mysql - 如何形成 SQL 查询

Question

这是我需要解决的问题:

这是我需要为每个用户获取的表格。

Date        | Working time (day) | Night shifts (23:00 - 06:00) | Total(day + night) |             | Notes
3.9.        |    7               |               2              |         9          |      1      | 
4.9.        |    8               |               0              |         8          |             |  
5.9.        |    6               |               2              |         8          |             |
6.9.        |    4               |               3              |         7          |             |
7.9.        |    0               |               0              |         0          |             |
8.9.        |    0               |               0              |         0          |             |
9.9.        |    0               |               0              |         0          |             |
10.9.       |    0               |               0              |         0          |             |
Total(week) |    25              |               7              |         32         |  Total(>48h)|    0    

所以让我解释一下。在此示例中，用户选择的星期是 3.9-10.9，用户具有唯一键。如果可能的话，这个表中的所有内容都需要在 sql 中生成。所以第一列是日期，第二列是(06:00-23:00)之间的工作时间，第三列是(23:00-06:00)之间的夜类，第四列是白天和晚上的总计移位，第五列仅在第四列大于 8 时才填充，即总数的 9h - 8 为 1，仅当底部的(日类和夜类)总数大于 48h 时才填充第六列，然后将差异写入最后一列。

我当前的数据表有以下列:user_key,id,time

现在这将是查询应该做什么的示例:

    1. user_key(100), id(1), time(15:00); 
    2. user_key(200), id(2), time(15:05); 
    3. user_key(100), id(3), time(16:00); 
    4. user_key(100), id(4), time(16:05); 
    5. user_key(100), id(5), time(17:05); 

    3.-1. = 1h 
    5.-4. = 1h 

    Result: 
    user_key(100), day_shift(2h)

所以我们 user_key 对于每个用户都是唯一的并且 id 是自动递增的，我们总是需要为同一个用户获取第一条和第二条记录并保存结果，然后为该用户获取第二条记录并节省工作时间等等...如何做到这一点？我需要使用一些sql变量吗？当然如果是白类和夜类就必须要有选择，因为我们不想把工作时间浪费在错误的栏目上。

编辑:

据我所知，你们中的许多人并不了解我真正需要的是什么。我将尝试对其进行更多解释。

当用户登录时，它会写入自己的 user_key。当他注销时，我在表中写了 user_key(0) 之后另一个用户可以唱歌等等。所以基本上我需要在例如白类中查看第一个唯一用户 key 并获取最后一个。我们可以获得的最后一个 user_key 只是为了检查下一个 user_key 是否不同。如果下一个用户 key 不同，那么我们有第一个和最后一个，它们之间的差异需要保存在变量中。然后例如我们有一些其他用户，稍后我们又有一对 user_key 然后我们只需要将它们之间的差异添加到同一个变量等等......

我的问题也可以用 PHP 解决。例如，我们只能获取特定 user_key 和特定时间的值。然后在 php 中我们可以遍历数据并检查。

请提供一些例子

编辑:

数据:

1447, 0, 2012-10-21 13:32:15.453
1448, 0, 2012-10-21 13:32:22.203
1449, 0, 2012-10-21 13:32:29.203
1450, , 2012-10-21 13:32:35.671
1451, , 2012-10-21 13:32:44.515
1452, , 2012-10-21 13:32:52.62
1453, 0010154108, 2012-10-21 13:32:59.203
1454, 0010154108, 2012-10-21 13:33:06.46
1455, 0010154108, 2012-10-21 13:33:13.171
1456, 0010154108, 2012-10-21 13:33:20.62
1457, 0010154108, 2012-10-21 13:33:27.62
1458, 0010154108, 2012-10-21 13:33:34.171
1459, 0010154108, 2012-10-21 13:33:40.234
1460, 0010154108, 2012-10-21 13:33:47 
1461, 0010154108, 2012-10-21 13:33:53.812
1462, 0010154108, 2012-10-21 13:33:59.828
1463, 0010154108, 2012-10-21 13:34:05.859
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1620, 0010154108, 2012-10-21 13:50:54.906
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Answer 1

我的答案基于下表结构，Rufo 爵士在另一个答案中提供:

CREATE TABLE `userstamp` (
  `user_key`  int(11) NOT NULL ,
  `id`  int(11) NOT NULL AUTO_INCREMENT ,
  `stamptime`  datetime NOT NULL ,
  PRIMARY KEY (`id`)
);

此 View 显示每个用户的每个“事件”间隔:

CREATE VIEW Intervals AS
SELECT
  userstamp.user_key,
  userstamp.stamptime as checkin,
  min(userstamp_2.stamptime) as checkout
FROM
  userstamp inner join userstamp userstamp_2
    on userstamp.user_key = userstamp_2.user_key
WHERE
  userstamp_2.id > userstamp.id
  AND Mod(
    (SELECT
       Count(*)
     FROM
       userstamp as userstamp_1
     WHERE
       userstamp.user_key = userstamp_1.user_key
       AND userstamp.id>userstamp_1.id)
     , 2 ) = 0
GROUP BY
  userstamp.user_key, userstamp.id

WHERE 子句中奇怪的 SELECT 子查询用于通过计算该用户的结帐时间是否在奇数行中来摆脱“非事件”间隔(如 16:00 - 16:05)。

现在我们必须处理一个事件间隔可能在一天开始并在第二天结束的事实。这也是将日期转换为秒的正确位置。我会使用这个基于 UNION 查询的 View :

CREATE VIEW intervals_2 AS
SELECT
  user_key,
  CAST(checkin AS DATE) AS day,
  TIME_TO_SEC(CAST(checkin AS TIME)) AS checkinsec,
  TIME_TO_SEC(CAST(checkout AS TIME)) AS checkoutsec
FROM intervals WHERE CAST(checkin AS date) = CAST(checkout AS date)
UNION
SELECT
   user_key,
   CAST(checkin AS date) AS day,
   TIME_TO_SEC(CAST(checkin AS TIME)) AS checkinsec,
   86400 AS checkoutsec
FROM intervals WHERE CAST(checkin AS date) < CAST(checkout AS date)
UNION
SELECT
  user_key,
  CAST(checkout AS DATE) AS day,
  0 AS checkinsec,
  TIME_TO_SEC(CAST(checkout AS TIME)) AS checkoutsec
FROM intervals WHERE CAST(checkin AS date) < CAST(checkout AS date)

好的!现在我们准备做一些基本的求和。这个是计算每一天的总和:

SELECT
  user_key,
  day,
  SEC_TO_TIME( SUM(IF(checkinsec<82800 and checkoutsec>21600,LEAST(82800, checkoutsec) - GREATEST(21600, checkinsec),0))) as WorkHours,
  SEC_TO_TIME( SUM(IF(checkinsec<21600, 21600-checkinsec, 0)+IF(checkoutsec>82800, checkoutsec-GREATEST(82800, checkinsec), 0))) as NightShift,
  SEC_TO_TIME( SUM(checkoutsec-checkinsec) ) as TotalDay
FROM
  intervals_2
WHERE
  day BETWEEN startdate AND enddate
GROUP BY
  user_key,
  day

但是如果您只需要整周的总和，您可以从该字段和 GROUP BY 中删除 DATE(checkin)。