Question

我可以在phpmyadmin中执行请求，以显示今天日期和用户选择的日期之间的天数： SELECT DATEDIFF（CURDATE（），date_instructed）ASDAYS from date_tester

..但我认为这不会更新所有表格，只会在表格中创建新记录时更新。

因此，任何人都可以建议使用用户输入的日期动态更新/计算列中的天数的最佳方法。

它甚至应该放在专栏中吗？或者我应该使用PHP来回显网页上每一行的结果。

如果我应该用脚本操作'date_instructed'，然后在html表格中单独输入它是一个链接，告诉我如何做到这一点？

非常感谢

<?php
include 'dbh.php';
?>
<table class="table table-sm table table-bordered 
  table table-hover table table-striped  "> <!--, table table-inverse-->
  <thead class="thead-inverse">
    <tr class="warning">
      <th>First Name</th>
      <th>Last</th>
      <th>Date Instructed</th>
      <th>Days On Market</th>
    </tr>      
  </thead>

    <?php

    $query  = "SELECT  first_name, last_name, date_instructed , days_onmarket 
                FROM date_tester";
    $result = mysqli_query($conn, $query);
    while ($date_tester = mysqli_fetch_assoc($result)) {
        echo "<tr>";
        echo "<td>" . $date_tester['first_name'] . "</td>";
        echo "<td>" . $date_tester['last_name'] . "</td>";
        echo "<td>" . $date_tester['date_instructed'] . "</td>";
    }
    ?>
    </tbody>
</table>
</div>

Answer 1

据我所知，这就是你所需要的：

 $query  = "SELECT  first_name, last_name, date_instructed ,    days_onmarket, DATEDIFF(CURDATE(), 'date_instructed') AS DAYS FROM date_tester";
   $result = mysqli_query($conn, $query);
while ($date_tester = mysqli_fetch_assoc($result)) {
    echo "<tr>";
    echo "<td>" . $date_tester['first_name'] . "</td>";
    echo "<td>" . $date_tester['last_name'] . "</td>";
    echo "<td>" . $date_tester['date_instructed'] . "</td>";
    echo "<td>" . $date_tester['DAYS'] . "</td>";
}

如何动态地将datediff插入网页以实时更新（现在）

1 个答案: