The path of this ajax request, right?
Is my cross-domain method correct? Why does the console keep reporting errors
迷茫2017-07-05 10:43:03
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.
我想大声告诉你2017-07-05 10:43:03
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',
....
});
Backendcallback (returned data)
高洛峰2017-07-05 10:43:03
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
PHP中文网2017-07-05 10:43:03
jsonp should not be introduced by src in js, and then called callback()
in js
滿天的星座2017-07-05 10:43:03
For jsonp in JQ, do you need to add a callback= to the query string? Field