Sql 与并集和计数不同_Sql_Sql Server

Sql 与并集和计数不同

sql sql-server

Sql 与并集和计数不同,sql,sql-server,Sql,Sql Server,我在数据库中有一个查询，在一个每天有可用时间的议程表中返回6个结果，例如：_08:10:00:000 | 08:30:00:000 uu和时段示例：T代表下午或M代表上午我已经康复了，但我有一个无法解决的问题。我只能有一天中每个时段的时间表示例：在2015-12-19，我上午和下午各有一段时间，但在2015-12-19，我上午和下午都不能有两个小时我的问题是： SELECT * FROM ( SELECT TOP(3) agendaidentificador,agendadata

我在数据库中有一个查询，在一个每天有可用时间的议程表中返回6个结果，例如：_08:10:00:000 | 08:30:00:000 uu和时段示例：T代表下午或M代表上午

我已经康复了，但我有一个无法解决的问题。我只能有一天中每个时段的时间表

示例：在2015-12-19，我上午和下午各有一段时间，但在2015-12-19，我上午和下午都不能有两个小时

我的问题是：

SELECT * FROM (
    SELECT TOP(3) agendaidentificador,agendadata, 'M' AS periodo
    FROM AGENDA
    WHERE 
        agendaconsumolocktempo IS NULL 
    AND 
        agendaconsumoidentificador IS NULL 
    AND 
        agendadata > GETDATE() 
    GROUP BY 
        agendaidentificador, 
        agendadata
    HAVING 
        CAST(DATEPART(HOUR,agendadata) AS INT) < 12 
    ORDER BY 
        NEWID(),
        agendadata asc
) A
UNION
SELECT * FROM (
    SELECT TOP(3) agendaidentificador,agendadata, 'T' AS periodo  
    FROM AGENDA
    WHERE 
        agendaconsumolocktempo IS NULL 
    AND 
        agendaconsumoidentificador IS NULL 
    AND 
        agendadata > GETDATE()
    GROUP BY 
        agendaidentificador,
        agendadata
    HAVING 
        CAST(DATEPART(HOUR,agendadata) AS INT) >= 12 
    AND 
        COUNT(CAST(agendadata AS DATE)) = 1 
    ORDER BY 
        NEWID(),
        agendadata asc
) B
GROUP BY 
    agendaidentificador, 
    agendadata,
    periodo
HAVING 
    COUNT(CAST(agendadata as DATE)) = 1
ORDER BY agendadata

不能发生相同的情况，如第1行和第2行。

您必须按AgendaData的datepart按最后一个group by进行分组。不能按整个日期+时间字段分组

我认为这应该是可行的：分组到“CHARtimestamp”字段，“YYYY-MM-DD”

让我们制作一些测试数据：

DECLARE @AGENDA TABLE
(
    agendaIdentificador int,
    agendaData  datetime,
    periodo varchar(1),
    agendaconsumolocktempo int,
    agendaconsumoidentificador int

)

INSERT INTO @AGENDA
( agendaIdentificador, agendaData, periodo, agendaconsumolocktempo, agendaconsumoidentificador )
VALUES
(173352, '2015-01-12 12:50:00.000', 'T', null, null),
(173353, '2015-01-12 12:50:00.000', 'T', null, null),
(173354, '2015-01-12 12:50:00.000', 'T', null, null),
(173355, '2015-01-12 12:50:00.000', 'T', null, null),
(173356, '2015-01-13 12:50:00.000', 'T', null, null),
(173363, '2015-01-12 14:40:00.000', 'T', null, null),
(175255, '2015-01-19 11:30:00.000', 'M', null, null),
(175520, '2015-01-26 14:50:00.000', 'T', null, null),
(125074, '2015-02-25 08:20:00.000', 'M', null, null),
(125076, '2015-02-25 08:40:00.000', 'M', null, null),
(125076, '2015-02-25 08:40:00.000', 'M', null, null),
(125076, '2015-02-25 08:40:00.000', 'M', null, null),
(125076, '2015-02-25 08:40:00.000', 'M', null, null),
(125076, '2015-02-26 08:40:00.000', 'M', null, null);

现在，我修复了测试数据，将上午和下午的所有记录分离出来，并对它们进行计数

WITH AGENDA AS
(
    SELECT *,
    ROW_NUMBER() OVER (PARTITION BY CAST(agendaData AS DATE), periodo ORDER BY agendaData) RN
    FROM @AGENDA
)

现在我们接受前面的查询，但是下午会话的更改是我只想要每个集合中的第一个，其中RN=1

    SELECT * FROM (
    SELECT TOP(3) agendaidentificador,agendadata, 'M' AS periodo
    FROM AGENDA
    WHERE agendaconsumolocktempo IS NULL 
    AND agendaconsumoidentificador IS NULL 
    AND agendadata > GETDATE() 
    AND DATEPART(HOUR,agendadata) < 12
    AND RN = 1
    GROUP BY 
        agendaidentificador, 
        agendadata
    ORDER BY 
        NEWID(),
        agendadata asc
) A
UNION
SELECT * FROM (
    SELECT TOP(3) agendaidentificador,agendadata, 'T' AS periodo  
    FROM AGENDA
    WHERE agendaconsumolocktempo IS NULL 
    AND agendaconsumoidentificador IS NULL 
    AND agendadata > GETDATE()
    and RN = 1
    and DATEPART(HOUR,agendadata) >= 12
    GROUP BY 
        agendaidentificador,
        agendadata
    ORDER BY 
        NEWID(),
        agendadata asc
) B
GROUP BY 
    agendaidentificador, 
    agendadata,
    periodo
ORDER BY agendadata

我添加了更多重复记录，清理了所有基于@Kevin Cook answer的无用having子句，我刚刚添加了其他CTE，这解决了我的问题

WITH CTE_AGENDA AS
(
    SELECT *,
    ROW_NUMBER() OVER (PARTITION BY CAST(agendaData AS DATE) ORDER BY newid(),agendaData) RN
    FROM AGENDA
)
,CTE_AGENDA2 AS
(
    SELECT *,
    ROW_NUMBER() OVER (PARTITION BY CAST(agendaData AS DATE) ORDER BY newid(),agendaData desc) RN2
    FROM AGENDA
)

SELECT * FROM (
    SELECT TOP(3) agendaidentificador,agendadata, 'M' AS periodo
    FROM AGENDA
    WHERE 
        agendaconsumolocktempo IS NULL 
    AND 
        agendaconsumoidentificador IS NULL 
    AND 
        agendadata > GETDATE() 
    GROUP BY 
        agendaidentificador, 
        agendadata
    HAVING 
        CAST(DATEPART(HOUR,agendadata) AS INT) < 12 
    ORDER BY 
        NEWID(),
        agendadata asc
) A
UNION
SELECT * FROM (
    SELECT TOP(3) agendaidentificador,agendadata, 'T' AS periodo  
    FROM AGENDA2
    WHERE     
        agendaconsumolocktempo IS NULL 
    AND 
        agendaconsumoidentificador IS NULL 
    AND 
        agendadata > GETDATE()
        and RN2 = 1
    GROUP BY 
        agendaidentificador,
        agendadata
    HAVING 
        CAST(DATEPART(HOUR,agendadata) AS INT) >= 12 
    ORDER BY 
        NEWID(),
        agendadata asc
) B
GROUP BY 
    agendaidentificador, 
    agendadata,
    periodo
HAVING 
    COUNT(CAST(agendadata as DATE)) = 1
ORDER BY agendadata,periodo

您是否在子选择中随机获得前三名？请使用UNION ALL而不是UNION，因为您确实希望返回3+3行。使用UNION ALL的服务器工作更少，因为不需要从结果中消除重复项。SpectralGhost是的，我。。。jarhl我和UNION试过但都失败了。。。再次获取副本您是否仅根据日期检查副本，而不检查时间和周期？因为您的groupby与日期和时间冲突。请将groupby中的agentdata转换为日期或在cte中取消时间。您真的不需要having子句，我没有修复其余的查询。根据你的解释，我猜你允许早上有多个记录，但只有一个下午的约会？如果一天中的每个时段只有一个，那么就在联合体的两侧使用RN=1。。。下午可以返回1个以上的约会，但不到同一天我添加了几个副本，清理了代码，修复了等式的两侧，我没有用此代码获得副本。你早上收到的旧密码是重复的吗？就这样，现在它被设置为您在上一篇评论中所说的基于您的答案，我创建了第二个代理2来获取下午的点评。。。当我按照你的方式行事时，我得到了唯一的但总是预约到12:00 pm。你使用Castagendata分区上的行数作为日期，periodo ORDER BY Agendata RN？periodo是分区的一部分吗？COUNTCASTagendadata为DATE=1为什么需要强制转换？

agendaidentificador agendadata  periodo
173352  2015-01-12 12:50:00.000 T
173356  2015-01-13 12:50:00.000 T
175255  2015-01-19 11:30:00.000 M
175520  2015-01-26 14:50:00.000 T
125074  2015-02-25 08:20:00.000 M
125076  2015-02-26 08:40:00.000 M

WITH CTE_AGENDA AS
(
    SELECT *,
    ROW_NUMBER() OVER (PARTITION BY CAST(agendaData AS DATE) ORDER BY newid(),agendaData) RN
    FROM AGENDA
)
,CTE_AGENDA2 AS
(
    SELECT *,
    ROW_NUMBER() OVER (PARTITION BY CAST(agendaData AS DATE) ORDER BY newid(),agendaData desc) RN2
    FROM AGENDA
)

SELECT * FROM (
    SELECT TOP(3) agendaidentificador,agendadata, 'M' AS periodo
    FROM AGENDA
    WHERE 
        agendaconsumolocktempo IS NULL 
    AND 
        agendaconsumoidentificador IS NULL 
    AND 
        agendadata > GETDATE() 
    GROUP BY 
        agendaidentificador, 
        agendadata
    HAVING 
        CAST(DATEPART(HOUR,agendadata) AS INT) < 12 
    ORDER BY 
        NEWID(),
        agendadata asc
) A
UNION
SELECT * FROM (
    SELECT TOP(3) agendaidentificador,agendadata, 'T' AS periodo  
    FROM AGENDA2
    WHERE     
        agendaconsumolocktempo IS NULL 
    AND 
        agendaconsumoidentificador IS NULL 
    AND 
        agendadata > GETDATE()
        and RN2 = 1
    GROUP BY 
        agendaidentificador,
        agendadata
    HAVING 
        CAST(DATEPART(HOUR,agendadata) AS INT) >= 12 
    ORDER BY 
        NEWID(),
        agendadata asc
) B
GROUP BY 
    agendaidentificador, 
    agendadata,
    periodo
HAVING 
    COUNT(CAST(agendadata as DATE)) = 1
ORDER BY agendadata,periodo