apache http客户端java程序出错

使用“ApacheHTTP客户端（4.5.2版本）”从url获取数据的简单程序有一个问题

请在下面查找代码和错误：

public static void main(String[] args) throws Exception {
        String username = "user";
        String password = "pwd";
        String urlString = "xyz.com?a=b&c=d";

        org.apache.http.client.HttpClient client = HttpClientBuilder.create().build();
        HttpGet request = new HttpGet(urlString);
        org.apache.http.auth.UsernamePasswordCredentials creds = new org.apache.http.auth.UsernamePasswordCredentials(
                username, password);
        request.addHeader(new BasicScheme().authenticate(creds, request));

        HttpResponse response = client.execute(request);
        System.out.println("Response Code : " + response.getStatusLine().getStatusCode());

        BufferedReader rd = new BufferedReader(new InputStreamReader(response.getEntity().getContent()));
        StringBuffer result = new StringBuffer();
        String line = "";
        while ((line = rd.readLine()) != null) {
            System.out.println(line);
            result.append(line);
        }
    }

错误：

<Error><Code>InvalidArgument</Code><Message>Only one auth mechanism allowed; only the X-Amz-Algorithm query parameter, Signature query string parameter or the Authorization header should be specified</Message>

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您能提供帮助吗？

我没有遇到上面的错误，但是在给定代码开头添加了“http://”之后，我能够毫无错误地运行此代码示例

String urlString = "xyz.com?a=b&c=d";

---

我没有遇到上面的错误，但是在给定代码开头添加了“http://”之后，我能够毫无错误地运行此代码示例

String urlString = "xyz.com?a=b&c=d";