15

我正在使用HttpUrlConnection从我的 android 应用程序发出网络请求。除了一件事 401 外，一切正常。每当服务器返回状态码为 401 的响应时，我的应用程序都会抛出IOException一条消息，说明"no authentication challenge found". 谷歌搜索后，我没有找到单一的解决方案，但只有解决方法（使用 try/catch 处理它，假设它是 401 响应）。

这是代码片段：

public Bundle request(String action, Bundle params, String cookie) throws FileNotFoundException, MalformedURLException, SocketTimeoutException,
        IOException {

    OutputStream os;

    String url = baseUrl + action;
    Log.d(TAG, url);
    HttpURLConnection conn = (HttpURLConnection) new URL(url).openConnection();
    conn.setConnectTimeout(30 * 1000);
    conn.setReadTimeout(30 * 1000);
    conn.setRequestProperty("User-Agent", System.getProperties().getProperty("http.agent") + "AndroidNative");
    conn.setRequestMethod("POST");
    conn.setRequestProperty("Content-Type", "multipart/form-data;boundary=" + boundary);
    conn.setRequestProperty("Connection", "Keep-Alive");
    conn.setDoOutput(true);
    conn.setDoInput(true);
    if (cookie != null) {
        conn.setRequestProperty("Cookie", cookie);
    }

    if (params != null) {
        os = conn.getOutputStream();
        os.write(("--" + boundary + endLine).getBytes());
        os.write((encodePostBody(params, boundary)).getBytes());
        os.write((endLine + "--" + boundary + endLine).getBytes());
        uploadFile(params, os);
        os.flush();
        os.close();
    }

    conn.connect();

    Bundle response = new Bundle();
    try {
        response.putInt("response_code", conn.getResponseCode());
        Log.d(TAG, conn.getResponseCode() + "");
        response.putString("json_response", read(conn.getInputStream()));
        List<String> responseCookie = conn.getHeaderFields().get("set-cookie");
        // Log.d(TAG, responseCookie.get(responseCookie.size() - 1));
        response.putString("cookie", responseCookie.get(responseCookie.size() - 1));
    } catch (SocketTimeoutException e) {
        throw new SocketTimeoutException(e.getLocalizedMessage());
    } catch (FileNotFoundException e) {
        throw new FileNotFoundException(e.getLocalizedMessage());
    } catch (IOException e) {
        e.printStackTrace();
        response.putInt("response_code", HttpURLConnection.HTTP_UNAUTHORIZED);
        response.putString("json_response", read(conn.getErrorStream()));
    }

    // debug
    Map<String, List<String>> map = conn.getHeaderFields();
    for (String key : map.keySet()) {
        List<String> values = map.get(key);
        for (String value : values) {
            Log.d(key, value);
        }
    }

    conn.disconnect();

    return response;
}

我真的很想知道，为什么会抛出这个异常？身份验证质询是什么意思？如何提供身份验证质询？为了克服这种情况，我必须对代码进行哪些更改？

请赐教.. :)

4

2 回答 2

14

IOException 是一个非常普遍的异常，您不能安全地假设每次抛出它时都会出现 401 状态代码。

如果您第一次请求状态码时恰好包含 401，则 HttpURLConnection 将抛出 IOException。此时，连接的内部状态将发生变化，它现在可以为您提供状态码，而不会出现任何错误。

int status = 0;
try {
    status = conn.getResponseCode();
} catch (IOException e) {
    // HttpUrlConnection will throw an IOException if any 4XX
    // response is sent. If we request the status again, this
    // time the internal status will be properly set, and we'll be
    // able to retrieve it.
    status = conn.getResponseCode();
}
if (status == 401) {
    // ...
}

更多详情，http://www.tbray.org/ongoing/When/201x/2012/01/17/HttpURLConnection

于 2013-08-27T10:15:45.847 回答
-2

尝试使用 HttpClient

private void setCredentials(String login, String password) {
    if (login != null && password != null)
        httpClient.getCredentialsProvider().setCredentials(
                new AuthScope(URL_HOST, 80), new UsernamePasswordCredentials(login, password));

}



private InputStream execute(String url, String login, String password) {
        setCredentials(login, password);
        HttpGet get = new HttpGet(url);
        try {
            HttpResponse response = httpClient.execute(get);
            int code = response.getStatusLine().getStatusCode();
            if (code == HttpURLConnection.HTTP_OK) {
                InputStream stream = response.getEntity().getContent();
                return stream;
            } else {
                Log.e("TAG", "Wrong response code " + code + " for request " + get.getRequestLine().toString());
                return null;
            }
        } catch (IOException e) {
            e.printStackTrace();
        }
        return null;
    }
于 2012-07-31T09:07:57.977 回答