1

这是我需要解决的问题：

这是我需要为每个用户获取的表。

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。用户在这个例子中选择的星期是从 3.9-10.9，用户有唯一的键。如果可能，此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）之间的夜班，第四列是白天和黑夜的总和shift，仅当第四列大于 8 时才填充第五列，即总计 9h - 8 为 1，仅当底部的（日夜班）总计大于 48h 时才填充第六列，然后将差异写入最后一栏。

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，如果是白班和夜班，它必须有选择，因为我们不想在错误的列中节省工作时间。

编辑：

正如我所见，你们中的很多人并不了解我真正需要什么。我将尝试多解释一下。

当用户登录时，它会写入自己的 user_key。当他退出时，我在表中写了 user_key(0)，之后另一个用户可以唱歌，依此类推。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。如果下一个用户密钥不同，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，我们可以通过数据进行检查。

请给我一些例子

编辑：

数据：

1447, 0, 2012-10-21 13:32:15.453
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4

3 回答 3

2

让我推荐这些东西：

  • 对于用户的每个时间戳，您还应该记录它是“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“结帐”而不是“签入”）。
  • 我还会考虑不要将所有评估逻辑都放入 SQL 查询中。使用编程语言（php、java、...）时，评估逻辑通常会变得更简单。您的任务似乎非常复杂，仅在 SQL 中解决。例如，为了获得一天的评估，您必须检查前一天的“签入”，其对应的“签出”是在您感兴趣的那一天等。
  • 使您的问题更简洁，例如，您的日子似乎是从早上 6 点到早上 6 点。因此，从凌晨 4 点开始到早上 8 点结束的工作时间段（即为同一用户匹配一对“签入”和“签出”）需要为您的摘要拆分吗？
  • 您应该添加健全性检查，例如单个工作期最多可以持续多长时间。
于 2012-10-20T11:08:55.243 回答
2

我的答案基于下表结构，Rufo 爵士在另一个答案中提供了该结构：

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

此视图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帐时间是否在奇数行中来摆脱“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。我会使用这个 VIEW，基于一个 UNION 查询：

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。这是计算每天的总和：

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，您可以从字段和 GROUP BY 中删除 DATE(checkin)。

于 2012-10-21T15:34:51.057 回答
0

我有一个存储过程供您处理并返回一个表格结果。恕我直言，普通的 SQL 查询无法处理此问题，因为您必须解析每条记录并切换状态“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 ;
于 2012-10-20T13:58:29.490 回答