Ajax 请求错误 404

再次问好，提前感谢。让我首先向您展示我试图解决的代码和问题：

<span class='dropdown' id='status'>
              <?php
                switch ($status) {
                  case "unassigned":
                  echo "<button type='button'  data-toggle='dropdown' id='edit' aria-haspopup='true' aria-expanded='true' class='btn btn-danger dropdown-toggle'> Status: " . $status . "<span class='caret'></span></button>";
                  echo "<ul class='dropdown-menu'   aria-labelledby='edit'>";
                  echo "<li><a class='editStatus' href='pending'>Pending</a></li>";
                  echo "</ul>";
                  break;
                  case "pending":
                  echo "<button type='button'  data-toggle='dropdown' id='edit' aria-haspopup='true' aria-expanded='true' class='btn btn-danger dropdown-toggle'> Status: " . $status . "<span class='caret'></span></button>";
                  echo "<ul class='dropdown-menu'   aria-labelledby='edit'>";
                  echo "<li><a class='editStatus' href='attending'>Attending</a></li>";
                  echo "<li><a class='editStatus' href='followup'>Follow Up</a></li>";
                  echo "<li><a class='editStatus' href='closed'>Closed</a></li>";
                  echo "</ul>";
                  break;
                  case "attending":
                  echo "<button type='button'  data-toggle='dropdown' id='edit' aria-haspopup='true' aria-expanded='true' class='btn btn-danger dropdown-toggle'> Status: " . $status . "<span class='caret'></span></button>";
                  echo "<ul class='dropdown-menu'   aria-labelledby='edit'>";
                  echo "<li><a class='editStatus' href='pending'>Pending</a></li>";
                  echo "<li><a class='editStatus' href='followup'>Follow Up</a></li>";
                  echo "<li><a class='editStatus' href='closed'>Closed</a></li>";
                  echo "</ul>";
                  break;
                  case "followup":
                  echo "<button type='button'  data-toggle='dropdown' id='edit' aria-haspopup='true' aria-expanded='true' class='btn btn-danger dropdown-toggle'> Status: " . $status . "<span class='caret'></span></button>";
                  echo "<ul class='dropdown-menu'   aria-labelledby='edit'>";
                  echo "<li><a class='editStatus' href='closed'>Closed</a></li>";
                  echo "</ul>";
                  break;
                  case "closed":
                  echo "<button type='button'  data-toggle='dropdown' id='edit' aria-haspopup='true' aria-expanded='true' class='btn btn-danger dropdown-toggle'> Status: " . $status . "<span class='caret'></span></button>";
                  echo "<ul class='dropdown-menu'   aria-labelledby='edit'>";
                  echo "<li><a class='editStatus' href='pending'>Pending</a></li>";
                  echo "<li><a class='editStatus' href='attending'>Attending</a></li>";
                  echo "<li><a class='editStatus' href='followup'>Follow Up</a></li>";
                  echo "</ul>";
                  break;
                }
              ?>        
                  </span> 

我的 php 脚本中有这个跨度，其中包括一个下拉按钮，其中包含表示此页面中显示的消息的不同状态的选项，并通过 ajax 将此信息发送到另一个 php 脚本

<script>
var phpvar1 = "<?php echo $frompost_id_sanitized; ?>";
var phpvar = "<?php echo $status; ?>";
$('a.editStatus').click(function(event) {
event.preventDefault();
    var statusJs = $(this).attr("href");
    alert('Change Status of ticket to: ' + statusJs)
    $.post('ajaxChangeStatus.php', {status: statusJs, id: phpvar1, initialStatus: phpvar}, function(data) {
        $('#status').html(data)
    });
});

最后，第二个 php 脚本将数据添加到 db 并将一些 html 发送回第一个 php 脚本，如下所示(省略了一些代码(

switch ($frompost_status_sanitized) {
    case "pending":
    echo "<button type='button'  data-toggle='dropdown' id='edit' aria-haspopup='true' aria-expanded='true' class='btn btn-danger dropdown-toggle'> Status: " . $frompost_status_sanitized . "&nbsp;<span class='caret'></span></button>";
    echo "<ul class='dropdown-menu'   aria-labelledby='edit'>";
    echo "<li><a class='editStatus' href='attending'>Attending</a></li>";
    echo "<li><a class='editStatus' href='followup'>Follow Up</a></li>";
    echo "<li><a class='editStatus' href='closed'>Closed</a></li>";
    echo "</ul>";
    break;
    case "attending":
    echo "<button type='button'  data-toggle='dropdown' id='edit' aria-haspopup='true' aria-expanded='true' class='btn btn-danger dropdown-toggle'> Status: " . $frompost_status_sanitized . "&nbsp;<span class='caret'></span></button>";
    echo "<ul class='dropdown-menu'   aria-labelledby='edit'>";
    echo "<li><a class='editStatus' href='pending'>Pending</a></li>";
    echo "<li><a class='editStatus' href='followup'>Follow Up</a></li>";
    echo "<li><a class='editStatus' href='closed'>Closed</a></li>";
    echo "</ul>";
    break;
    case "followup":
    echo "<button type='button'  data-toggle='dropdown' id='edit' aria-haspopup='true' aria-expanded='true' class='btn btn-danger dropdown-toggle'> Status: " . $frompost_status_sanitized . "&nbsp;<span class='caret'></span></button>";
    echo "<ul class='dropdown-menu'   aria-labelledby='edit'>";
    echo "<li><a class='editStatus' href='closed'>Closed</a></li>";
    echo "</ul>";
    break;
    case "closed":
    echo "<button type='button'  data-toggle='dropdown' id='edit' aria-haspopup='true' aria-expanded='true' class='btn btn-danger dropdown-toggle'> Status: " . $frompost_status_sanitized . "&nbsp;<span class='caret'></span></button>";
    echo "<ul class='dropdown-menu'   aria-labelledby='edit'>";
    echo "<li><a class='editStatus' href='pending'>Pending</a></li>";
    echo "<li><a class='editStatus' href='attending'>Attending</a></li>";
    echo "<li><a class='editStatus' href='followup'>Follow Up</a></li>";
    echo "</ul>";
    break;
}

一切似乎都正常：

这个想法是我(在第一个 php 脚本中(有一个带有选项的下拉按钮(表示显示的消息的状态(。当选择一个选项(例如状态已更改(时(单击事件(，我通过 ajax 将所选选项(新状态(发送到第二个 php 脚本，然后执行一些操作并将适当的 html(具有不同操作的下拉按钮(返回到第一个 php 脚本。

问题所在如果然后我选择另一个选项(例如，如果我尝试在不重新加载页面的情况下再次更改状态(，那么我会收到 404 错误，因为当单击该选项而不是触发点击事件然后event.preventdefault......(应该发生(，应用程序读取 href 属性[用于将数据发送到第二个 PHP 脚本，如下所示：var statusJs = $(this).attr("href");]并尝试加载导致 404 错误的页面http://localhost:8888/ticketing/pending

通常 js 脚本应该从 href 属性中读取值，防止默认操作(例如导致 404 的加载页面(，将数据发送到第二个 php 脚本并将 span contants 替换为返回的 html

怎么了？