由于某种原因,我无法绕过这个。我试图从多个列不同的表中获取ID ...基本上类似于 -
Select ID from table where ID in (Select distinct ID, Card, PunchTime, PunchDate)
虽然这显然不起作用。我想获得所有这些字段唯一的ID作为标准。我似乎无法提出有效的语法。我不确定我还能说些什么呢,看起来它看起来应该很简单...但是我从昨天起就一直在反弹,没有什么工作。谁知道我应该走哪条路?提前谢谢!
编辑:发布的内容有效,但结果不符合我的预期。这是一些saple数据:
ID Card PunchDate PunchTime In/Out
================================
1 00123 3/17/2012 13:00 1
2 00123 3/17/2012 17:00 2
3 00123 3/17/2012 17:00 1
4 00123 3/17/2012 20:00 2
5 00456 3/17/2012 14:00 1
6 00456 3/17/2012 17:00 2
我试图这样做的原因是计时软件决定任何卡片,打卡和打卡时间与另一张相同的内容都是重复的,无论是打入还是打出并删除一个。我唯一的解决方案是消除重复,并基本上从第一次冲击到最后一次重复的冲击。所以我的目标是只选择基于卡片,打卡日期和打卡时间的唯一值。然而,我所拥有的并不是将ID排除在使其成为独特价值的事物中。我有一个解决方法,所以时间不是特别的问题,但我更愿意弄清楚如何获得正确的数据。
再次感谢大家的快速回复!
答案 0 :(得分:2)
有新信息更新的答案:
SELECT *
FROM TABLE
WHERE NOT EXISTS
(
SELECT 1
FROM TABLE AS Duplicates
WHERE Duplicates.Card = TABLE.Card
AND Duplicates.PunchDate = TABLE.PunchDate
AND Duplicates.PunchTime = TABLE.PunchTime
AND Duplicates.ID != TABLE.ID
)
基本上,这就是说,获取所有不具有相同card, punchdate, punchtime
的记录(确保不对自己计算相同的行。)
答案 1 :(得分:2)
假设没有第二次转移,从一天开始,到下一次结束......
表格强>
DECLARE @table TABLE
(
[ID] INT IDENTITY,
[Card] INT,
[PunchDate] DATETIME,
[PunchTime] DATETIME,
[In/Out] TINYINT
)
INSERT INTO @table
(
[Card],
[PunchDate],
[PunchTime],
[In/Out]
)
SELECT 00123,
'3/17/2012',
'3/17/2012 13:00',
1
UNION ALL
SELECT 00123,
'3/17/2012',
'3/17/2012 17:00',
2
UNION ALL
SELECT 00123,
'3/17/2012',
'3/17/2012 17:00',
1
UNION ALL
SELECT 00123,
'3/17/2012',
'3/17/2012 20:00',
2
UNION ALL
SELECT 00456,
'3/17/2012',
'3/17/2012 14:00',
1
UNION ALL
SELECT 00456,
'3/17/2012',
'3/17/2012 17:00',
2
查询:
SELECT [Card],
[PunchDate],
MIN([PunchTime]) [PunchTime],
[In/Out]
FROM @table
WHERE [In/Out] = 1
GROUP BY [Card],
[PunchDate],
[In/Out]
UNION
SELECT [Card],
[PunchDate],
MAX([PunchTime]) [PunchTime],
[In/Out]
FROM @table
WHERE [In/Out] = 2
GROUP BY [Card],
[PunchDate],
[In/Out]
ORDER BY [Card],
[PunchDate]
<强>结果:强>
Card PunchDate PunchTime In/Out
123 2012-03-17 00:00:00.000 2012-03-17 13:00:00.000 1
123 2012-03-17 00:00:00.000 2012-03-17 20:00:00.000 2
456 2012-03-17 00:00:00.000 2012-03-17 14:00:00.000 1
456 2012-03-17 00:00:00.000 2012-03-17 17:00:00.000 2
接下来他会想要这个:
SELECT a.[Card],
a.[PunchDate],
a.[PunchTime],
b.[PunchTime],
DATEDIFF(hour, a.[PunchTime], b.[PunchTime]) TotalTime
FROM (
SELECT [Card],
[PunchDate],
MIN([PunchTime]) [PunchTime]
FROM @table
WHERE [In/Out] = 1
GROUP BY [Card],
[PunchDate]
) a
INNER JOIN (
SELECT [Card],
[PunchDate],
MAX([PunchTime]) [PunchTime]
FROM @table
WHERE [In/Out] = 2
GROUP BY [Card],
[PunchDate]
) b
ON a.[Card] = b.[Card]
AND a.[PunchDate] = b.[PunchDate]
ORDER BY a.[Card],
a.[PunchDate]
<强>结果强>
Card PunchDate PunchTime PunchTime TotalTime
123 2012-03-17 00:00:00.000 2012-03-17 13:00:00.000 2012-03-17 20:00:00.000 7
456 2012-03-17 00:00:00.000 2012-03-17 14:00:00.000 2012-03-17 17:00:00.000 3
答案 2 :(得分:1)
Select
*
FROM
table
WHERE
NOT EXISTS (
SELECT
*
FROM
table AS lookup
WHERE
ID <> table.ID
AND Card = table.Card
AND PunchTime = table.PunchTime
AND PunchDate = table.PunchDate
)
答案 3 :(得分:0)
Select ID
from table
where ID
in
(SELECT A.ID FROM (Select distinct ID, Card, PunchTime, PunchDate) A);
在你的查询中你写过..你应该在IN子句之外和IN子句中只有相同数量的列。如果您使用单个列,则不需要IN子句之外的任何括号,但如果您有多个列,则需要将它们包含在括号中。
Thumb规则:SELECT Col1,Col2 ..Coln FROM TABLE WHERE Col1 IN (SELECT Col1 FROM TABLE ...)
(对于单列)
SELECT Col1,Col2 ..Coln FROM TABLE WHERE (Col1,Col2..Coln) IN (SELECT Col1,Col2..Coln FROM TABLE ...)
(对于多列)
答案 4 :(得分:0)
Select ID from table where ID in (
select ID from (
Select distinct ID,
Card,
PunchTime,
PunchDate
FROM
OTHER_TABLE
) x
)
IN子句中只能有一列;因此,你需要别名(x
- 在我的回答中)结果,然后从那里选择ID列。另请注意,在子选择内部,您需要指定从中选择额外列的表(请参阅我的答案中的CAPS)。