测试了这个查询 MS-Access不支持两件事：

MS Access不支持CROSS JOIN个关键字它可以通过以下方式转换：How to use cross join in access?
在表格之间使用“旧”语法和逗号：

SELECT DISTINCT u.user, g.game
FROM game g, user u

MS-Access不支持案例表达式
： CASE WHEN condition THEN value1 ELSE value2 END
但你可以改用IIF功能：
IIF( condition, value1, value2 )
更多相关信息：Does MS Access support "CASE WHEN" clause if connect with ODBC?

整个查询转换为MS-Access方言：

SELECT x.user, x.game, 
       sum( IIF( u.score IS NOT NULL, u.score, 0)) as score
FROM (
    SELECT DISTINCT u.user, g.game
    FROM game g, user u
) x
LEFT JOIN user u ON x.user = u.user and x.game = u.game
GROUP BY x.user, x.game
ORDER BY x.user, x.game

查询结果（在MS-Access中）：

查询显示另一个表中的每个值

1 个答案:

===编辑 - 如何将此查询转换为MS-Access方言===