Question

如果URL连接超过5秒，我想返回false - 如何使用Java？这是我用来检查URL是否有效的代码

HttpURLConnection.setFollowRedirects(false);
HttpURLConnection con = (HttpURLConnection) new URL(url).openConnection();
con.setRequestMethod("HEAD");
return (con.getResponseCode() == HttpURLConnection.HTTP_OK);

Answer 1

HttpURLConnection有setConnectTimeout方法。

只需将超时设置为5000毫秒，然后捕获java.net.SocketTimeoutException

您的代码应如下所示：


try {
   HttpURLConnection.setFollowRedirects(false);
   HttpURLConnection con = (HttpURLConnection) new URL(url).openConnection();
   con.setRequestMethod("HEAD");

   con.setConnectTimeout(5000); //set timeout to 5 seconds

   return (con.getResponseCode() == HttpURLConnection.HTTP_OK);
} catch (java.net.SocketTimeoutException e) {
   return false;
} catch (java.io.IOException e) {
   return false;
}

Answer 2

你可以像这样设置超时，

con.setConnectTimeout(connectTimeout);
con.setReadTimeout(socketTimeout);

Answer 3

如果HTTP连接没有超时，你可以在后台线程本身（AsyncTask，Service等）中实现超时检查器，下面的类是Customize AsyncTask的一个例子，它在一段时间后超时

public abstract class AsyncTaskWithTimer<Params, Progress, Result> extends
    AsyncTask<Params, Progress, Result> {

private static final int HTTP_REQUEST_TIMEOUT = 30000;

@Override
protected Result doInBackground(Params... params) {
    createTimeoutListener();
    return doInBackgroundImpl(params);
}

private void createTimeoutListener() {
    Thread timeout = new Thread() {
        public void run() {
            Looper.prepare();

            final Handler handler = new Handler();
            handler.postDelayed(new Runnable() {
                @Override
                public void run() {

                    if (AsyncTaskWithTimer.this != null
                            && AsyncTaskWithTimer.this.getStatus() != Status.FINISHED)
                        AsyncTaskWithTimer.this.cancel(true);
                    handler.removeCallbacks(this);
                    Looper.myLooper().quit();
                }
            }, HTTP_REQUEST_TIMEOUT);

            Looper.loop();
        }
    };
    timeout.start();
}

abstract protected Result doInBackgroundImpl(Params... params);
}

此示例

public class AsyncTaskWithTimerSample extends AsyncTaskWithTimer<Void, Void, Void> {

    @Override
    protected void onCancelled(Void void) {
        Log.d(TAG, "Async Task onCancelled With Result");
        super.onCancelled(result);
    }

    @Override
    protected void onCancelled() {
        Log.d(TAG, "Async Task onCancelled");
        super.onCancelled();
    }

    @Override
    protected Void doInBackgroundImpl(Void... params) {
        // Do background work
        return null;
    };
 }

Answer 4

我可以通过添加简单的线

来解决类似的问题

HttpURLConnection hConn = (HttpURLConnection) url.openConnection();
hConn.setRequestMethod("HEAD");

我的要求是知道响应代码，因为只是获取元信息就足够了，而不是获得完整的响应体。

默认请求方法是GET，并且需要花费大量时间才能返回，最后抛出了SocketTimeoutException。当我将请求方法设置为HEAD时，响应非常快。

HttpURLConnection超时设置

4 个答案: