从下表中：
timestamp inout Name
2018-04-01 14:00 0 Tom
2018-04-02 06:00 1 Tom
2018-04-02 14:00 0 Tom
2018-04-03 06:00 1 Tom
2018-04-01 22:00 0 Rob
2018-04-02 14:00 1 Rob
2018-04-02 22:00 0 Rob
2018-04-03 13:00 1 Rob
2018-04-01 12:55 0 John
2018-04-02 06:05 1 John
2018-04-03 06:10 1 John
2018-04-01 14:05 0 Anna
2018-04-02 14:10 0 Anna
2018-04-02 14:15 1 Anna
2018-04-02 14:20 0 Anna
2018-04-03 14:05 0 Anna
2018-04-01 22:00 1 Mary
2018-04-02 06:00 0 Mary
2018-04-02 22:00 1 Mary
2018-04-03 06:00 0 Mary
其中1 = in 0 = out
我需要收集2018-04-02的“进入和退出记录”的数据，并将其显示在这样的表中：
d1-in-timestamp d0-out-timestamp Name
2018-04-02 07:00 2018-04-02 15:00 Tom
2018-04-02 14:00 2018-04-02 22:00 Rob
2018-04-02 06:05 - John
- 2018-04-02 14:10 Anna
2018-04-02 14:15 2018-04-02 14:20 Anna
2018-04-02 00:00 2018-04-02 06:00 Mary
2018-04-02 22:00 2018-04-02 00:00 Mary
在一个完美的世界中
汤姆通过“DOOR”进入建筑物，然后经过“DOOR”一次。汤姆很完美！就像汤姆！ :)
罗布也很完美，但他很瞌睡，所以他来到了下午班。 ：P
安娜来和汤姆一起工作。汤姆为她保持“DOOR”开放，所以没有关于她进入的记录。此外，她一直回来，因为她忘记了什么
约翰是道奇！他上班迟到了，所以他应该解决这个问题，但是当别人离开时，他总是和别人一起通过“DOOR”滑倒。
最后结婚。她是夜班，所以她需要在一张桌子上看到一天分成两条记录
是否可以通过一个SQL查询在一个表中获得此类结果？
到目前为止，我管理这样的SQL查询：
select timestamp as d1, (select timestamp from DOOR where timestamp>m1.timestamp and inout=0 and name=m1.name) as d0, name from DOOR as m1 where substring(timestamp,1,10)='2018-04-02' and inout=1 order by name, timestamp
查询适用于来自“完美世界”（Tom＆amp; Rob）的人，而对于约翰则更多/更少 不幸的查询对Anna和Marry不起作用。
PS：抱歉我的英文
答案 0 :(得分：1)
这是一个艰难的，但我想出了一个查询来做到这一点。它使用一些连接，子查询和联合，但会产生您想要的输出。我从@RajatJaiswals小提琴开始，但创建了一个全新的查询。
SELECT * FROM (
SELECT
IF(inA.timestamp < '2018-04-02', '2018-04-02 00:00:00', inA.timestamp) AS `d1-in-timestamp`,
IFNULL(IF(outA.timestamp > '2018-04-02 23:59:59', CAST(DATE_ADD('2018-04-02', INTERVAL 1 DAY) AS DATETIME), outA.timestamp), '-') AS `d0-out-timestamp`,
inA.name AS `Name`
FROM
attendance AS inA
LEFT JOIN attendance AS outA ON (
inA.name = outA.name
AND outA.inout = 0
AND inA.timestamp < outA.timestamp
AND NOT EXISTS(
SELECT betweenA.name
FROM attendance AS betweenA
WHERE
betweenA.name = inA.name
AND betweenA.timestamp > inA.timestamp
AND betweenA.timestamp < outA.timestamp
)
)
WHERE
inA.inout = 1
AND (
CAST(inA.timestamp AS DATE) = '2018-04-02'
OR CAST(outA.timestamp AS DATE) = '2018-04-02'
)
UNION
SELECT
'-' AS `d1-in-timestamp`,
IF(outA.timestamp > '2018-04-02 23:59:59', CAST(DATE_ADD('2018-04-02', INTERVAL 1 DAY) AS DATETIME), outA.timestamp) AS `d0-out-timestamp`,
outA.name AS `Name`
FROM
attendance AS outA
LEFT JOIN attendance AS inA ON (
inA.name = outA.name
AND inA.inout = 1
AND inA.timestamp < outA.timestamp
AND NOT EXISTS(
SELECT betweenA.name
FROM attendance AS betweenA
WHERE
betweenA.name = inA.name
AND betweenA.timestamp > inA.timestamp
AND betweenA.timestamp < outA.timestamp
)
)
WHERE
outA.inout = 0
AND CAST(outA.timestamp AS DATE) = '2018-04-02'
AND inA.name IS NULL
) AS a
ORDER BY `Name`, `d1-in-timestamp`
这是一个复杂的查询，起初可能看起来令人生畏，但我尝试将其分解成一小部分来解释它的作用：
外部SELECT只是为了对整个结果进行排序。由于UNION语句，这是必需的。
SELECT子句只处理一些输出转换
IF(inA.timestamp < '2018-04-02', '2018-04-02 00:00:00', inA.timestamp)仅用于格式化，并替换当天00:00之前的时间戳，IFNULL(IF(outA.timestamp > '2018-04-02 23:59:59', CAST(DATE_ADD('2018-04-02', INTERVAL 1 DAY) AS DATETIME), outA.timestamp), '-')再次进行格式化，但会做两件事：如果此人没有离开建筑物，则将{null替换为-，并替换当天的00:00之后的时间戳。 在FROM子句中，我使用JOIN将进入建筑物（inA）的记录与离开建筑物的记录（outA）连接在一起。有趣的部分是ON子句：
name列仅加入同一个人的记录outA表应该只关注离开的人（inout = 0）inA.timestamp < outA.timestamp如果此人在离开这两个条目之前没有进入，则不应该联合起来加入的两个记录之间不应有任何活动记录。这由NOT EXISTS子查询处理。它搜索
betweenA.name = inA.name）inA有问题的记录outA记录之前如果存在任何此类记录，则NOT EXISTS子句的计算结果为false，并且记录未加入。这样，只有后续的进出条目才能连接在一起。
WHERE子句很简单：
inA CAST(inA.timestamp AS DATE)将时间戳转换为日期，从而删除时间部分并简化比较）选择记录进入建筑物的人的所有记录。我们现在仍然怀念没有记录进入大楼的安娜的情况。这就是UNION到位的地方，并将此信息添加到结果中。
SELECT再次只是输出逻辑：
ON子句执行以下操作：
name仅加入一个人的记录outA.inout = 1因为联接表应该只使用输入inA.timestamp < outA.timestamp）WHERE子句再次执行一些重要限制：
outA.inout = 0因为我们需要将表限制在离开建筑物的记录中CAST(outA.timestamp AS DATE) = '2018-04-02'这次只检查outA的日期，因为没有进入记录。inA.name IS NULL 最后一件事是ORDER BY条款应该是自我解释的。
| d1-in-timestamp | d0-out-timestamp | Name |
|---------------------|---------------------|------|
| - | 2018-04-02 14:10:00 | Anna |
| 2018-04-02 14:15:00 | 2018-04-02 14:20:00 | Anna |
| 2018-04-02 06:05:00 | - | John |
| 2018-04-02 00:00:00 | 2018-04-02 06:00:00 | Mary |
| 2018-04-02 22:00:00 | 2018-04-03 00:00:00 | Mary |
| 2018-04-02 14:00:00 | 2018-04-02 22:00:00 | Rob |
| 2018-04-02 06:00:00 | 2018-04-02 14:00:00 | Tom |
您可以在以下SQL小提琴中尝试：http://sqlfiddle.com/#!9/e618bb/7/0
答案 1 :(得分：0)
SELECT t.Name,t1.TimeStamp asintime,t.timestamp as outtime
FROM tmpAttendance t
LEFT OUTER JOIN tmpAttendance t1 ON t.Name = t1.Name
AND t1.inout = 1
AND CAST(t.timestamp AS DATE) = CAST(t1.timestamp AS DATE)
WHERE t.inout = 0
ORDER BY t.Name ,t1.TimeStamp
答案 2 :(得分：0)
只需添加一个简单的内容：为Mary添加一行最长时间（2018-04-02 23:59）
http://sqlfiddle.com/#!9/bb41b1/6
然后你可以使用下面的逻辑：
select a.name,
case when a.name != 'Mary' and (b.out_time is null or a.in_time <= b.out_time) then a.in_time
when a.name != 'Mary' and b.out_time is not null and a.in_time > b.out_time then null
when a.name = 'Mary' and a.in_time > b.out_time then '2018-04-02 00:00:00'
else a.in_time
end as in_time,
b.out_time
from
(select name,time1 as in_time from have1
where inout1 = 1 and (substring(time1,1,10)='2018-04-02')) a
left join
(select name,time1 as out_time from have1
where inout1 = 0 and (substring(time1,1,10)='2018-04-02')) b
on a.name = b.name
<强>输出：
name in_time out_time
Tom 2018-04-02 06:00:00 2018-04-02 14:00:00
Rob 2018-04-02 14:00:00 2018-04-02 22:00:00
John 2018-04-02 06:05:00 (null)
Anna (null) 2018-04-02 14:10:00
Anna 2018-04-02 14:15:00 2018-04-02 14:20:00
Mary 2018-04-02 00:00:00 2018-04-02 06:00:00
Mary 2018-04-02 22:00:00 2018-04-02 23:59:59
如果需要澄清，请告诉我。
答案 3 :(得分：0)
玛丽和其他夜班人员
首先让我们谈谈玛丽和其他夜班人员。当我们过滤所有邮票时，我们可以找到它们的邮票：
对于外面的印章，我们使用印章的第二天00:00。
SELECT d.`timestamp` `timestamp_in`,
timestampadd(second, -1 * second(d.`timestamp`), timestampadd(minute, -1 * minute(d.`timestamp`), timestampadd(hour, -1 * hour(d.`timestamp`), timestampadd(day, 1, (d.`timestamp`))))) `timestamp_out`,
d.`name`
FROM `door` d
WHERE d.`inout` = 1
AND NOT EXISTS (SELECT *
FROM `door` di
WHERE di.`name` = d.`name`
AND di.`inout` = 0
AND di.`timestamp` > d.`timestamp`
AND date(di.`timestamp`) = date(d.`timestamp`))
AND EXISTS (SELECT *
FROM `door` di
WHERE di.`name` = d.`name`
AND di.`inout` = 0
AND date(di.`timestamp`) = date_add(date(d.`timestamp`), INTERVAL 1 DAY)
AND NOT EXISTS (SELECT *
FROM `door` dii
WHERE dii.`name` = di.`name`
AND date(dii.`timestamp`) = date_add(date(d.`timestamp`), INTERVAL 1 DAY)
AND dii.`timestamp` < di.`timestamp`
AND dii.`inout` = 1))
为了获得夜间移位器的标记，我们可以使用相同的逻辑，但切换所有内容：
>在日期比较中变为<，反之亦然。0，则标记为1，反之亦然。对于印章，我们使用印章日当天00:00。
如果我们UNION ALL两个查询，我们都会获得夜间移位器的部分。
该团伙的其余成员也称为日间移民
现在可以轻松获得日间移位器的印章。我们只需要否定我们用来寻找夜班者的条件。获得印章：
SELECT d.`timestamp`,
d.`name`
FROM `door` d
WHERE d.`inout` = 1
AND (EXISTS (SELECT *
FROM `door` di
WHERE di.`name` = d.`name`
AND di.`inout` = 0
AND di.`timestamp` > d.`timestamp`
AND date(di.`timestamp`) = date(d.`timestamp`))
OR NOT EXISTS (SELECT *
FROM `door` di
WHERE di.`name` = d.`name`
AND di.`inout` = 0
AND date(di.`timestamp`) = date_add(date(d.`timestamp`), INTERVAL 1 DAY)
AND NOT EXISTS (SELECT *
FROM `door` dii
WHERE dii.`name` = di.`name`
AND date(dii.`timestamp`) = date_add(date(d.`timestamp`), INTERVAL 1 DAY)
AND dii.`timestamp` < di.`timestamp`
AND dii.`inout` = 1)))
对于外面的邮票，我们可以再次切换，>变为<等，如上所述。
当然，在这里我们不能为邮票生成印章，反之亦然。我们必须FULL OUTER JOIN邮票和日间变速器的邮票。但MySQL至少在较低版本中不支持此操作。因此，我们按UNION两个LEFT JOIN进行操作，将第二个LEFT JOIN中的角色与第一个角色进行比较。
全部放在一起
现在我们可以简单地UNION ALL夜间移民和日间移位器来获取所有日子的完整结果。我们可以SELECT FROM，以获得特定日期的结果：
SELECT coalesce(x.`timestamp_in`, '-') `d1-in-timestamp`,
coalesce(x.`timestamp_out`, '-') `d0-out-timestamp`,
x.`name`
FROM (SELECT r.`timestamp` `timestamp_in`,
s.`timestamp` `timestamp_out`,
r.`name`
FROM (SELECT d.`timestamp`,
d.`name`
FROM `door` d
WHERE d.`inout` = 1
AND (EXISTS (SELECT *
FROM `door` di
WHERE di.`name` = d.`name`
AND di.`inout` = 0
AND di.`timestamp` > d.`timestamp`
AND date(di.`timestamp`) = date(d.`timestamp`))
OR NOT EXISTS (SELECT *
FROM `door` di
WHERE di.`name` = d.`name`
AND di.`inout` = 0
AND date(di.`timestamp`) = date_add(date(d.`timestamp`), INTERVAL 1 DAY)
AND NOT EXISTS (SELECT *
FROM `door` dii
WHERE dii.`name` = di.`name`
AND date(dii.`timestamp`) = date_add(date(d.`timestamp`), INTERVAL 1 DAY)
AND dii.`timestamp` < di.`timestamp`
AND dii.`inout` = 1)))) r
LEFT JOIN (SELECT d.`timestamp`,
d.`name`
FROM `door` d
WHERE d.`inout` = 0
AND (EXISTS (SELECT *
FROM `door` di
WHERE di.`name` = d.`name`
AND di.`inout` = 1
AND di.`timestamp` < d.`timestamp`
AND date(di.`timestamp`) = date(d.`timestamp`))
OR NOT EXISTS (SELECT *
FROM `door` di
WHERE di.`name` = d.`name`
AND di.`inout` = 1
AND date(di.`timestamp`) = date_add(date(d.`timestamp`), INTERVAL -1 DAY)
AND NOT EXISTS (SELECT *
FROM `door` dii
WHERE dii.`name` = di.`name`
AND date(dii.`timestamp`) = date_add(date(d.`timestamp`), INTERVAL -1 DAY)
AND dii.`timestamp` > di.`timestamp`
AND dii.`inout` = 0)))) s
ON s.`name` = r.`name`
AND date(s.`timestamp`) = date(r.`timestamp`)
AND s.`timestamp` > r.`timestamp`
UNION
SELECT u.`timestamp` `timestamp_in`,
t.`timestamp` `timestamp_out`,
t.`name`
FROM (SELECT d.`timestamp`,
d.`name`
FROM `door` d
WHERE d.`inout` = 0
AND (EXISTS (SELECT *
FROM `door` di
WHERE di.`name` = d.`name`
AND di.`inout` = 1
AND di.`timestamp` < d.`timestamp`
AND date(di.`timestamp`) = date(d.`timestamp`))
OR NOT EXISTS (SELECT *
FROM `door` di
WHERE di.`name` = d.`name`
AND di.`inout` = 1
AND date(di.`timestamp`) = date_add(date(d.`timestamp`), INTERVAL -1 DAY)
AND NOT EXISTS (SELECT *
FROM `door` dii
WHERE dii.`name` = di.`name`
AND date(dii.`timestamp`) = date_add(date(d.`timestamp`), INTERVAL -1 DAY)
AND dii.`timestamp` > di.`timestamp`
AND dii.`inout` = 0)))) t
