在儀表板中顯示數字，每年從 1 開始

Question

我有一個表格，我希望在我的 .php 文件中顯示，因為它會顯示每行的每個數字，但是當談到全新的一年時，它將再次從 1 開始。

.php 文件 html 代碼：

<table id="example">
      <thead>
             <tr>
             <th style="width: 40px">Year</th>
             <th style="width: 40px">Id</th>
             </tr>
     </thead>
<tbody>
<?php
include('include/config1.php');
$query = "SELECT * from table ORDER BY Id DESC";
$result = mysqli_query($db, $query);
while ($row = mysqli_fetch_assoc($result)) {
?>
<tr>
<td><?php
     if($row['status']!=3){ //display the year from the date according to status
       echo date("Y",strtotime($row['sdate'])); }
     else{
       echo date("Y",strtotime($row['ssdate']));
     } ?>
     //the years are the same for both columns in each row
</td>

<td>//The ID that auto starts from 1 every year with an increment of 1 to the newest (not the same as the unique id from the table </td>

<?php}?>
</tbody>
</table>

我必須實現什么樣的 javascript 函數才能使其工作？

注意：插入到MySQL表中的所有數據都不會從數據庫中刪除，狀態只會更改為刪除，僅此而已。

Answer 1

首先，如果你想讓你的記錄年復一年地排序，你還必須按date字段排序！

"SELECT * from table ORDER BY sdate,Id DESC";

其次，您必須循環記住您的前一年，並將其與當前進行比較。 如果不同，則將您的 ID 重置為1

<?php
    include('include/config1.php');
    $query = "SELECT * from table ORDER BY sdate, Id DESC";
    $result = mysqli_query($db, $query);

    $currentYear = null;
    $data = [];
    while ($row = mysqli_fetch_assoc($result)) {
        $year = ($row['status'] == 3) ? date("Y", strtotime($row['ssdate'])) : date("Y", strtotime($row['sdate']));

        $data[$year][] = $row;                     
    }
?>

<table id="example">
<thead>
    <tr>
        <th style="width: 40px">Year</th>
        <th style="width: 40px">Id</th>
        <th style="width: 80px">Machine No</th>
        <th style="width: 80px">Project Name</th>
        <th style="width: 80px">PIC</th>
        <th style="width: 80px">Purpose of Service</th>
    </tr>
</thead>
<tbody>
<?php 
    foreach ($data as $year => $oneYear) {
        for ($i = count($oneYear); $i >= 1; $i--) {
?>
    <tr>
        <td><?= $year ?></td>
        <td><?= $i ?></td>
        <td><?= $oneYear[$i]['Machine_No']; ?></td>
        <td><?= $oneYear[$i]['projectName']; ?></td>
        <td><?= $oneYear[$i]['pic']; ?></td>
        <td><?= substr(str_replace('\r\n', "\r\n", $row['Purpose_of_Service']), 0, 50); ?></td>
    </tr>
<?php
        }
    }
?>
</tbody>
</table>

Answer 2

基於@Dmitriy Gritsenko 的幫助，我設法做到了，但問題是無法顯示表列中的其他數據

<table id="example" >
        <thead>
            <tr>
                <th style="width: 40px">Year</th>
                <th style="width: 40px">Id</th>
                <th style="width: 80px">Machine No</th>
                <th style="width: 80px">Project Name</th>
                <th style="width: 80px">PIC</th>
                <th style="width: 80px">Purpose of Service</th>
            </tr>
        </thead>

        <tbody>
            <?php

                include('include/config1.php');
                $query = "SELECT * from table ORDER BY sdate DESC, Id DESC";
                $result = mysqli_query($db, $query);

                $currentYear = null;
                $data = [];
                while ($row = mysqli_fetch_assoc($result)) {
                    $year = ($row['status'] == 3) ? date("Y", strtotime($row['ssdate'])) : date("Y", strtotime($row['sdate']));

                    $data[$year][] = $row;                     
                }
                foreach ($data as $year => $oneYear) {
                    for ($i = count($oneYear); $i >= 1; $i--) {
            ?>
            <tr>
                <td><?= $year ?></td>
                <td><?= $i ?></td>
                <td><?php echo $row['Machine_No']; ?></td>
                <td><?php echo $row['projectName']; ?></td>
                <td><?php echo $row['pic']; ?></td>

                <td><?php 
                   $row['Purpose_of_Service'] =substr( $row['Purpose_of_Service'],0,50);
                    echo (str_replace('\r\n', "\r\n", $row['Purpose_of_Service'])); 
                    ?>
                </td>

             <?php
                }}
            ?>
        </tbody>
</table>