Question

我正在开发一个在线时间卡系统并遇到了障碍。我需要做的是在每行计算timeIn和timeOut，以确定员工每天可以进出几次的日期。我有一个工作查询来查找整天的总数，但不是每个单独的行。我已经包含了数据库表的结构以及我希望看到的一些示例。

+----+------------+----------+---------+------------------------------+-------------+------------+-----------+----------+
| id | employeeID | date     | timeIn  | jobDescription               | equipType   | unitNumber | unitHours | timeOut  |
+----+------------+----------+---------+------------------------------+-------------+------------+-----------+----------+
|  8 |          1 | 01/15/13 | 7:00 AM | Loaded sand in Jefferson     | Excavator   | 345        |      NULL | 9:30 PM  |
|  9 |          1 | 01/15/13 | 10:00 AM | Loaded sand in Jefferson     | Excavator   | 345        |      NULL | 12:00 PM |
| 10 |          1 | 01/16/13 | 7:00 AM | Loaded sand in Jefferson     | Excavator   | 345        |      NULL | 5:30 PM  |
| 11 |          1 | 01/17/13 | 7:00 AM | Loaded sand in Jefferson     | Excavator   | 345        |      NULL | 5:30 PM  |
| 12 |          1 | 01/18/13 | 8:00 AM | Backfill in Whispering Creek | Skid Loader | 297        |      NULL | 5:30 PM  |
| 13 |          1 | 01/19/13 | 8:00 AM | Backfill in Whispering Creek | Skid Loader | 297        |      NULL | 3:30 PM  |
|  1 |          1 | 01/20/13 | 6:00    |                              | Excavator   | 01E        |      7238 | 17:00    |
|  2 |          1 | 01/21/13 | 6:00    | Worked in Jefferson          | Excavator   | 01E        |      7238 | 17:00    |
|  3 |          1 | 01/22/13 | 6:00    | Worked in Jefferson          | Excavator   | 02E        |      7238 | 17:30    |
|  4 |          1 | 01/23/13 | 6:00    | Worked in Whispering Creek   | Skid Loader | 32SL       |      2338 | 18:30    |
|  5 |          1 | 01/24/13 | 8:00    | Worked in Hubbard            | Scraper     | 54C        |      9638 | 11:30    |
|  6 |          1 | 01/25/13 | 8:00    | Worked in Jefferson          | Dozer       | 4D         |       941 | 19:30    |
|  7 |          1 | 01/26/13 | 8:00    | Pushed Snow                  | Loader      | 950H       |       342 | 20:30    |
+----+------------+----------+---------+------------------------------+-------------+------------+-----------+----------+

好的，我希望看到的是，例如日期01/15/13出现两次，其中timeIn和timeOut值不同我希望得到一个值，然后说明在行中id为8的时间是2:30，然后再获得另一个值，说id在9的行中有2:00小时。如果可能，我想用一个查询来完成这个。

这是我的查询以获得一整天的时间：

SELECT `employeeID`, SEC_TO_TIME(SUM(TIME_TO_SEC(TIMEDIFF(`timeOut`, `timeIn`))))   AS `totalTime` FROM `timeRecords` WHERE   `date` = "01/15/13"  AND `employeeID` = 1 GROUP BY `employeeID`;

Answer 1

如果我理解你，你需要这个吗？

SELECT `employeeID`, TIMEDIFF(`timeOut`, `timeIn`)  AS `totalTime` FROM `timeRecords`;

您可以使用PDO query查询数据库，该数据库将结果集作为关联数组的数组返回。为了获得更好的性能，您可以使用PDO prepare/execute。

<?php
function getTimeByRow($connection) {
    $sql = "SELECT `employeeID`, TIMEDIFF(`timeOut`, `timeIn`)  AS `totalTime` FROM `timeRecords`";
    foreach ($connection->query($sql) as $row) {
        print $row['employeeID'] . "\t";
        print $row['totalTime'] . "\n";
    }
}
?>

Answer 2

所以你想做这样的事情：

从timeRecords中选择employeeID，date，TIMEDIFF（超时，timein），其中date = x和employeeID = y

获取表格中每行的时差

2 个答案: