使用 Google Places API，Location 类在使用时抛出空指针异常

Question

我希望程序能够使用 Location location.getLatitude()/Location location.getLongitude 方法返回上述 a 和 b 值的值，但每当我用返回双精度值的方法替换 a 和 b 时getLongitude 和 getLatitude 提供的值或实际的 get 函数本身，我得到一个错误。

public void onClick(View v) {

    requesturl = "https://maps.googleapis.com/maps/api/place/search/json?" + 
    "location=" + a + "," + b + "&radius=6000&" + "types=bank&sensor=false&key=AIzaSyDXNlHRDZnWW0T0tvBUjpyA8k2K9sjS2cM";    

    HHand(requesturl);


}

我怎样才能让 a 和 b 可以从 getLatitude() 和 getLongitude() 函数中获取它们的值？

已编辑：代码

public class Freedom extends ActivityGroup implements LocationListener {
/** Called when the activity is first created. 
 * @return */

String keystring="";
String requesturl;
LocationManager locman;
Location location;

//Awesome      
public String convertStreamToString(InputStream is) { 
    return new Scanner(is).useDelimiter("\\A").next();
} 



public void HHand(String a) {


    //HTTP Request Processing for URL
       HttpClient client=new DefaultHttpClient();
       StringBuilder builder=new StringBuilder(a);
       builder.append(URLEncoder.encode(keystring));
       HttpPost post=new HttpPost(a);

       try {
           org.apache.http.HttpResponse response=client.execute(post);
           HttpEntity entity=response.getEntity();


           if (entity != null) {
               InputStream is = entity.getContent();
               String val = convertStreamToString(is);
               Log.e("", val);

              }


           }


           catch (ClientProtocolException e) {   
               e.printStackTrace();
           }

           catch (IOException e) {
               e.printStackTrace();
           }


}


@Override
public void onCreate(Bundle savedInstanceState) {


    super.onCreate(savedInstanceState);        
    setContentView(R.layout.main);
    locman = (LocationManager) this.getSystemService(LOCATION_SERVICE);



}



protected void onResume() {
    super.onResume();
    locman.requestLocationUpdates(LocationManager.GPS_PROVIDER, 1000, 10, this);
  }


public boolean onCreateOptionsMenu(Menu menu)  {
    super.onCreateOptionsMenu(menu);
    MenuInflater inflater = getMenuInflater();
    inflater.inflate(R.menu.btmmnu, menu);
    return true;
}

protected Dialog onCreateDialog(int id) {

        final Dialog dialog = new Dialog(this);
        OnClickListener txtlstn = new OnClickListener() {
            public void onClick(View v) {
                dismissDialog(0);
            }
        };

        switch(id) {
        case 0:                         

            ...
        }
        return dialog;
}




public boolean onOptionsItemSelected(MenuItem item) {

    switch (item.getItemId()) {
    case R.id.About:                
        showDialog(0);
        return true;
    case R.id.Guidance:

         final Button Travel = (Button) findViewById(R.id.travel);  
         Travel.setOnClickListener(new View.OnClickListener() {

            public void onClick(View v) {
                requesturl = "https://maps.googleapis.com/maps/api/place/search/json?" + 
                          "location=" + location.getLatitude() + "," + location.getLongitude() + "&radius=6000&" +
                          "types=car_rental&sensor=false&key=AIzaSyDXNlHRDZnWW0T0tvBUjpyA8k2K9sjS2cM";  
                        HHand(requesturl);


            }
        });

Answer 1

如果您从location.getLatitude() 或location.getLongitude() 获得NullPointerException，则表明location 为空。

您还没有显示location 的来源，或者您已经看到方法调用返回a 和b，但基本上这就是原因。也许您有一个局部变量，在某处隐藏了一个实例变量？例如：

public class Foo
{
    private Location location;

    public Foo(double latitude, double longitude)
    {
        // Careful - this declares a new *local* variable, so it's not
        // changing the value of this.location
        Location location = new Location();
        location.setLatitude(latitude);
        location.setLongitude(longitude);
        ...
    }

    public void bang()
    {
        // This will throw, as location is still null
        double x = location.getLatitude();
    }
}