Javascript 如何声明PHP变量并将其传递到AJAX url?
当我调用Javascript 如何声明PHP变量并将其传递到AJAX url?,javascript,php,ajax,Javascript,Php,Ajax,当我调用/comment.php时,它返回我请求的数据。其中post_id是固定的 /comment.php "ajax": { "url": "http://localhost/fb-callback.php?post_id=12345_67890", "type" : "GET", "dataSrc": function (response) { // console.log(response);
/comment.php
时,它返回我请求的数据。其中post_id是固定的
/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;
},
我想在这里声明一个php变量,而不是固定id(12345_67890),
类似这样的东西::-
<?php
$uid = $_GET['uid'];
?>
"ajax":
{
"url": "http://localhost/fb-callback.php?post_id="+"&uid",
"type" : "GET",
................
“ajax”:
{
“url”:”http://localhost/fb-callback.php?post_id=“+”&uid“,
“类型”:“获取”,
................
因此,当我调用url/comment.php?uid=12345_67890**
时,它将返回结果。但它似乎不是这样工作的。
我该怎么做,或者有其他方法打这个电话吗
以下是完整的代码::您可以
在ajax url上回送它。试试这个
<?php
$uid = $_GET['uid'];
?>
"ajax":
{
"url": "http://localhost/fb-callback.php?post_id=<?php echo $uid; ?>",
"type" : "GET",
................
“ajax”:
{
“url”:”http://localhost/fb-callback.php?post_id=",
“类型”:“获取”,
................
试试这个
<?php
$uid = $_GET['uid'];
?>
var uid = <?php echo $uid; ?>;
"ajax":
{
"url": "http://localhost/fb-callback.php?post_id="+uid,
"type" : "GET",
................
变量uid=;
“ajax”:
{
“url”:”http://localhost/fb-callback.php?post_id=“+uid,
“类型”:“获取”,
................
“url”:http://localhost/fb-callback.php?post_id=“,
另请参见此处:谢谢,这真的帮了我很大的忙……感谢:)这是一种糟糕的编码实践。