MySQL where 语句中的 where 语句

Question

现有 MySQL 循环中的 WHERE 语句似乎有问题，这让我认为这是不可能的。如果可行，我将如何执行？

这个例子使用了它现有的连接，我也试过它自己的连接但没有成功。

我有 2 个表，booking 和 jobs.. Booking 是主要描述数据库的一部分和 jobs 可能有几行与来自 booking

的同一预订相关

<?php
    $conn = mysqli_connect($servername,$username,$password,$dbname);

    if ($conn->connect_error) {
    die("Connection failed: " . $conn->connect_error);
    }

    $sql = "SELECT * FROM rwxoq_Bookings WHERE user_id = $user_id";
    $result = mysqli_query($conn, $sql);
    while($rowb = mysqli_fetch_assoc($result)) { 
?>
    <tr>
      <td valign="top">
        <?php echo $rowb{'Booking_ID'};?>
      </td>
      <td>
       <?php echo date("D, d-m-Y g:ia", strtotime($rowb{'created'}));?>
      </td>
      <td>
        <?php
         $sqlj = "SELECT * FROM rwxoq_jobs WHERE Booking_ID = $rowb{'Booking_ID'}";
         $resultj = mysqli_query($conn, $sqlj);
         while($rowj = mysqli_fetch_assoc($resultj)) {
             $count++;
         }
        echo $count;
?>
      </td>
      <td id="<?php echo $row{'id'};?>">PENDING</td>
    </tr>
<?php } ?>

Answer 1

我知道你想做什么，但这不是一个好的解决方案；
我建议你在你们之间使用连接table： </p> <pre><code>$conn = mysqli_connect($servername,$username,$password,$dbname); if ($conn->connect_error) { die("Connection failed: " . $conn->connect_error); } $sql = "SELECT b.*, COUNT(*) AS cnt FROM rwxoq_Bookings b LEFT JOIN rwxoq_jobs j ON j.Booking_ID = b.Booking_ID WHERE user_id = $user_id GROUP BY b.Booking_ID"; $result = mysqli_query($conn, $sql); $table = ""; while($rowb = mysqli_fetch_assoc($result)) { $table .= " <tr> <td valign='top'> {$rowb['Booking_ID']} </td> <td> ". date('D, d-m-Y g:ia', strtotime($rowb['created'])) ." </td> <td> {$rowb['cnt']} </td> <td id='{$row['id']}'>PENDING</td> </tr>"; } echo $table;

MySQL where 语句中的 where 语句

MySQL where statement in a where statement

php

mysqli

where