IntentService不会显示Toast

时间:2011-03-18 01:01:37

标签: android intentservice android-toast

我创建的这个IntentService将在onStartCommand()和onDestroy()中显示Toasts,但不在onHandleIntent()中显示。我错过了一些关于IntentService限制的内容吗?

public class MyService extends IntentService {

private static final String TAG = "MyService";

public MyService(){
    super("MyService");
}

@Override
protected void onHandleIntent(Intent intent) {
    cycle();
}

@Override
public int onStartCommand(Intent intent, int flags, int startId) {
    Toast.makeText(this, "service starting", Toast.LENGTH_SHORT).show(); //This happens!
    return super.onStartCommand(intent,flags,startId);
}

@Override
public void onCreate() {
    super.onCreate();

}

@Override
public void onDestroy() {
    Toast.makeText(this, "service stopping", Toast.LENGTH_SHORT).show(); //This happens!
    super.onDestroy();
}

private void cycle(){
      Toast.makeText(this, "cycle done", Toast.LENGTH_SHORT).show();  //This DOESN'T happen!
      Log.d(TAG,"cycle completed"); //This happens!
}
}

4 个答案:

答案 0 :(得分:65)

接受的答案不正确。

以下是您如何通过onHandleIntent()展示吐司:

Create a DisplayToast class

public class DisplayToast implements Runnable {
    private final Context mContext;
    String mText;

    public DisplayToast(Context mContext, String text){
        this.mContext = mContext;
        mText = text;
    }

    public void run(){
        Toast.makeText(mContext, mText, Toast.LENGTH_SHORT).show();
    }
}

在服务的构造函数中实例化Handler,并使用post对象调用DisplayToast方法。

public class MyService extends IntentService {
Handler mHandler;

public MyService(){
    super("MyService");
    mHandler = new Handler();
}

@Override
protected void onHandleIntent(Intent intent) {
    mHandler.post(new DisplayToast(this, "Hello World!"));

}
}

答案 1 :(得分:23)

你应该在主线程上启动Toast:

new Handler(Looper.getMainLooper()).post(new Runnable() {
        @Override
        public void run() {
            Toast.makeText(getApplicationContext(), message, Toast.LENGTH_LONG).show();
        }
});

这是因为否则IntentService的线程在吐出之前退出,导致IllegalStateException:

  

java.lang.IllegalStateException:Handler(android.os.Handler){12345678}在死线程上向处理程序发送消息

答案 2 :(得分:19)

onHandleIntent()是从后台线程调用的(这就是IntentService的全部内容),所以你不应该从那里做UI。

答案 3 :(得分:2)

另一个选项是RxJava,例如:

private void showToast(final String text) {
    Observable.just(text)
        .observeOn(AndroidSchedulers.mainThread())
        .subscribe(new Action1<String>() {
            @Override
            public void call(String s) {
                Toast.makeText(getApplicationContext(), s, Toast.LENGTH_LONG).show();
            }
        });
}

警告:我是Android的新手。