LEFT JOIN (SELECT d.`timestamp`,
d.`name`
FROM `door` d
WHERE d.`inout` = 1
AND (EXISTS (SELECT *
FROM `door` di
WHERE di.`name` = d.`name`
AND di.`inout` = 0
AND di.`timestamp` > d.`timestamp`
AND date(di.`timestamp`) = date(d.`timestamp`))
OR NOT EXISTS (SELECT *
FROM `door` di
WHERE di.`name` = d.`name`
AND di.`inout` = 0
AND date(di.`timestamp`) = date_add(date(d.`timestamp`), INTERVAL 1 DAY)
AND NOT EXISTS (SELECT *
FROM `door` dii
WHERE dii.`name` = di.`name`
AND date(dii.`timestamp`) = date_add(date(d.`timestamp`), INTERVAL 1 DAY)
AND dii.`timestamp` < di.`timestamp`
AND dii.`inout` = 1)))) u
ON u.`name` = t.`name`
AND date(u.`timestamp`) = date(t.`timestamp`)
AND u.`timestamp` < t.`timestamp`
UNION
SELECT d.`timestamp` `timestamp_in`,
timestampadd(second, -1 * second(d.`timestamp`), timestampadd(minute, -1 * minute(d.`timestamp`), timestampadd(hour, -1 * hour(d.`timestamp`), timestampadd(day, 1, (d.`timestamp`))))) `timestamp_out`,
d.`name`
FROM `door` d
WHERE d.`inout` = 1
AND NOT EXISTS (SELECT *
FROM `door` di
WHERE di.`name` = d.`name`
AND di.`inout` = 0
AND di.`timestamp` > d.`timestamp`
AND date(di.`timestamp`) = date(d.`timestamp`))
AND EXISTS (SELECT *
FROM `door` di
WHERE di.`name` = d.`name`
AND di.`inout` = 0
AND date(di.`timestamp`) = date_add(date(d.`timestamp`), INTERVAL 1 DAY)
AND NOT EXISTS (SELECT *
FROM `door` dii
WHERE dii.`name` = di.`name`
AND date(dii.`timestamp`) = date_add(date(d.`timestamp`), INTERVAL 1 DAY)
AND dii.`timestamp` < di.`timestamp`
AND dii.`inout` = 1))
UNION ALL
SELECT timestampadd(second, -1 * second(d.`timestamp`), timestampadd(minute, -1 * minute(d.`timestamp`), timestampadd(hour, -1 * hour(d.`timestamp`), d.`timestamp`))) `timestamp_in`,
d.`timestamp` `timestamp_out`,
d.`name`
FROM `door` d
WHERE d.`inout` = 0
AND NOT EXISTS (SELECT *
FROM `door` di
WHERE di.`name` = d.`name`
AND di.`inout` = 1
AND di.`timestamp` < d.`timestamp`
AND date(di.`timestamp`) = date(d.`timestamp`))
AND EXISTS (SELECT *
FROM `door` di
WHERE di.`name` = d.`name`
AND di.`inout` = 1
AND date(di.`timestamp`) = date_add(date(d.`timestamp`), INTERVAL -1 DAY)
AND NOT EXISTS (SELECT *
FROM `door` dii
WHERE dii.`name` = di.`name`
AND date(dii.`timestamp`) = date_add(date(d.`timestamp`), INTERVAL -1 DAY)
AND dii.`timestamp` > di.`timestamp`
AND dii.`inout` = 0))) x
WHERE date(x.`timestamp_in`) = '2018-04-02'
OR date(x.`timestamp_out`) = '2018-04-02'
ORDER BY x.`name`,
x.`timestamp_in`;