javascript - Cross-domain, is my jsonp method correct? Why do I keep getting errors?
高洛峰
高洛峰 2017-07-05 10:41:03
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The path of this ajax request, right?

Is my cross-domain method correct? Why does the console keep reporting errors

高洛峰
高洛峰

拥有18年软件开发和IT教学经验。曾任多家上市公司技术总监、架构师、项目经理、高级软件工程师等职务。 网络人气名人讲师,...

reply all(6)
迷茫

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.

黄舟

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

我想大声告诉你

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)

小葫芦

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

洪涛

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

滿天的星座

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

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