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我正在创建一个反馈页面,允许用户提供反馈并使用 php 和 mysqli 将此反馈存储在数据库中,而无需使用 jquery 和 ajax 刷新页面,但问题是我没有得到任何插入的数据,尽管我得到了成功如果有人可以帮助我,我将不胜感激

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">
<br />
<br />

  <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" />



<?php require_once('footer.php'); ?>

</body>
</html> 

反馈过程.php

<?php

session_start();
if($_SESSION['login'] != 'true'){
        header("location:index.php");
    }


   $login = ($_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', '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 = $conn->stmt_init();
  var_dump($stmt);

  if($stmt->prepare($query))
  {

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

  }
  $query2 = "UPDATE feedback SET feedback_text = ?, user_name = ? WHERE user_name = ? ";

  $stmt = $conn->stmt_init();
  if($stmt->prepare($query2))
  {
     $stmt->bind_param('sss', $comments, $login_user, $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";
?>

表字段是:feedback_id feedback_text、user_name

4

2 回答 2

0

您的查询中有一个额外的 , 。

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

//new
$query = "INSERT into feedback (feedback_text, user_name) VALUES(?,?)";
于 2013-03-27T08:05:43.980 回答
0

您需要检查是否存在具有相同用户名的行:

//Check a row with same username exists 
$queryChecking = "Select COUNT(*) FROM feedback WHERE user_name = ?"; 

如果结果等于一,则需要更新行而不是插入新行;)

于 2013-03-27T08:28:28.870 回答