Uncaught SyntaxError: Unexpected token: An error like this is caused by the returned json data not being enclosed in "(" and ")" parentheses or not adding a callback value in front.

There is a problem with the server program, js execution error

Set the jsonp parameters, and the data returned by the background needs to be wrapped in the jsonp variable you passed
Front-end:

$.ajax({
    url: 'xx',
    dataType: 'jsonp',
    jsonp: 'callback',
    ....
});

Backend
callback (returned data)

Add a callback parameter to url: jsoncallback=?, the question mark program will automatically generate the corresponding parameters

The backend program accepts this parameter and wraps the returned data in this callback function

eg:
PHP后端写法

$jsoncallback = $_GET['jsoncallback'];

$result = json_encode($data);

echo  $jsoncallback."(".$result.")";//后端要以这种格式返回数据才能实现跨域
exit;

Your return value is not wrapped in a callback function

jsonp should not be introduced by src in js, and then called callback() in js

For jsonp in JQ, do you need to add a callback= to the query string? Field