发布于2021-07-24 21:21 阅读(1132) 评论(0) 点赞(6) 收藏(1)
我正在尝试从我的 API POST 数据,但我无法通过基本身份验证。
我尝试:
$.ajax({
type: 'POST',
url: http://theappurl.com/api/v1/method/,
data: {},
crossDomain: true,
beforeSend: function(xhr) {
xhr.setRequestHeader('Authorization', 'Basic [REDACTED]');
}
});
我的服务器配置响应是:
response["Access-Control-Allow-Origin"] = "*"
response["Access-Control-Allow-Methods"] = "POST"
response["Access-Control-Max-Age"] = "1000"
response["Access-Control-Allow-Headers"] = "*"
我得到的标题是:
请求头
OPTIONS /api/v1/token-auth/ HTTP/1.1
Host: theappurl.com
Connection: keep-alive
Access-Control-Request-Method: POST
Origin: http://127.0.0.1:8080
User-Agent: Mozilla/5.0 (X11; Linux x86_64) AppleWebKit/537.31 (KHTML, like Gecko) Chrome/26.0.1410.63 Safari/537.31
Access-Control-Request-Headers: origin, authorization, content-type
Accept: */*
Referer: http://127.0.0.1:8080/
Accept-Encoding: gzip,deflate,sdch
Accept-Language: es,en;q=0.8
Accept-Charset: ISO-8859-1,utf-8;q=0.7,*;q=0.3
响应头
HTTP/1.1 401 Unauthorized
Server: nginx/1.1.19
Date: Fri, 16 Aug 2013 01:29:21 GMT
Content-Type: text/html
Content-Length: 597
Connection: keep-alive
WWW-Authenticate: Basic realm="Restricted"
我猜服务器配置很好,因为我可以从高级 REST 客户端(Chrome 扩展程序)访问 API
有什么建议么?
PD:我从 Advanced REST 客户端获得的标头是:
User-Agent: Mozilla/5.0 (X11; Linux x86_64) AppleWebKit/537.31 (KHTML, like Gecko) Chrome/26.0.1410.63 Safari/537.31
Origin: chrome-extension://hgmloofddffdnphfgcellkdfbfbjeloo
Authorization: Basic [REDACTED]
Content-Type: application/x-www-form-urlencoded
Accept: */*
Accept-Encoding: gzip,deflate,sdch
Accept-Language: es,en;q=0.8
Accept-Charset: ISO-8859-1,utf-8;q=0.7,*;q=0.3
和
Server: nginx/1.1.19
Date: Fri, 16 Aug 2013 01:07:18 GMT
Content-Type: application/json; charset=utf-8
Transfer-Encoding: chunked
Connection: keep-alive
Vary: Accept, Cookie
Allow: POST, OPTIONS
X-Robots-Tag: noindex
发送 OPTION 方法
您可以将用户和密码包含在 URL 中:
http://user:passwd@www.server.com/index.html
查看此 URL,了解更多信息
当然,您需要用户名密码,而不是'Basic hashstring
。
希望这可以帮助...
作者:官方问答小能手
链接:http://www.pythonpdf.com/blog/article/247/e232e7f2f9c422407ef0/
来源:编程知识网
任何形式的转载都请注明出处,如有侵权 一经发现 必将追究其法律责任
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