从AppWidget单击后应用程序加载缓慢

Question

当您单击应用程序中的一个按钮时，浮动应用程序小部件就会打开。

我写了一个函数，当点击Widget时应该打开应用程序屏幕。（代码如下）

我在下面列出了两种方法

问题：

点击后，应用程序打开需要几秒钟的时间。

我想在点击后打开内存栏中的应用程序

MotionEvent.ACTION_UP -> {
    if (System.currentTimeMillis() - clickStartTimer < FloatingWidgetView.CLICK_DELTA) {
        Toast.makeText(context, "clicked floating widget", Toast.LENGTH_SHORT).show()

        val configIntent = Intent(context, MainActivity::class.java)
        configIntent.flags = Intent.FLAG_ACTIVITY_NEW_TASK
        configIntent.data = Uri.parse(configIntent.toUri(Intent.URI_INTENT_SCHEME))

        val pendIntent = PendingIntent.getActivity(
            context,
            0,
            configIntent,
            PendingIntent.FLAG_UPDATE_CURRENT
        )

        // Second method
        // val intent = Intent(context, MainActivity::class.java)
        // intent.addFlags(Intent.FLAG_ACTIVITY_NEW_TASK)
        // startActivity(context, intent, null)
    }
}

Answer 1

试用一下我的片段，首先检查一下来自评论的链接（对不起Java，它只是从工作项目复制粘贴的）。代码在服务中执行，当所有活动被删除到后台（也称为“homed”）时，该服务将启动

            Intent i = new Intent(this, mainActivityClass);
            // brings back whole activitys stack
            i.setAction(Intent.ACTION_MAIN);
            i.addCategory(Intent.CATEGORY_LAUNCHER);
            PendingIntent pendingIntent = PendingIntent.getActivity(this, 0, i, 0);

            // http://code.google.com/p/android/issues/detail?id=4536
            boolean showToastBecauseOfDelay;
            try { // works without delay till Android P, on Q also delayed...
                pendingIntent.send();
                showToastBecauseOfDelay = Build.VERSION.SDK_INT >= 27;
            } catch (PendingIntent.CanceledException e) {
                // startActivity may cause delay!!
                // when called less that 5 sec after Home button press...
                startActivity(i);
                showToastBecauseOfDelay = true;
            }

            // service starts when app/activity goes to background ("homed")
            // Toast will be shown only if user want to go back quickly after that
            // (just for some distration and feedback that click was registered)
            long diff = System.currentTimeMillis() - serviceStartedAt;
            if (showToastBecauseOfDelay && diff < 5000) {
                // long - 3.5 sec, short 2 - sec, no other option
                int toastDuration = diff < 2000 ? Toast.LENGTH_LONG : Toast.LENGTH_SHORT;
                Toast.makeText(this, R.string.resuming_app, toastDuration).show();
            }