在位置获取空指针异常

时间:2014-06-13 07:02:58

标签: android android-layout android-intent android-fragments nullpointerexception

我是android.i的新手需要获得当前位置的格度和逻辑。但是它在位置显示空指针异常。在下面我添加了我的代码。

LocationManager lm;
Location location;

@Override
protected void onCreate(Bundle savedInstanceState) {
    super.onCreate(savedInstanceState);
    setContentView(R.layout.activity_main);

    lm=(LocationManager)getSystemService(Context.LOCATION_SERVICE);
    location=lm.getLastKnownLocation(LocationManager.GPS_PROVIDER);
    SaveLocation(location);


}

private void SaveLocation(Location location) {

    Long time=new GregorianCalendar().getTimeInMillis()+1*60*1000;
    Intent intent=new Intent(MainActivity.this,AlarmReceiver.class);

    intent.putExtra("lat",location.getLatitude());
    intent.putExtra("lon",location.getLongitude());

    AlarmManager alarmManager=(AlarmManager)getSystemService(Context.ALARM_SERVICE);

    alarmManager.set(AlarmManager.RTC_WAKEUP,time,PendingIntent.getBroadcast(this,1,intent,PendingIntent.FLAG_UPDATE_CURRENT));


}

我的androidmanifasted文件就像

<?xml version="1.0" encoding="utf-8"?>
<manifest xmlns:android="http://schemas.android.com/apk/res/android"
package="com.example.gps"
android:versionCode="1"
android:versionName="1.0" >

<uses-sdk
    android:minSdkVersion="8"
    android:targetSdkVersion="17" />

<uses-permission android:name="com.android.alarm.permission.SET_ALARM" />
<uses-permission android:name="android.permission.INTERNET"/>
<uses-permission android:name="android.permission.ACCESS_FINE_LOCATION" />
<uses-permission android:name="android.permission.ACCESS_LOCATION" />
<uses-permission android:name="android.permission.ACCESS_GPS" />

<application
    android:allowBackup="true"
    android:icon="@drawable/ic_launcher"
    android:label="@string/app_name"
    android:theme="@style/AppTheme" >
    <activity
        android:name="com.example.gps.MainActivity"
        android:label="@string/app_name" >
        <intent-filter>
            <action android:name="android.intent.action.MAIN" />

            <category android:name="android.intent.category.LAUNCHER" />
        </intent-filter>
    </activity>
</application>

任何人都可以帮助我?

3 个答案:

答案 0 :(得分:1)

您的位置为空,因为您在创建LocationManager后立即尝试获取最后的已知位置。在这段时间(几毫秒),您的设备无法找到任何位置。 在LocationManager的初始化和getLastKnownLocation方法之间需要一个时间间隔。我建议将getLastKnownLocation方法放在按钮上的OnClick事件中以创建时间间隔。

@Override
protected void onCreate(Bundle savedInstanceState) {
     super.onCreate(savedInstanceState);
     setContentView(R.layout.activity_main);
     lm=(LocationManager)getSystemService(Context.LOCATION_SERVICE);
     /.... 
}

button.setOnClickListener(new OnClickListener(){
      @Override
      public void onClick(View v){
           location=lm.getLastKnownLocation(LocationManager.GPS_PROVIDER);
           SaveLocation(location);
      }
});

答案 1 :(得分:0)

您的Location对象为空,因为没有存储上一个已知位置。

location=lm.getLastKnownLocation(LocationManager.GPS_PROVIDER);

// Check if last known location exists.
if(location != null) {
    SaveLocation(location);
}

答案 2 :(得分:0)

//to find the current latitude and longitude of user
public String  getMyLocation(){

    locationManager=(LocationManager) context.getSystemService(Context.LOCATION_SERVICE);
    Location location=locationManager.getLastKnownLocation(LocationManager.GPS_PROVIDER);


    if(location!=null){

        loc[0]=location.getLatitude();
        loc[1]=location.getLongitude();
        Log.d("STATUS n ", loc[0]+"entered 1"+loc[1]);

    }
    else {
        location=locationManager.getLastKnownLocation(LocationManager.NETWORK_PROVIDER);
        if(location!=null)
        {

            loc[0]=location.getLatitude();
            loc[1]=location.getLongitude();
            Log.d("STATUS n","else"+loc[0]+"entered 1"+loc[1] );
        }

    }
    return getPlace(loc);
}
相关问题