Question

我这里有一个奇怪的错误。

我正在尝试将以下查询的结果限制为仅一行/结果：

SELECT userEmail
    FROM [LunchUsersAvailable]
    WHERE LunchGroupsID = '301' AND availableDate = '2015-01-01' AND userEmail != 'userEmail@company.com' EXCEPT
    (
        SELECT DISTINCT (COALESCE([user1], [user2])) AS matchedEmail
        FROM
        (
                SELECT [user1], [user2]
                FROM [LunchMatched]
            UNION ALL
                SELECT [user2], [user1]
                FROM [LunchMatched]
        ) AS tmp
        WHERE (user1 = 'userEmail@company.com' OR user2 = 'userEmail@company.com')
    )

输出：

test2@company.com

test3@company.com

test4@company.com

我尝试使用 TOP 1 。没工作。根本没有结果。

SELECT TOP 1 userEmail
    FROM [LunchUsersAvailable]
    WHERE LunchGroupsID = '301' AND availableDate = '2015-01-01' AND userEmail != 'userEmail@company.com' EXCEPT
    (
        SELECT DISTINCT (COALESCE([user1], [user2])) AS matchedEmail
        FROM
        (
                SELECT [user1], [user2]
                FROM [LunchMatched]
            UNION ALL
                SELECT [user2], [user1]
                FROM [LunchMatched]
        ) AS tmp
        WHERE (user1 = 'userEmail@company.com' OR user2 = 'userEmail@company.com')
    )

我也用 MIN（）进行了测试。没有结果。

SELECT MIN (userEmail)  as email
    FROM [LunchUsersAvailable]
    WHERE LunchGroupsID = '301' AND availableDate = '2015-01-01' AND userEmail != 'userEmail@company.com' EXCEPT
    (
        SELECT DISTINCT (COALESCE([user1], [user2])) AS matchedEmail
        FROM
        (
                SELECT [user1], [user2]
                FROM [LunchMatched]
            UNION ALL
                SELECT [user2], [user1]
                FROM [LunchMatched]
        ) AS tmp
        WHERE (user1 = 'userEmail@company.com' OR user2 = 'userEmail@company.com')
    )

我用 MAX（）测试以防万一。 有效！

SELECT MAX (userEmail)  as email
    FROM [LunchUsersAvailable]
    WHERE LunchGroupsID = '301' AND availableDate = '2015-01-01' AND userEmail != 'userEmail@company.com' EXCEPT
    (
        SELECT DISTINCT (COALESCE([user1], [user2])) AS matchedEmail
        FROM
        (
                SELECT [user1], [user2]
                FROM [LunchMatched]
            UNION ALL
                SELECT [user2], [user1]
                FROM [LunchMatched]
        ) AS tmp
        WHERE (user1 = 'userEmail@company.com' OR user2 = 'userEmail@company.com')
    )

输出：

test4@company.com

总之，我的问题是：如何获得 test2@company.com ？

更新测试更新 我做了两个不同的测试。一个工作，另一个不工作。我会详细展示每一个。首先，测试一切正常：

我想分别解释每个SELECT和每个部分的结果：

-- this outter SELECT lists all available users who are available
SELECT userEmail FROM [LunchUsersAvailable]
WHERE LunchGroupsID = '301' AND availableDate = '2015-01-01' AND userEmail != 'userEmail@company.com'

输出：

test1@company.com

test2@company.com

test3@company.com

test4@company.com

（4行（s）受影响）

-- this inner SELECT lists all the users who has been paired with userEmail@company.com before
SELECT DISTINCT (COALESCE([user1], [user2])) AS matchedEmail
FROM
(
        SELECT [user1], [user2]
        FROM [LunchMatched]
    UNION ALL
        SELECT [user2], [user1]
        FROM [LunchMatched]
) AS tmp
WHERE (user1 = 'userEmail@company.com' OR user2 = 'userEmail@company.com')

输出：

userEmail@company.com

test2@company.com

（2行（s）受影响）

运行完整代码，没有 TOP 1，MIN（）或MAX（）：

test1@company.com

test3@company.com

test4@company.com

（3行受影响）

到目前为止所有正确的。现在我只需要获得 test1@company.com 部分，所以我改变了代码的第一行 SELECT userEmail FROM [LunchUsersAvailable] 至 SELECT TOP 1 userEmail FROM [LunchUsersAvailable]

结果是：

test1@company.com

（1行受影响）

绝对正常。

但现在，如果 userEmail@company.com 先前已与 test1@company.com而非test2@company.com 匹配（[LunchMatched]表）结果不同。

INNER，OUTTER和FULL代码（没有TOP 1，MIN或MAX）的工作方式与上面说明的相同，很好。

但是，如果我更改代码的第一行 SELECT userEmail FROM [LunchUsersAvailable] 至 SELECT TOP 1 userEmail FROM [LunchUsersAvailable]

结果是 null ，而不是正确的结果： test2@company.com

（1行受影响）

Answer 1

这表明userEmail接受多个值，其中一个是空字符串（'' - 而不是NULL）。

userEmail返回的第一行具有此空值。 min()会抓住它。 max()没有。

选中userEmail <> ''或只使用max()。

如果您想要min()，您还可以使用带有case的条件聚合：

select min(case when userEmail > '' then userEmail end)

编辑：

我看到了问题。 top 1，min()和max()引用except之前的第一个查询。我错过了except，因为它在行尾滚动了。 except的作用类似于union，将查询连接在一起。它不是where子句的一部分。

Answer 2

我明白了！ TOP 1或MIN正在处理之前 EXCEPT，这意味着它将首先从可用用户结果获得TOP 1，并且只有他们应用EXCEPTION。

解决方案：获取可用用户的完整列表，运行例外，然后获取TOP 1

SELECT TOP 1 useremail FROM ( SELECT userEmail FROM [LunchUsersAvailable] WHERE LunchGroupsID = '301' AND availableDate = '2015-01-01' AND userEmail != 'userEmail@company.com' EXCEPT -- from a specific cafeteria and date, EXCEPT ( SELECT DISTINCT (COALESCE([user1], [user2])) AS matchedEmail -- select all matched users in one column FROM ( SELECT [user1], [user2] FROM [LunchMatched] UNION ALL SELECT [user2], [user1] FROM [LunchMatched] ) AS tmp WHERE (user1 = 'userEmail@company.com' OR user2 = 'userEmail@company.com') -- who has been previously matched with the user who is registering now ) ) as tmp

Answer 3

这不是答案。这只是为了说明如何简化原始查询。以下是三种可能性：

SELECT userEmail
FROM LunchUsersAvailable
WHERE LunchGroupsID = '301' 
AND availableDate = '2015-01-01' 
AND userEmail != 'userEmail@company.com' 
EXCEPT
(
  SELECT user1
  FROM LunchMatched 
  WHERE user2 = 'userEmail@company.com'
  UNION ALL
  SELECT user2
  FROM LunchMatched 
  WHERE user1 = 'userEmail@company.com'
);

或者：

SELECT userEmail
FROM LunchUsersAvailable
WHERE LunchGroupsID = '301' 
AND availableDate = '2015-01-01' 
AND userEmail != 'userEmail@company.com' 
AND userEmail NOT IN
(
  SELECT user1
  FROM LunchMatched 
  WHERE user2 = 'userEmail@company.com'
)
AND userEmail NOT IN
(
  SELECT user2
  FROM LunchMatched 
  WHERE user1 = 'userEmail@company.com'
);

或者：

SELECT userEmail
FROM LunchUsersAvailable
WHERE LunchGroupsID = '301' 
AND availableDate = '2015-01-01' 
AND userEmail != 'userEmail@company.com' 
AND NOT EXISTS
(
  SELECT *
  FROM LunchMatched 
  WHERE LunchUsersAvailable.userEmail IN (LunchMatched.user1, LunchMatched.user2)
  AND 'userEmail@company.com' IN (LunchMatched.user1, LunchMatched.user2)
);

在第二个和第三个查询中，可以轻松地将userEmail替换为MIN（userEmail）。在第一个中，这将通过SELECT MIN(userEmail) FROM (first query here)完成。

SQL - TOP 1和MIN不起作用但MAX工作

3 个答案: