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我正在创建一个反馈表单，允许用户编写他们的反馈并使用 php 和 mysqli 我根据用户的用户名将这些反馈存储在数据库中我确实成功插入数据但没有用户的用户名所以问题是：当我写了一个更新查询我变得无法插入任何数据有人可以帮助我吗？

反馈表格.php

<?php

session_start();
 $login = ($_SESSION['login']);
   $userid = ($_SESSION['user_id']);
   $login_user = ($_SESSION['username']);
   $fname = ($_SESSION['first_name']);
   $lname = ($_SESSION['last_name']);
   $sessionaddres =($_SESSION['address']);
?>
<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">
<html xmlns="http://www.w3.org/1999/xhtml">
<head>
<meta http-equiv="Content-Type" content="text/html; charset=utf-8" />
<title>feedback page</title>
    <script type = "text/javascript" src = "http://ajax.googleapis.com/ajax/libs/jquery/1.9.1/jquery.min.js"></script>
    <link href="style/stylesheet.css"rel="stylesheet" type="text/css"/>
    <script type = "text/javascript">
    $(function(){
       $('#submit').click(function(){
         $('#container').append('<img src = "images/loading.gif" alt="Currently loading" id = "loading" />');
             var comments = $('#comments').val();
             $.ajax({
                url: 'feedback_process.php',
                type: 'POST',
                data: {"comments": comments},
                success: function(result){
                     $('#response').remove();
                     $('#container').append('<p id = "response">' + result + '</p>');
                     $('#loading').fadeOut(500, function(){
                         $(this).remove();
                     });
                }
             });         
            return false;
       });
    });
    </script>
    </head>
<?php require_once('header.php'); ?>
<body>
<form action = "feedback_form.php" method = "post">
  <div id = "container">
            <h2><?php echo $login_user ?></h2>

          <label for = "comments">Comments</label>
          <textarea rows = "5"cols = "35" name = "comments" id = "comments"></textarea>
          <br />
  </div>
   </form>
       <input type = "submit" name = "submit" id = "submit" value = "send feedBack" />
</body>
</html>

反馈过程.php

<?php

session_start();

 $login = ($_SESSION['login']);
   $userid = ($_SESSION['user_id']);
   $login_user = ($_SESSION['username']);
   $fname = ($_SESSION['first_name']);
   $lname = ($_SESSION['last_name']);
   $sessionaddres =($_SESSION['address']);


$conn = new mysqli('localhost', 'root', '', 'lam_el_chamel_db');

  echo"<pre>";
  print_r($_POST);
  echo"</pre>";

  if(isset($_POST['comments'])){

  $comments = $_POST['comments'];



  $query = "INSERT into feedback (feedback_text user_name,) VALUES(?,?)";

  $stmt = $conn->stmt_init();
  if($stmt->prepare($query))
  {

     $stmt->bind_param('ss', $comments, $login_user);
     //$stmt->execute();

  }
  $query2 = "UPDATE feedback SET (feedback = ?, user_name = ?) WHERE user_name = '$login_user' ";
  $stmt = $conn->stmt_init();
  if($stmt->prepare($query))
  {
     $stmt->bind_param('ss', $comments, $login_user);
     $stmt->execute();

  }



  if($stmt){

  echo "thank you .we will be in touch soon <br />";

  }
  else{
   echo "there was an error. try again later.";
   }  

}

else
   echo"it is a big error";
?>
4

3 回答 3

0

You can also try

  $query2 = "UPDATE feedback SET (feedback = ?, user_name = ?) WHERE user_name = '$login_user' ";
  $stmt = $conn->query($query2);
于 2013-03-24T06:35:43.017 回答
0

有不止一个错误/错误。

请参阅以下代码段：

第一的

$query = "INSERT into feedback (feedback_text user_name,) VALUES(?,?)";

您没有,（逗号）分隔字段。利用：

$query = "INSERT into feedback (feedback_text, user_name) VALUES(?, ?)";

下一个

 //$stmt->execute();

您没有执行此语句。利用

 $stmt->execute();

其他

$query2 = "UPDATE feedback SET (feedback = ?, user_name = ?) WHERE user_name = '$login_user' ";

查询中的字段名称在这里和上面不同INSERT。也许试试

$query2 = "UPDATE feedback SET feedback_text = ?, user_name = ? WHERE user_name = ? ";
$stmt = $conn->stmt_init();
if($stmt->prepare($query)) {
    $stmt->bind_param('sss', $comments, $login_user, $login_user);
    $stmt->execute();
}
于 2013-03-24T06:43:43.077 回答
0 
  $query2 = "UPDATE feedback SET (feedback = ?, user_name = ?) WHERE user_name = '$login_user' ";
  $stmt = $conn->stmt_init();
  if($stmt->prepare($query2))
于 2013-03-23T23:56:41.260 回答