【问题标题】:Activity has leaked window com.android.internal.policy.impl.PhoneWindow$DecorView@46368a28 that was originally added hereActivity 泄露了最初添加在这里的窗口 com.android.internal.policy.impl.PhoneWindow$DecorView@46368a28
【发布时间】:2023-12-22 04:12:01
【问题描述】:

这个问题让我发疯。我错过了一些基本但非常重要的知识,关于如何在活动中创建的新线程中处理长操作以及如何在长操作完成后修改文本等视图组件。

首先让我向您展示我的代码中发生此问题的部分:

mProgressDialog = ProgressDialog.show(mContext, "Tripplanner", "please wait...", true, false);
connectAndGetRoute();


private void connectAndGetRoute(){

    new Thread(){
        @Override
        public void run() {
            try {
            if(!connectTo9292ov()) return;// conncetto9292ov() connects to a website, parses the reasult into an arraylist. The arraylist contains route.

            } catch(UnknownHostException e){
                Toast.makeText(mContext, "failed to connect to server", Toast.LENGTH_LONG).show();
            }catch (ClientProtocolException e) {
               Toast.makeText(mContext, "failed to connect to server", Toast.LENGTH_LONG).show();                   
        } catch (IOException e) {
                Toast.makeText(mContext, "failed to connect to server", Toast.LENGTH_LONG).show();
            }

            handler.post(runConnection);
        }

    }.start();

    handler = new Handler();
    runConnection = new Runnable(){
        @Override
        public void run() {
            mProgressDialog.dismiss();
            showOnScreen();
        }   
    };
}

这是我得到的错误:

ERROR/WindowManager(8297): Activity mp.tripplanner.OvPlanner has leaked window com.android.internal.policy.impl.PhoneWindow$DecorView@46368a28 that was originally added here
ERROR/WindowManager(8297): android.view.WindowLeaked: Activity mp.tripplanner.OvPlanner has leaked window com.android.internal.policy.impl.PhoneWindow$DecorView@46368a28 that was originally added here
ERROR/WindowManager(8297): at android.view.ViewRoot.<init>(ViewRoot.java:251)
ERROR/WindowManager(8297): at android.view.WindowManagerImpl.addView(WindowManagerImpl.java:148)
ERROR/WindowManager(8297): at android.view.WindowManagerImpl.addView(WindowManagerImpl.java:91)
ERROR/WindowManager(8297): at android.view.Window$LocalWindowManager.addView(Window.java:424)
ERROR/WindowManager(8297): at android.app.Dialog.show(Dialog.java:241)
ERROR/WindowManager(8297): at android.app.ProgressDialog.show(ProgressDialog.java:107)
ERROR/WindowManager(8297): at android.app.ProgressDialog.show(ProgressDialog.java:95)
ERROR/WindowManager(8297): at mp.tripplanner.OvPlanner$3.onClick(OvPlanner.java:351)
ERROR/WindowManager(8297): at android.view.View.performClick(View.java:2408)
ERROR/WindowManager(8297): at android.view.View$PerformClick.run(View.java:8817)
ERROR/WindowManager(8297): at android.os.Handler.handleCallback(Handler.java:587)
ERROR/WindowManager(8297): at android.os.Handler.dispatchMessage(Handler.java:92)
ERROR/WindowManager(8297): at android.os.Looper.loop(Looper.java:144)
ERROR/WindowManager(8297): at android.app.ActivityThread.main(ActivityThread.java:4937)
ERROR/WindowManager(8297): at java.lang.reflect.Method.invokeNative(Native Method)
ERROR/WindowManager(8297): at java.lang.reflect.Method.invoke(Method.java:521)
ERROR/WindowManager(8297): at com.android.internal.os.ZygoteInit$MethodAndArgsCaller.run(ZygoteInit.java:868)
ERROR/WindowManager(8297): at com.android.internal.os.ZygoteInit.main(ZygoteInit.java:626)
ERROR/WindowManager(8297): at dalvik.system.NativeStart.main(Native Method)

但是在上面一条之前的日志中写入了另外一条错误信息,即:

ERROR/AndroidRuntime(8297): FATAL EXCEPTION: Thread-9
ERROR/AndroidRuntime(8297): java.lang.RuntimeException: Can't create handler inside thread that has not called Looper.prepare()
ERROR/AndroidRuntime(8297): at android.os.Handler.<init>(Handler.java:121)
ERROR/AndroidRuntime(8297): at android.widget.Toast.<init>(Toast.java:68)
ERROR/AndroidRuntime(8297): at android.widget.Toast.makeText(Toast.java:231)
ERROR/AndroidRuntime(8297): at mp.tripplanner.OvPlanner$4.run(OvPlanner.java:371)

感谢您的帮助。

【问题讨论】:

    标签: android multithreading window memory-leaks


    【解决方案1】:

    只是一个快速的想法。

    我认为,这是您在线程中的祝酒词。尝试将它们注释掉。

    如果您仍想显示消息,请保存线程的状态并在您的处理程序中处理它。从 finally 块中调用你的处理程序

    mProgressDialog = ProgressDialog.show(mContext, "Tripplanner", "please wait...", true, false);
    connectAndGetRoute();
    
    
    private void connectAndGetRoute(){
    
        new Thread(){
            @Override
            public void run() {
                try {
                if(!connectTo9292ov()) return;// conncetto9292ov() connects to a website, parses the reasult into an arraylist. The arraylist contains route.
    
                } catch(UnknownHostException e){
                    // an int member of your activity
                    threadStatus = 404;
                    // Toast.makeText(mContext, "failed to connect to server", Toast.LENGTH_LONG).show();
                }catch (ClientProtocolException e) {
                    threadStatus = 404;
                   // Toast.makeText(mContext, "failed to connect to server", Toast.LENGTH_LONG).show();                   
            } catch (IOException e) {
                    threadStatus = 404;
                    // Toast.makeText(mContext, "failed to connect to server", Toast.LENGTH_LONG).show();
                }
                finally {
                handler.post(runConnection);}
            }
    
        }.start();
    
        handler = new Handler();
        runConnection = new Runnable(){
            @Override
            public void run() {
                if (threadStatus == 404) { Toast.makeText(mContext, "failed to connect to server", Toast.LENGTH_LONG).show();}
                else {
                mProgressDialog.dismiss();
                showOnScreen();}
            }   
        };
    }
    

    【讨论】:

    • 您猜对了,也猜对了,但不确定,仍然不知道如何解决。因为我不明白你提供的解决方案:D。你能告诉我你保存线程状态的意思吗?我该怎么做?我实际上需要对您的解决方案进行更多解释:D。对不起。
    • 我稍微修改了你的代码。如果你问我原因,那么你不能从线程更新你的 ui。你应该在外面做。
    • 看看@Olsavage的回答
    【解决方案2】:

    您的Handler 需要在您的 UI 线程中创建,它才能更新 UI。 然后我会使用处理程序的sendMessage 方法,而不是发布可运行的:

    private static final int HANDLER_MESSAGE_ERROR = 0;
    private static final int HANDLER_MESSAGE_COMPLETED = 1;
    ...
    private void connectAndGetRoute(){
        new Thread(){
            @Override
            public void run() {
                try {
                    if(!connectTo9292ov()) return;
    
                } catch(UnknownHostException e){
                    sendMessage(HANDLER_MESSAGE_ERROR);
                } catch (ClientProtocolException e) {
                    sendMessage(HANDLER_MESSAGE_ERROR);
                } catch (IOException e) {
                    sendMessage(HANDLER_MESSAGE_ERROR);
                } finally {
                    sendMessage(HANDLER_MESSAGE_COMPLETED);
                }
            }
            private void sendMessage(int what){
                Message msg = Message.obtain();
                msg.what = what;
                mHandler.sendMessage(msg);
            }
        }.start();
    
    }
    ...
    private Handler mHandler = new Handler(){
        @Override
        public void handleMessage(Message msg) {
            switch(msg.what){
            case HANDLER_MESSAGE_ERROR:
                Toast.makeText(mContext, "failed to connect to server", Toast.LENGTH_LONG).show();
                break;
            case HANDLER_MESSAGE_COMPLETED:
                mProgressDialog.dismiss();
                showOnScreen();
                break;
            default:
                Log.w("MyTag","Warning: message type \""+msg.what+"\" not supported");
            }
        }
    }
    

    【讨论】:

    • 非常感谢,这解决了问题。我只是想知道您是否建议使用 Handler 的 sendMessage 而不是发布可运行文件?
    • @mnish 这两种方法都应该有效,我只是发现如果它们都在一个地方,跟踪你的 UI 更新会更容易。使用sendMessage更容易看到职责分离,所以我觉得更容易调试。
    【解决方案3】:

    示例代码:

    new Thread(new Runnable() {                 
                    @Override
                    public void run() {
                        doNotUIthings();
                        updateUI.sendEmptyMessage(0);   //Update ui in UI thread
                    }
                    private Handler updateUI = new Handler(){
                        @Override
                        public void dispatchMessage(Message msg) {
                            super.dispatchMessage(msg);
                            updateUI();
                        }
                    };
                }).start();
    

    但我建议使用 AsyncTask 而不是 Java 线程。

    【讨论】:

    • 非常感谢。实际上从来没有想过这个。虽然我不知道这是如何工作的,但如果当前问题再次发生,我肯定会考虑在未来使用它。但现在我将继续使用 Dave.c 提供的解决方案,因为它对我来说效果很好。
    最近更新 更多