javascript - 如何在php中获取脚本变量值

Question

我正在使用弹出窗口根据点击功能的 id 显示消息。所以我想在 php 中打开弹出窗口时获取 id。提前致谢

<p>Link 1</p>
<a data-toggle="modal" data-id="ISBN564541" title="Add this item" class="open-AddBookDialog btn btn-primary" href="#addBookDialog">test</a>

<p>&nbsp;</p>

<div class="modal hide" id="addBookDialog">
 <div class="modal-header">
    <button class="close" data-dismiss="modal">×</button>
    <h3>Modal header</h3>
  </div>
    <div class="modal-body">
        <p>some content</p>
        <?php //here i want to getting script variable value for database purpose?>

    </div>
</div>

我的脚本是

 $(document).on("click", ".open-AddBookDialog", function () {
     var myBookId = $(this).data('id');
     $(".modal-body #bookId").val( myBookId );
});

在 ajax 调用中，我从下一页获取 ID，但我想在同一页中获取值

function viwpost(id) { 

    var pid=id; 

    $.ajax({  
        type: "POST",  
        url: "view_more.php",  
        data: "pid=" + pid ,  
        success: function(data)
        {  
            $("#myModal21").html(data);         
        }  
    }); 
}

Answer 1

因为您已经在使用 jQuery 框架/API:

你可以简单地使用

$.get('phpfile.php' { id: myBookId }).done(function (response) { alert(response); });

您的 PHP 文件可以像这样获取请求:

$_GET['id'];

If you're trying to make a live pop-up to load the data, this is an incorrect method of doing so. PHP script is only loaded once on server running and sending a request will only give you a response so ensure you research.

我建议您创建 PHP 文件来回显要弹出的数据。

将响应发送回请求的 PHP 文件演示(按要求):

if(isset($_GET['id']))
{
    echo 'what ever is outputted to the client will be received in the response';
}