Question

如何将具有相同日期的同一行分组在一起..

我想要这样的桌子具有相同日期分组的每一行都被尝试了group_concat和group_by，但是它不起作用，我希望php中的一些过滤器将具有相同日期的行合并在一起

我的代码：

   while ($row = mysqli_fetch_array($result)) {
                            $start_res = strtotime($row['from_date']);
                            $end_res = strtotime($row['to_date']);
                            for ($res = $start_res; $res <= $end;$res+=86400) {
                                $res_date = date("m/d/Y", $res);
                                $startres = strtotime($res_date);

                                if ($startres >= $start_res && $startres<=$end_res) {
                      ?>
                                    <tr>
                                        <td><?= $res_date ?> </td>
                                        <td><?= $row['1'] ?> </td>
                                        <td><?= $row['2'] ?> </td>
                                        <td><?= $row['3'] ?> </td>
                                        <td><?= $row['4'] ?> </td>
                                        <td><?= $row['5'] ?> </td>
                                        <td><?= $row['6'] ?> </td>
                                        <td>0 </td>
                                        <td style="background-color:gold"> 0</td>
                                        <td>0 </td>        
                                        <td>0</td>
                                        <td>0</td>
                                        <td>0</td>
                                    </tr>
        <?php }}}?>

我的桌子： sql查询：

       $sql = "
         SELECT * FROM  mytable
        WHERE hotel_id = '" . $hotel_id . "'          
        and from_date  between '" . $date . "' and '" . $end_date . "'
        or to_date  between '" . $date . "' and '" . $end_date . "'
        ORDER BY from_date asc

        ";
    $result = mysqli_query($con, $sql);

Answer 1

根据您的表结构，用PHP进行此操作可能最简单。要在SQL中执行此操作，需要一个日历表将JOIN连接到每一行。因此，您需要汇总每个日期的值，然后输出它们。注意，我更喜欢将列名作为数组索引使用，因为这样更容易查看正在执行的操作：

$data = array();
// collate the data
while ($row = mysqli_fetch_array($result)) {
    $start_res = strtotime($row['from_date']);
    $end_res = strtotime($row['to_date']);
    for ($res = $start_res; $res <= $end && $res <= $end_res; $res += 86400) {
        if (!isset($data[$res])) $data[$res] = array('totals' => 0, 'agent' => 0, 'direct' => 0, 'over_agent' => 0, 'over_hotel' => 0, 'allotment' => 0);
        $data[$res]['totals'] += $row['totals'];
        $data[$res]['agent'] += $row['agent'];
        $data[$res]['direct'] += $row['direct'];
        $data[$res]['over_agent'] += $row['over_agent'];
        $data[$res]['over_hotel'] += $row['over_hotel'];
        $data[$res]['allotment'] += $row['allotment'];
    }
}
// sort the data by date
ksort($data);
// now display the combined data
foreach ($data as $res_date => $row) {
?>
<tr>
    <td><?= date('m/d/Y', $res_date) ?> </td>
    <td><?= $row['totals'] ?> </td>
    <td><?= $row['agent'] ?> </td>
    <td><?= $row['direct'] ?> </td>
    <td><?= $row['over_agent'] ?> </td>
    <td><?= $row['over_hotel'] ?> </td>
    <td><?= $row['allotment'] ?> </td>
    <td>0</td>
    <td style="background-color:gold"> 0</td>
    <td>0 </td>        
    <td>0</td>
    <td>0</td>
    <td>0</td>
</tr>
<?php
}

如何合并具有相同日期的行

1 个答案: