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

解决方案

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

在 URL 和加密中传递的 HTTP 基本身份验证凭据

当然，您需要用户名密码，而不是'Basic hashstring。

希望这可以帮助...