Question

我意识到这可能是一个简单的执行问题，但我被困住了。我有记录 DELETE 就好了，但它没有显示在第二个表中。

 <?php
   if( $_SERVER['REQUEST_METHOD']=='GET' && isset( $_GET['job_numb'] ) ){
   $job_numb = filter_input( INPUT_GET, 'job_numb', FILTER_SANITIZE_STRING );
   }
 ?>

请注意上述网址中的服务器请求。这就是我通过$job_numb的方式。现在，我将在下面的SELECT查询中重新选择：

 $servername = "localhost";
 $username = "ccccc";
 $password = "xxxxx";
 $dbname = "jobs_users";

     // Create connection
     $conn = mysqli_connect($servername, $username, $password, $dbname);

     // Check connection
     if (!$conn) {
          die("Connection failed: " . mysqli_connect_error());
     }

 $sql = "SELECT * FROM jobs_canjobs WHERE job_numb = $job_numb";
 $results = mysqli_query($conn, $sql);

 if ($row = mysqli_fetch_array($results)){
      $job_name = $row['job_name'];
      $comments = $row['comments'];
      $due_date = $row['due_date'];
      $show_date = $row['show_date'];
      $attachment1 = $row['attachment1'];
      $requestor = $row['requestor'];
      $status = $row['status'];
      $req_email = $row['req_email'];
      $Property = $row['Property'];
      $assignee = $row['assignee'];
      $assign_email = $row['assign_email'];
      $AE = $row['AE'];
 }else{
      echo 'no records found';
 }
      mysqli_close($conn);
?>

我还可以将这些结果放入一个数组中并从每个数据中选择，但在撰写本文时，这就是代码存在的方式所以我没有编辑该部分。

现在，所有这些变量都是从MySQL中加载的，与查询的$job_numb相对应，您可以将它们添加到页面中以获得结果。

Answer 1

尝试

  $sql = "INSERT INTO `archived_jobs` SELECT * from `jobs_canjobs` where job_numb = $job_numb";
  $results = mysqli_query($conn, $sql);
  $sql = "DELETE FROM `jobs_canjobs` WHERE job_numb = $job_numb";
  $results = mysqli_query($conn, $sql);

从一个SQL表中删除并添加到另一个SQL表中

1 个答案: