HttpClient抛出ssl错误，但webview和HttpsURLConnection不抛出

Question

因此，我使用服务器（google-app-engine）在客户端应用程序上进行身份验证。当我使用JavaFX、Webview或HttpsURLConnection时，一切都很好，它们连接起来了，但是当我使用HttpClient时，它会抛出以下错误

javax.net.ssl.sslpeerunverifiedexception：peer未在sun.security.ssl.sslsessionimpl.getPeerCertificates（来源未知）在org.apache.http.conn.ssl.abstractVerifier.verific(abstractVerier.java:128)在org.apache.http.conn.ssl.sslactVerier.verific(abstractVerier.java:128)在java:147)在org.apache.http.impl.conn.abstractPooledConnAdapter.open(abstractPooledConnAdapter.java:108)在org.apache.http.impl.client.defaultRequestDirector.execute(DefaultRequestDirector.java:415)在org.apache.http.impl.client.abstractHttpClient.execute(abstractHttpClient.java:641)在

完整的ssl报告在这里

我正在使用的代码

System.out.println("I started");
        String https_url = "https://app.quagrum.com";
          URL url;
          try {

             url = new URL(https_url);
             HttpsURLConnection con = (HttpsURLConnection)url.openConnection();//Works 
             System.out.println("I ran?");
    }
          catch(Exception a)
          {

          }
    try {
        HttpGet get = new HttpGet("https://app.quagrum.com");
        HttpClient httpclient = new DefaultHttpClient();
        HttpResponse response = httpclient.execute(get);//Throws error
        System.out.println(response.getStatusLine().getStatusCode());
        if(response.getStatusLine().getStatusCode()!=200)
        {
            HttpEntity entity = response.getEntity();

            if (entity != null) {
                InputStream instream = entity.getContent();
                try {
                    Scanner s = new Scanner(instream).useDelimiter("\\A");
                    String result = s.hasNext() ? s.next() : "";
                    System.out.println(result);
                    result =result.substring(result.indexOf("\"",7));
                    result=result.substring(0, result.indexOf("\""));
                    System.out.println(result);
                } finally {
                    instream.close();
                }
            }
        }
        }catch(Exception a)
        {
            a.printStackTrace();
        }
    }

Answer 1

现在才管用。这是好的java让我完全重新设计我的程序的一部分，然后只是让它在以后的版本中工作