json 跨域 $http 请求 AngularJS
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Cross-domain $http request AngularJS
提问by Tom
I have simple website I'm building with AngularJS which calls an API for json data.
我有一个简单的网站，我正在使用 AngularJS 构建它，它为 json 数据调用 API。
However I am getting Cross domain origin problem is there anyway around this to allow for cross domain requests?
但是，我遇到了跨域源问题，无论如何，是否允许跨域请求？
Error:
错误：
XMLHttpRequest cannot load http://api.nestoria.co.uk/api?country=uk&pretty=1&action=search_listings&place_name=soho&encoding=json&listing_type=rent. No 'Access-Control-Allow-Origin' header is present on the requested resource. Origin 'http:// localhost' is therefore not allowed access.
XMLHttpRequest 无法加载http://api.nestoria.co.uk/api?country=uk&pretty=1&action=search_listings&place_name=soho&encoding=json&listing_type=rent。请求的资源上不存在“Access-Control-Allow-Origin”标头。因此，不允许访问 Origin 'http:// localhost'。
searchByPlaceName: function() {
var url = baseurl+'country=uk&pretty=1&action=search_listings&place_name=london'+encoding+type;
return $http.get(url);
}
回答by Explosion Pills
It seems that api.nestoria.co.ukdoes not allow CORS. It has to set the Access-Control-Allow-Originheader itself -- you have no direct control over that.
似乎api.nestoria.co.uk不允许CORS。它必须设置Access-Control-Allow-Origin标头本身——您无法直接控制它。
However, you can use JSONP. That site allows it via the callbackquery parameter.
但是，您可以使用 JSONP。该站点通过callback查询参数允许它。
$http.jsonp(baseurl+'country=uk&pretty=1&action=search_listings&place_name=london'
+encoding+type + "&callback=JSON_CALLBACK")
回答by uiwarrior
Install Fiddler. Add a custom rule to it:
安装提琴手。为其添加自定义规则：
static function OnBeforeResponse(oSession: Session)
{
oSession.oResponse.headers.Add("Access-Control-Allow-Origin", "*");
}
This would allow you to make cross domain requests from localhost. If the API is HTTPS make sure you enable 'decrypt HTTPS traffic' in fiddler.
这将允许您从本地主机发出跨域请求。如果 API 是 HTTPS，请确保在 fiddler 中启用“解密 HTTPS 流量”。
-------------------- UPDATE
- - - - - - - - - - 更新
The response you are getting is JSON. Specifying JSONP as datatype would not work. When you do specify JSONP the return should be a function not JSON object.
您得到的响应是 JSON。将 JSONP 指定为数据类型将不起作用。当您指定 JSONP 时，返回应该是一个函数而不是 JSON 对象。
