Question

以下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

让我解释一下。用户在本例中选择周是3.9-10.9，用户有唯一密钥。如果可能的话，需要在sql中生成此表中的所有内容。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）之间的夜班，第四列是白天和晚上的总和移位，第五列仅在第四列大于8时填充，即总共9小时 - 8为1，如果底部的（白天和夜班）总数大于48小时，则仅填充第六列，然后写入差异最后一栏。

我当前的数据表包含以下列：user_key,id,time

现在这将是查询应该做什么的一个例子：

    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对每个用户都是唯一的，而id是自动递增的，我们总是需要为同一个用户取第一个和第二个记录并保存结果，然后为该用户输入第二个记录并节省工作时间等等... 怎么做到这一点？我需要使用一些sql变量？当然，如果是白班和夜班，它必须有选项，因为我们不想在错误的栏目中节省工作时间。

编辑：

我看到很多人都不理解我真正需要的东西。我会尝试解释一下。

当用户登录时，它自己写了user_key。当他注销然后我在表中写了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班中看到第一个唯一的用户密钥并获得最后一个。我们可以得到的最后一个user_key来检查下一个user_key是否不同。如果下一个用户密钥不同，那么我们有第一个用户密钥和最后一个用户密钥，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......

我的问题也可以在PHP中解决。例如，我们只能获得某个user_key和特定时间的值。然后在php中我们可以通过数据检查。

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
1464, 0010154108, 2012-10-21 13:34:11.828
1465, 0010154108, 2012-10-21 13:34:18.843
1466, 0010154108, 2012-10-21 13:34:24.828
1467, 0010154108, 2012-10-21 13:34:30.781
1468, 0010154108, 2012-10-21 13:34:36.765
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1471, 0010154108, 2012-10-21 13:34:55.593
1472, 0010154108, 2012-10-21 13:35:03.156
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1476, 0010154108, 2012-10-21 13:35:31.750
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1575, 0010154108, 2012-10-21 13:46:09.906
1576, 0010154108, 2012-10-21 13:46:15.906
1577, 0010154108, 2012-10-21 13:46:22.921
1578, 0010154108, 2012-10-21 13:46:28.859
1579, 0010154108, 2012-10-21 13:46:34.796
1580, 0010154108, 2012-10-21 13:46:40.703
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1582, 0010154108, 2012-10-21 13:46:54.687
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1603, 0010154108, 2012-10-21 13:49:07.843
1604, 0010154108, 2012-10-21 13:49:13.890
1605, 0010154108, 2012-10-21 13:49:19.890
1606, 0010154108, 2012-10-21 13:49:28.468
1607, 0010154108, 2012-10-21 13:49:34.500
1608, 0010154108, 2012-10-21 13:49:40.484
1609, 0010154108, 2012-10-21 13:49:46.453
1610, 0010154108, 2012-10-21 13:49:52.859
1611, 0010154108, 2012-10-21 13:49:58.984
1612, 0010154108, 2012-10-21 13:50:04.875
1613, 0010154108, 2012-10-21 13:50:10.875
1614, 0010154108, 2012-10-21 13:50:17.796
1615, 0010154108, 2012-10-21 13:50:23.812
1616, 0010154108, 2012-10-21 13:50:29.796
1617, 0010154108, 2012-10-21 13:50:35.812
1618, 0010154108, 2012-10-21 13:50:42.812
1619, 0010154108, 2012-10-21 13:50:48.828
1620, 0010154108, 2012-10-21 13:50:54.906
1621, 0, 2012-10-21 13:51:00.828
1622, 0, 2012-10-21 13:51:07.890
1623, 0, 2012-10-21 13:51:13.796
1624, 0, 2012-10-21 13:51:19.796
1625, 0, 2012-10-21 13:51:25.812
1626, 0, 2012-10-21 13:51:32.828
1627, 0, 2012-10-21 13:51:38.828
1628, 0, 2012-10-21 13:51:44.796
1629, 0, 2012-10-21 13:51:50.781
1630, 0, 2012-10-21 13:51:57.750
1631, 0, 2012-10-21 13:52:03.718
1632, 0, 2012-10-21 13:52:12.718
1633, 0, 2012-10-21 13:52:18.765
1634, 0, 2012-10-21 13:52:25.46
1635, 0, 2012-10-21 13:52:32.796
1636, 0, 2012-10-21 13:52:39.765
1637, 0, 2012-10-21 13:52:45.781
1638, 0, 2012-10-21 13:52:52.812
1639, 0, 2012-10-21 13:52:59.218
1640, 0, 2012-10-21 13:53:05.250
1641, 0, 2012-10-21 13:53:12.765
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1644, 0, 2012-10-21 13:53:30.890
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1663, 0, 2012-10-21 13:55:30.718
1664, 0, 2012-10-21 13:55:37.703
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1666, 0, 2012-10-21 13:55:49.671
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1669, 0, 2012-10-21 13:56:08.625
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1690, 0, 2012-10-21 13:58:28.703
1691, 0, 2012-10-21 13:58:35 
1692, 0, 2012-10-21 13:58:42.656
1693, 0, 2012-10-21 13:58:49.468
1694, 0, 2012-10-21 13:58:55.453
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1776, 0, 2012-10-21 14:07:35.328
1777, 0, 2012-10-21 14:07:42.343
1778, 0, 2012-10-21 14:07:48.296
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1805, 0, 2012-10-21 14:10:57.265
1806, 0, 2012-10-21 14:11:03.125
1807, 0, 2012-10-21 14:11:09.125
1808, 0, 2012-10-21 14:11:15.125
1809, 0, 2012-10-21 14:11:22.125
1810, 0, 2012-10-21 14:11:28.109
1811, 0, 2012-10-21 14:11:35.390
1812, 0, 2012-10-21 14:11:42.15
1813, 0, 2012-10-21 14:11:48.31
1814, 0, 2012-10-21 14:11:54.46
1815, 0, 2012-10-21 14:12:00.78
1816, 0, 2012-10-21 14:12:08.140
1817, 0, 2012-10-21 14:12:14.140
1818, 0, 2012-10-21 14:12:20.171
1819, 0, 2012-10-21 14:12:27.218
1820, 0, 2012-10-21 14:12:33.171
1821, 0, 2012-10-21 14:12:39.125
1822, 0, 2012-10-21 14:12:46 
1823, 0, 2012-10-21 14:12:52.93
1824, 0, 2012-10-21 14:12:58.593
1825, 0, 2012-10-21 14:13:05.484
1826, 0, 2012-10-21 14:13:12.78
1827, 0, 2012-10-21 14:13:18.93
1828, 0, 2012-10-21 14:13:24.109
1829, 0, 2012-10-21 14:13:30.140
1830, 0, 2012-10-21 14:13:37.140
1831, 0, 2012-10-21 14:13:43.203
1832, 0, 2012-10-21 14:13:49.281
1833, 0, 2012-10-21 14:13:57.296
1834, 0, 2012-10-21 14:14:03.203
1835, 0, 2012-10-21 14:14:09.671
1836, 0, 2012-10-21 14:14:17.265
1837, 0, 2012-10-21 14:14:23.171
1838, 0, 2012-10-21 14:14:29.187
1839, 0, 2012-10-21 14:14:35.109
1840, 0, 2012-10-21 14:14:43.484
1841, 0, 2012-10-21 14:14:49.515
1842, 0, 2012-10-21 14:14:57.203
1843, 0, 2012-10-21 14:15:03.187
1844, 0, 2012-10-21 14:15:09.125
1845, 0, 2012-10-21 14:15:17.156
1846, 0, 2012-10-21 14:15:23.140
1847, 0, 2012-10-21 14:15:29.109
1848, 0, 2012-10-21 14:15:35.109
1849, 0, 2012-10-21 14:15:42.125
1850, 0, 2012-10-21 14:15:48.93
1851, 0, 2012-10-21 14:15:54.125
1852, 0, 2012-10-21 14:16:00.109
1853, 0, 2012-10-21 14:16:07.546
1854, 0, 2012-10-21 14:16:14.78
1855, 0, 2012-10-21 14:16:20.156
1856, 0, 2012-10-21 14:16:28.93
1857, 0, 2012-10-21 14:16:34.546
1858, 0, 2012-10-21 14:16:42.62
1859, 0, 2012-10-21 14:16:50.437
1860, 0, 2012-10-21 14:17:13.15
1861, 0, 2012-10-21 14:17:20.593
1862, 0, 2012-10-21 14:17:28.125
1863, 0, 2012-10-21 14:17:35.578
1864, 0, 2012-10-21 14:17:43.125
1865, 0, 2012-10-21 14:17:52.484

Answer 1

让我推荐一下这些：

对于用户的每个时间戳，您还应记录它是“签入”还是“签出”。否则你很可能会遇到麻烦。例如，缺少时间戳（可能有原因......）将完全混淆您的评估，因为该用户的下一个时间戳将被错误地解释（例如，作为“结账”而不是“签到”）。 / LI>
我还考虑不将所有评估逻辑放入SQL查询中。在编程语言（php，java，...）中执行此操作时，评估逻辑通常会变得更加简单。您的任务似乎非常复杂，只能在SQL中解决它。例如。为了获得一天的评估，你必须检查前一天的“签到”，其相应的“结账”是你感兴趣的那天，等等。
让您的问题更简洁，例如你的日子似乎从早上6点到早上6点。因此工作时段（即，对于同一用户匹配“签到”和“结账”对）从例如开始。凌晨4点，早上8点结束需要拆分为您的摘要？
您应该添加完整性检查，例如单个工作期最长可以达到多长时间。

Answer 2

我的答案基于下表结构，Rufo爵士提供了另一个答案：

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“活动”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子查询是通过计算该用户的结账时间是否在奇数行中来消除“非活动”间隔（如16:00 - 16:05）。

