Question

当我致电/comment.php时，它将返回我已请求的数据。固定 post_id 的位置。

/comment.php

 "ajax":
{
 "url": "http://localhost/fb-callback.php?post_id=12345_67890",
   "type" : "GET",
"dataSrc":  function (response) { 
                // console.log(response);
                            if(response.whaterver == 0){
                               //DO YOUR THING HERE
                            }
                            //return back the response
                            return response; 
                        },

我想在那里声明一个php变量，而不是固定ID（ 12345_67890 ），像这样的东西::-

<?php
$uid = $_GET['uid'];
?>

 "ajax":
{
 "url": "http://localhost/fb-callback.php?post_id="+"&uid",
"type" : "GET",
 ................

这样，当我调用网址/comment.php?uid=12345_67890**时，它将返回结果。但是看来这行不通。我该怎么办？或者还有其他方法可以打电话吗？

以下是完整代码::: https://pastebin.com/iy3htv63

Answer 1

您可以echo在ajax网址上。试试这个

 <?php
  $uid = $_GET['uid'];
?>

 "ajax":
{
 "url": "http://localhost/fb-callback.php?post_id=<?php echo $uid; ?>",
 "type" : "GET",
 ................

Answer 2

尝试一下

<?php
$uid = $_GET['uid'];
?>
var uid = <?php echo $uid; ?>;
 "ajax":
{
 "url": "http://localhost/fb-callback.php?post_id="+uid,
"type" : "GET",
 ................

如何声明一个PHP变量并将其传递给AJAX url？

2 个答案: