feedback_form.php

Question

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：当我写一个更新查询，我变得无法插入任何数据，任何人都可以帮助我吗？

feedback_form.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>

feedback_process.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";
?>

Answer 1

  $query2 = "UPDATE feedback SET (feedback = ?, user_name = ?) WHERE user_name = '$login_user' ";
  $stmt = $conn->stmt_init();
  if($stmt->prepare($query2))

Answer 2

您也可以尝试

  $query2 = "UPDATE feedback SET (feedback = ?, user_name = ?) WHERE user_name = '$login_user' ";
  $stmt = $conn->query($query2);

Answer 3

有多个错误/错误。

请参阅以下代码段：

第一

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

您没有,（逗号）分隔字段。使用：

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

下一步

 //$stmt->execute();

您没有执行此声明。使用

 $stmt->execute();

另一个

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

INSERT查询中的字段名称与此处不同。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();
}

php mysqli插入和更新查询

feedback_form.php

feedback_process.php

3 个答案:

第一

下一步

另一个