Java WiFi开关小部件…无法访问的代码

Java WiFi开关小部件…无法访问的代码,java,android,widget,Java,Android,Widget,我第一次尝试使用一些在线教程和我在网上找到的一些反编译代码制作一个简单的小部件。但我在下面的方法中遇到了问题。我得到了错误的不可访问代码 private void Update(Context paramContext, AppWidgetManager paramAppWidgetManager, int[] paramArrayOfInt) { RemoteViews localRemoteViews = new RemoteViews(paramContext.

我第一次尝试使用一些在线教程和我在网上找到的一些反编译代码制作一个简单的小部件。但我在下面的方法中遇到了问题。我得到了错误的不可访问代码

private void Update(Context paramContext, AppWidgetManager paramAppWidgetManager, int[] paramArrayOfInt)
      {
        RemoteViews localRemoteViews = new RemoteViews(paramContext.getPackageName(), R.xml.widget);
        WifiManager localWifiManager = (WifiManager)paramContext.getSystemService("wifi");
        try
        {
          int i = localWifiManager.getWifiState();
          if (i == 3)
          {
            localRemoteViews.setImageViewResource(R.xml.widget, R.drawable.off);
            link(localRemoteViews, paramContext, paramAppWidgetManager);
            if ((paramArrayOfInt == null) || (paramArrayOfInt.length <= 0))
              break;
            paramAppWidgetManager.updateAppWidget(paramArrayOfInt, localRemoteViews);
          }
          while (true)
          {
            return;
            if (i == 1)
            {
              localRemoteViews.setImageViewResource(R.xml.widget, R.drawable.on);
              break;
            }
            localRemoteViews.setImageViewResource(R.xml.widget, R.drawable.off);
            break;
            paramAppWidgetManager.updateAppWidget(this.THIS_WIDGET, localRemoteViews);
          }
        }
        catch (Exception localException)
        {
          break;
        }
      }
private void更新(Context paramContext,AppWidgetManager paramAppWidgetManager,int[]paramArrayOfInt)
{
RemoteViews localRemoteViews=新的RemoteViews(paramContext.getPackageName(),R.xml.widget);
WifiManager localWifiManager=(WifiManager)paramContext.getSystemService(“wifi”);
尝试
{
int i=localWifiManager.getWifiState();
如果(i==3)
{
setImageViewResource(R.xml.widget,R.drawable.off);
链接(LocalRemoteView、paramContext、paramAppWidgetManager);

如果((paramArrayOfInt==null)|(paramArrayOfInt.length一个问题是
onUpdate
方法中的break语句。break语句将中断循环或switch语句。在代码中,您需要从该方法返回。因此,将break更改为return

另一个问题是
onUpdate
onReceive
方法。如果返回(true)时,这将中断循环并返回,因此无法到达下一个语句。您需要删除此处的return语句或根据任何条件将逻辑更改为return

while (true)
{
  return;
  if ("com.wifiswitch.bestwifiswitch.toggle".equals(str))
  {
            Toggle(paramContext, "com.roundedlabs.widgets.wifi.toggle");
            Update(paramContext, AppWidgetManager.getInstance(paramContext), null);
  }
while (true)
{
  return;
  if ("com.wifiswitch.bestwifiswitch.toggle".equals(str))
  {
            Toggle(paramContext, "com.roundedlabs.widgets.wifi.toggle");
            Update(paramContext, AppWidgetManager.getInstance(paramContext), null);
  }