Android NTLM 获取 HTTP/1.1 401 Unauthorized答案

【问题标题】：Android NTLM getting HTTP/1.1 401 UnauthorizedAndroid NTLM 获取 HTTP/1.1 401 Unauthorized
【发布时间】：2016-11-23 09:18:02
【问题描述】：

我正在尝试从共享点服务器获取数据。以下是我的代码。

    DefaultHttpClient httpclient = new DefaultHttpClient();

    httpclient.getAuthSchemes().register("ntlm", new NTLMSchemeFactory());
    httpclient.getCredentialsProvider().setCredentials(new AuthScope("masconsult.eu", -1),
            new NTCredentials(username, password, "", ""));
    HttpGet httpGet = new HttpGet(webserviceUrl);
    httpGet.addHeader("Content-type", "application/json");
    httpGet.getParams().setBooleanParameter(CoreProtocolPNames.USE_EXPECT_CONTINUE, false);
    // HttpResponse response="";
    String responseXML = "";
    HttpResponse response = httpclient.execute(httpGet);
    response.getStatusLine().getReasonPhrase();
    responseXML = EntityUtils.toString(response.getEntity());
    Toast.makeText(this, responseXML, Toast.LENGTH_LONG).show();

我收到响应 HTTP/1.1 401 Unauthorized。甚至我都添加了所有正确的凭据。在 Chrome 浏览器中使用相同的凭据可以正常工作。请建议我对代码进行任何更改。

【问题讨论】：

您可能不需要对代码进行任何更改。在 IE 中，在工具 > 高级下，确保您已启用“集成 Windows 身份验证”，在工具 > 安全 > 本地 Intranet 区域 > 站点 > 高级下，确保列出了目标网站。清除 cookie 并重新启动 IE。

标签： java android authentication sharepoint ntlm

【解决方案1】：

现在它的工作。我用标题修改了我的代码。

 DefaultHttpClient httpclient = new DefaultHttpClient();
    httpclient.getAuthSchemes().register("ntlm", new NTLMSchemeFactory());
    httpclient.getCredentialsProvider().setCredentials(new AuthScope(AuthScope.ANY_HOST, AuthScope.ANY_PORT), new NTCredentials(username, password, "", ""));
    HttpGet httpGet = new HttpGet(webserviceUrl);
    httpGet.addHeader("accept", "application/json;odata=verbose");
    httpGet.addHeader("content-Type", "application/json;odata=verbose");
    httpGet.getParams().setBooleanParameter(CoreProtocolPNames.USE_EXPECT_CONTINUE, false);
    HttpResponse response = httpclient.execute(httpGet);
    System.out.println("Responseeee" + response.getStatusLine());
    responseXML = EntityUtils.toString(response.getEntity());
    new JSONObject(responseXML).toString();

【讨论】：