现在我们必须处理这样一个事实，即活动间隔可以在一天内开始，并在第二天结束。这也是将日期转换为秒的正确位置。我会根据UNION查询使用此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！现在我们准备做一些基本的总和。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

但如果您只需要整周的总和，则可以从字段和GROUP BY中删除DATE（签入）。

Answer 3

我有一个存储过程可以处理这个并返回表结果。恕我直言，一个普通的SQL查询无法处理这个，因为你必须解析每个记录并切换状态“来”和“去”。

这是SP的表格布局

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

调用SP看起来像

CALL get_user_result( 100, '2012-09-03', '2012-09-10' );

这是SP

DROP PROCEDURE IF EXISTS `get_user_result`;
DELIMITER ;;
CREATE PROCEDURE `get_user_result`(IN Auser_key INT, IN AStartDate DATE, IN AEndDate DATE)
BEGIN

DECLARE done INT DEFAULT 0;
DECLARE LKnownUser INT DEFAULT 0;
DECLARE LCstampdate DATE;
DECLARE LCuser_key INT;
DECLARE Lday_shift_inc, Lnight_shift_inc DECIMAL( 9, 6 ) DEFAULT 0; -- CHANGE
DECLARE LCstamptime, Lstamptime DATETIME;

DECLARE c_ustamp CURSOR FOR SELECT user_key, stamptime FROM userstamp WHERE user_key = Auser_key AND stamptime >= ADDDATE( AStartDate, INTERVAL 6 HOUR ) AND stamptime < ADDDATE( DATE_ADD( AEndDate, INTERVAL 6 HOUR ), 1 ) ORDER BY user_key,stamptime;
DECLARE CONTINUE HANDLER FOR SQLSTATE '02000' SET done = 1;

DROP TEMPORARY TABLE IF EXISTS user_result;

CREATE TEMPORARY TABLE user_result (
`stampdate`  date NULL DEFAULT NULL ,
`user_key`  int(11) NOT NULL ,
`day_shift`  decimal(9,6) NOT NULL DEFAULT 0.00 , -- CHANGE
`night_shift`  decimal(9,6) NOT NULL DEFAULT 0.00 , -- CHANGE
`stamptime`  datetime NULL DEFAULT NULL ,
PRIMARY KEY (`stampdate`,`user_key`)
);

OPEN c_ustamp;

REPEAT

    FETCH c_ustamp INTO LCuser_key, LCstamptime;

    IF NOT done THEN

        IF ( TIME( LCstamptime ) >= MAKETIME(00,00,00) ) AND ( TIME( LCstamptime ) < MAKETIME(06,00,00) ) THEN
            SET LCstampdate = DATE_SUB( DATE( LCstamptime ), INTERVAL 1 DAY );
        ELSE
            SET LCstampdate = DATE( LCstamptime );
        END IF;

        SELECT COUNT(*)
        INTO LKnownUser
        FROM user_result
        WHERE stampdate = LCstampdate AND user_key = LCuser_key;

        IF NOT LKnownUser THEN

            INSERT INTO user_result
                SET stampdate = LCstampdate, user_key = LCuser_key, stamptime = LCstamptime, day_shift = 0, night_shift = 0;

        ELSE

            SELECT stamptime
            INTO Lstamptime
            FROM user_result
            WHERE stampdate = LCstampdate AND user_key = LCuser_key;

            IF Lstamptime IS NULL THEN

                UPDATE user_result
                SET stamptime = LCstamptime
                WHERE stampdate = LCstampdate AND user_key = LCuser_key;

            ELSE

                IF ( TIME( Lstamptime ) >=  MAKETIME(06,00,00) ) AND ( TIME( Lstamptime ) < MAKETIME(23,00,00) )
                THEN -- day_shift 06:00:00 til 23:00:00
                    SET Lday_shift_inc = TIME_TO_SEC( TIMEDIFF( LCstamptime, Lstamptime ) ) / 3600.0; -- CHANGE
                    SET Lnight_shift_inc = 0;
                ELSE -- night_shift
                    SET Lday_shift_inc = 0;
                    SET Lnight_shift_inc = TIME_TO_SEC( TIMEDIFF( LCstamptime, Lstamptime ) ) / 3600.0; -- CHANGE
                END IF;

                UPDATE user_result
                SET 
                    day_shift = day_shift + Lday_shift_inc,
                    night_shift = night_shift + Lnight_shift_inc,
                    stamptime = NULL
                WHERE stampdate = LCstampdate AND user_key = LCuser_key;

            END IF;

        END IF;

    END IF;

UNTIL done END REPEAT;

CLOSE c_ustamp;

SELECT stampdate, user_key, day_shift, night_shift
FROM user_result
WHERE day_shift + night_shift > 0;

DROP TEMPORARY TABLE IF EXISTS user_result;

END;;
DELIMITER ;

如何形成SQL查询

3 个答案: