javascript - '意外的标记 ;'当尝试在 jQuery AJAX 请求中传递 PHP 变量时

Question

因此，我尝试运行一个 PHP 脚本，如果您将某个文本元素拖动到可放置区域，该脚本将数据库中已删除的字段设置为填充。

目前我有这个可放置区域:

<div class="dropbin" id="dropbin" >
    <span class="fa fa-trash-o noSelect hover-cursor" style="font-size: 20pt; line-height: 225px;">&nbsp;</span>
</div>

以及这个可拖动文本:

<div id='dragme' data-toggle='modal' data-target='#editNoteNameModal' class='display-inline'>" . $data['NoteName'] . "</div>

我遇到问题的地方是:

$("#dropbin").droppable
  ({
    accept: '#dragme', 
    hoverClass: "drag-enter",
    drop: function(event) 
    {
      var noteid = <?php if(isset($_POST['noteid'])){ echo $_POST['noteid'];} ?>;
      var deletedby = <? if(isset($_SESSION['username'])){ echo $_SESSION['username'];} ?>
      var data = {noteid1: noteid, deletedby1: deletedby};

      if (confirm('Delete the note?')==true) 
      {
        $('#dragme').hide();
        debugger
        $.ajax({
          type: 'POST',
          url: 'deleteNote.php',
          datatype: 'json',
          data: data,
          success: function(result)
              {
                  alert("Success");
              }
        });

        window.location = "http://discovertheplanet.net/general_notes.php";
      }
      else
      {
        window.location = "http://discovertheplanet.net/general_notes.php";
      }
    }
  });

编辑:我收到错误的行是:

var noteid = <?php if(isset($_POST['noteid'])){ echo $_POST['noteid'];} ?>;

我目前收到一个“意外 token ；”，它阻止了 droppable 的工作。

只是一个旁注，如果我在没有变量的情况下运行它，它会击中除以下内容之外的所有内容:

url: 'deleteNote.php',

deleteNote.php 中还有这个，以防它有帮助:

<?php

include "connectionDetails.php";

?>

<?php

if (isset($_POST['noteid1'], $_POST['deletedby1'])) 
{
    $noteid2 = $_POST['noteid1'];
    $deletedby2 = $_POST['deletedby1'];

    // echo "Hello...". $noteid;

    $stmt = "UPDATE Notes SET Deleted = GETDATE() WHERE NoteID = (?)";
    $params = array($noteid2);

    $stmt = sqlsrv_query($conn, $stmt, $params);

    if ($stmt === false) 
    {
        die( print_r(sqlsrv_errors(), true));
    }
}

else
{
    echo "No Data";
}


?>

(我故意在数据库中还没有deleteby，忽略它)

有人可以帮助我让它发挥作用吗？

Answer 1

尝试在这些行中添加引号并添加 php之后<?第二行:

var noteid = "<?php if(isset($_POST['noteid'])){ echo $_POST['noteid'];} ?>";
var deletedby = "<?php if(isset($_SESSION['username'])){ echo $_SESSION['username'];} ?>";

或

var noteid = "<?=isset($_POST['noteid']) ? $_POST['noteid'] : "" ?>";
var deletedby = "<?=isset($_SESSION['username']) ? $_SESSION['username'] : "" ?>";