Question

在以下问题上，我需要您的帮助：提交表单时，我希望在用户侧显示一条消息。该代码运行正常，但是成功后不会向用户显示警告消息。我想念什么吗？我认为我的脚本位置有问题。我必须将其放置在代码中的其他位置还是出现一些错误？

<html>

<head>
  <script src="https://code.jquery.com/jquery-3.3.1.min.js" integrity="sha256-FgpCb/KJQlLNfOu91ta32o/NMZxltwRo8QtmkMRdAu8=" crossorigin="anonymous"></script>
  <script src="https://ajax.googleapis.com/ajax/libs/jquery/3.3.1/jquery.min.js"></script>

  <script>
    function save3() {
      var pn = 4; 
      $("#user_id").val(prompt("Give the User Id:"))
      $("#book_id").val(prompt("Give the Book Id:"))
      $("#game_id").val(prompt("Give the Game Id:"))
      $("#site_id").val(pn)
    }
    function load2() {

    }

  </script>

</head>

<body>

  <p align="center">example</p>
  <table align="center" width="730">
    <tr>
      <td align="center">
        <div>
          <table class="blueTable" style="float: left">
            <thead>
              <tr>
                <th colspan="1"><u>Menu</u></th>
              </tr>
            </thead>
            <tbody>
              <tr>
                <td><input type="button" value="New" id="new" onclick="new1()" class="button12" /></td>
              </tr>
              <tr>
                <td>
                </td>
              </tr>
              <tr>
                <td>
                  <form name="SaveGame" id="SaveGame" method="post" action="http://127.0.0.1/PHP/mine2.php" enctype="multipart/form-data">
                    <input type="submit" value="Save" id="save" onclick="save3()" class="button12" />
                    <input type="hidden" name="user_id" id="user_id">
                    <input type="hidden" name="book_id" id="book_id">
                    <input type="hidden" name="game_id" id="game_id">
                    <input type="hidden" name="site_id" id="site_id">
                  </form>
                  <script>
                    $("#SaveGame").submit(function(e) {


                      var form = $(this);
                      var url = form.attr('action');

                      $.ajax({
                        type: "POST",
                        url: url,
                        data: form.serialize(), // serializes the form's elements.
                        success: function(data) {
                          alert("The game has been saved!"); // show response from the php script.
                        }
                      });

                      e.preventDefault(); // avoid to execute the actual submit of the form.
                    });
                    </script>
                    </td>
                </tr>
            </tbody>
        </table>    
        </div>
        </td>
        </tr>
        </table>

</body>
</html>

和我的php代码：

<?php
/* Attempt MySQL server connection. Assuming you are running MySQL
server with default setting (user 'root' with no password) */
$link = mysqli_connect("127.0.0.1", "root", "", "mysql3");
// Check connection
if($link === false) {
    die("ERROR: Could not connect. " . mysqli_connect_error());
}

$user_id =$_POST['user_id'];
$book_id =$_POST['book_id'];
$game_id =$_POST['game_id'];
$site_id =$_POST['site_id'];

    // Attempt insert query execution
if ($result = $link->query("INSERT INTO `components` (`user_id`, `book_id`, `game_id`, `site_id`) VALUES ('$user_id','$book_id','$game_id', '$site_id')") == true)
  echo json_encode($result);
else {
  echo "Error";
}
            //$qry ="INSERT INTO `components` (`user_id`, `book_id`, `game_id`, `site_id`) VALUES ('".$user_id."','".$book_id."','".$game_id."', '".$site_id."')";
  //      if($qry){
    //    mysqli_query($link,$qry);
      //  }

// Close connection
mysqli_close($link);
?>

提交表单时出现确认消息问题

0 个答案: