Question

如何使用PHP找出MySQL中缺少的日期？

echo "<tr> <th>Username</th><th>Date</th><th>Check In</th><th>Check Out</th> </tr>";

// get results1 from database 
$result1 = mysql_query("SELECT * FROM attend WHERE user_name='ali' AND date BETWEEN '2015-07-01' AND '2015-07-15' order by date");

while($row = mysql_fetch_array($result1))
{   
    // echo out the contents of each row into a table
    echo "<tr>";
    echo '<td>' . $row['user_name'] . '</td>';
    echo '<td>' . $row['date'] . '</td>';
    echo '<td>' . $row['checkin'] . '</td>';
    echo '<td>' . $row['checkout'] . '</td>';
    echo "</tr>";    
}
echo "</table>";

目前的结果

Username    Date    Check In    Check Out
ali     2015-07-01  11:30:34    17:23:47
ali     2015-07-02  10:11:34    17:15:15
ali     2015-07-03  09:32:34    18:16:27
ali     2015-07-06  10:41:34    16:56:13
ali     2015-07-07  08:51:34    17:36:01
ali     2015-07-08  05:61:34    17:16:26
ali     2015-07-09  04:11:34    17:14:12
ali     2015-07-10  02:81:34    17:25:25
ali     2015-07-13  11:71:34    17:02:29
ali     2015-07-14  10:81:34    17:04:20
ali     2015-07-15  09:31:34    17:00:43

我想要这样的结果

Username    Date    Check In    Check Out
ali     2015-07-01  11:30:34    17:23:47
ali     2015-07-02  10:11:34    17:15:15
ali     2015-07-03  09:32:34    18:16:27
        2015-07-04
        2015-07-05
ali     2015-07-06  10:41:34    16:56:13
ali     2015-07-07  08:51:34    17:36:01
ali     2015-07-08  05:61:34    17:16:26
ali     2015-07-09  04:11:34    17:14:12
ali     2015-07-10  02:81:34    17:25:25
        2015-07-11
        2015-07-12
Ali     2015-07-13  11:71:34    17:02:29
ali     2015-07-14  10:81:34    17:04:20
ali     2015-07-15  09:31:34    17:00:43

Answer 1

所有遗失的签到都未分配给任何用户。如果要获取分配给用户ali或未分配的记录，则在两者之间给出日期：

SELECT * FROM attend WHERE (user_name='ali' OR user_name = '') AND date BETWEEN '2015-07-01' AND '2015-07-15' order by date

如果您向我们展示您的架构，那将会更加清晰。

Answer 2

您可以使用函数中的范围执行此操作，这将允许您重用代码：

function getAttendInInterval($begin,$end) {

    $begin = new DateTime( $begin );
    $end = new DateTime( $end );

    $interval = DateInterval::createFromDateString('1 day');
    $period = new DatePeriod($begin, $interval, $end);

    foreach ( $period as $dt ) {
        $result = mysql_query('SELECT * FROM attend WHERE user_name="ali" AND date = '".$dt->format( 'Y-m-d' )."' order by date');

      ...
}

这个功能允许在一个时间间隔内显示每一天的参与者。

然后：

 getAttendInInterval('2015-07-01','2015-07-15');

Answer 3

我会在该日期列中使用时间戳。如果你这样做，你可以简单地“检查”预计接下来会是哪一天。如果它不是那一天，那么按照你想要的那天打印然后继续，如果你有预期的日期。

这样做，您应该在当天的给定时间保存所有时间戳。总是像12:00:00。这使得处理这些数字变得更加容易。

echo "<tr> <th>Username</th><th>Date</th><th>Check In</th><th>Check Out</th> </tr>";

// get results1 from database
$iLastTimestamp = 0 
$result1 = mysql_query("SELECT * FROM attend WHERE user_name='ali' AND date BETWEEN '2015-07-01' AND '2015-07-15' order by date");

while($row = mysql_fetch_array($result1))
{   
    if ($iLastTimestamp != 0)
    {
        while ($iLastTimestamp != $row['date'])
        {
            // echo out the contents of each row into a table
            echo "<tr>";
            echo '<td>&nbsp;</td>';
            echo '<td>' . date("Y-m-d", $iLastTimestamp) . '</td>';
            echo '<td>&nbsp;</td>';
            echo '<td>&nbsp;</td>';
            echo "</tr>";    
            $iLastTimestamp = strtotime("+1 Day 12:00:00", $iLastTimestamp)
        }
    }
    // echo out the contents of each row into a table
    echo "<tr>";
    echo '<td>' . $row['user_name'] . '</td>';
    echo '<td>' . $row['date'] . '</td>';
    echo '<td>' . $row['checkin'] . '</td>';
    echo '<td>' . $row['checkout'] . '</td>';
    echo "</tr>";    
}
echo "</table>";

另一种方法是在使用我发布的代码之前，首先将日期转换为时间戳。你也可以在这里使用“strtotime”-Func。只是添加时间，只是为了确定。：）

Answer 4

您可以在LOOP中使用日期，如果日期在数据库中匹配，则显示所有信息。带有日期的数据库，否则只显示日期。

如何使用php找出mysql中缺少的日期

4 个答案: