Android Context没有参加活动?和其他无活动的编程? [英] Android Context without being in an activity? And other activity-less programming?

问题描述

我会尽力把这个问题变成一个全面的问题:

我正在编写一种获取String的方法，该字符串包含由LocationManager和getLastKnownLocation()以及所有其他内容确定的Android设备城市的名称.

然后我意识到我需要在另一个活动中再次做同样的事情，所以为什么不制作一个可以在整个程序中使用的完全独立的类(LocationFinder)，而不是在各处编写重复的代码? /p>

但是我遇到了使我感到困惑的问题.例如，如果我创建此类(LocationFinder)，即使从未真正可视化它，它也应该扩展Activity吗?该类所要做的就是拥有各种getLastKnownCity()或getCurrentCity()这样的getter并返回字符串.我以为不必扩展Activity类，因为它实际上不是活动.

但是接下来我要使用什么上下文:

Geocoder geocoder = new Geocoder(Context context, Locale locale)

?

这使我认为它必须是一项活动.所以我扩展了Activity，并用

替换了构造函数

@Override
protected void onCreate(..............

但是由于某些原因，即使我放了，它也永远不会被呼叫

String city = new LocationFinder().getLastKnownCity();

LocationFinder的onCreate()的第一行是

System.out.println("HEY!")

，甚至都没有做到这一点.我在android.internal.os.LoggingPrintStream.println()和其他地方得到了空指针.

此外，有许多来自Activity类的系统常量.例如，我需要获取LOCATION_SERVICE，它是一个字符串，如果不扩展Activity则无法获取.当然，我可以作弊，只需输入文字字符串，但这感觉是错误的.

解决方案

构造类时，可以有一个构造函数，该构造函数接受Context并将其分配给您的类中的本地Context对象.

public class LocationFinder {
     private Context myContext;
     private Geocoder geocoder;

     public LocationFinder(Context context)
     {
         myContext = context;
         geocoder = new Geocoder(myContext);
     }

}

然后，当您尝试访问此类时，请确保将其初始化为:

public class TestActivity extends Activity {
     protected void onCreate(Bundle savedInstanceState)
     {
          super.onCreate(savedInstanceState);
          setContentView(R.layout.main);
          LocationFinder lFinder = new LocationFinder(getApplication());
     }
}

当然，您不能从将要运行的每个类中访问上下文.因此，对View的引用就足够了.

LocationFinder lFinder = new LocationFinder(anyView.getApplication());

I'll try really hard to turn this into one comprehensive question:

I'm writing a method to get a String that contains the name of an Android device's city, as determined by the LocationManager and getLastKnownLocation() and all that.

Then I realized I'd need to do the same thing again in another activity, so why not just make an entirely separate class (LocationFinder) that I could use across my program, instead of writing duplicate code everywhere?

But I've run into problems that confuses me. For instance, if I make this class (LocationFinder), should it extend Activity, even though it is never actually visualized? All this class would do is have a variety of getters like getLastKnownCity() or getCurrentCity() and return strings. I assumed it wouldn't HAVE to extend the Activity class, since it's really not an activity.

But then what Context do I use for:

Geocoder geocoder = new Geocoder(Context context, Locale locale)

?

This made me assume it MUST be an activity. So I extended Activity, and replaced the constructor with

@Override
protected void onCreate(..............

but for some reason, that never ends up getting called, even when I put

String city = new LocationFinder().getLastKnownCity();

My very first line of LocationFinder's onCreate() is

System.out.println("HEY!")

and it never even gets to that. I get a null pointer at android.internal.os.LoggingPrintStream.println() and other stuff.

Plus, there's a bunch of system constants that come from Activity classes. For instance, I need to get at LOCATION_SERVICE, which is a String, which I can't get without extending Activity. Sure, I could cheat and just put in the literal string, but that feels wrong.

解决方案

When constructing your class, you can have a constructor that takes in a Context and assigns it a local Context object within your class.

public class LocationFinder {
     private Context myContext;
     private Geocoder geocoder;

     public LocationFinder(Context context)
     {
         myContext = context;
         geocoder = new Geocoder(myContext);
     }

}

And then when you try to access this class, make sure you initialise it like:

public class TestActivity extends Activity {
     protected void onCreate(Bundle savedInstanceState)
     {
          super.onCreate(savedInstanceState);
          setContentView(R.layout.main);
          LocationFinder lFinder = new LocationFinder(getApplication());
     }
}

Of course, you can't access a context from every class that you will be running. So a reference to a View can suffice.

LocationFinder lFinder = new LocationFinder(anyView.getApplication());

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