Question

嘿伙计们，我有这个问题。我不知道发生了什么，但它是我第一次看到这个错误。我使用这个代码，它永远不会导致任何问题。我不知道问题是因为代码工作没有错误突然它给了我那个错误并关闭了应用程序。

守则

    public void run() {
    try {
        send();
    } catch (ClientProtocolException e) {
        e.printStackTrace();
    } catch (IOException e) {
        e.printStackTrace();
    }

}



public void send() throws ClientProtocolException, IOException
{
    OutputStream mOutputStream = null;
    BufferedWriter mWriter = null;

    List<NameValuePair> mParameters = req.getParameters();

    URL url = null;
    HttpURLConnection connection = null;
    try {
        Looper.prepare();
        url = new URL(req.returnRequestUrl());
        connection = (HttpURLConnection) url.openConnection();
        connection.setReadTimeout(READ_TIMEOUT);
        connection.setConnectTimeout(CONNECTION_TINEOUT);
        connection.setRequestMethod(Params.HTTP_REQUEST_METHOD_POST);
        connection.setDoOutput(true);
        connection.setDoInput(true);
        mOutputStream = connection.getOutputStream();
        mWriter = new BufferedWriter(new OutputStreamWriter(mOutputStream, Params.UTF8));
        String sparams = URLEncodedUtils.format(mParameters, Params.UTF8);
        mWriter.write(sparams);
        mWriter.flush();

        mResponseCode = connection.getResponseCode();
        if (mResponseCode > 203) {
            mData = readWebData(connection.getErrorStream());
            this.req.getResponse().notGoodServerEroorr();
        } else {
            mData = readWebData(connection.getInputStream());

        }

    } catch (IOException e) {
        // TODO Auto-generated catch block
        e.printStackTrace();
    } finally {
        if (connection != null) {
            try {
                if (mOutputStream != null)
                    mOutputStream.close();
                if (mWriter != null)
                    mWriter.close();

            } catch (IOException e) {
                 // TODO Auto-generated catch block
                e.printStackTrace();
            }

            connection.disconnect();
            evaluateDataAndRespondToFragment(mData);
            myLooper = Looper.myLooper();
            Looper.loop();
            myLooper.quit();
        }

    }
}

private void evaluateDataAndRespondToFragment(String mData) {
    Listen lis = this.req.getResponse();
    if (mData.equals("1"))
        lis.good();
    else
        lis.notGood();
}

Answer 1

创建一个处理程序并从新线程调用它以更新视图

final Hanlder hanlder=new Handler(){
    @Override
    public void handleMessage(Message msg){
         //update views here
    }
}
//from the thread
new Thread(new Runnable{
    @Override
    public void run(){
       //do your async task
       //call hanlder
       handler.obtainMessage().sendToTarget()
    }
}).start();

请参阅此处以在runOnUiThread和handler之间进行选择。 https://stackoverflow.com/a/7452913/5819589

Answer 2

此问题是由连接请求引起的。通常，这些类型的请求不适用于主线程。因此，创建一个新线程来运行此代码或使用runOnUiThread()方法。这可以解决您的问题。试试这个

    public void run() {
    runOnUiThread(new Runnable() {
            @Override
            public void run() {
                send();
            }
        });
}

Answer 3

如错误消息所述，只有主线程可以触及VAny类型的视图，似乎

evaluateDataAndRespondToFragment(mData);

上面的方法触及Android不允许的主线程组件，从Main Thread调用此方法就像这样 -

getActivity().runOnUiThread(new Runnable() {
   @Override
   public void run() {
     evaluateDataAndRespondToFragment(mData);
   }
});

Android错误：只有创建视图层次结构的原始线程才能触及其视图

3 个答案: