Android 出现的警告(Service Intent must be explicit)解决办法详解
有些时候我们使用Service的时需要采用隐私启动的方式,但是Android 5.0一出来后,其中有个特性就是Service Intent  must be explitict,也就是说从Lollipop开始,service服务必须采用显示方式启动。
而android源码是这样写的(源码位置:sdk/sources/android-21/android/app/ContextImpl.java):
private void validateServiceIntent(Intent service) {  if (service.getComponent() == null && service.getPackage() == null) {   if (getApplicationInfo().targetSdkVersion >= Build.VERSION_CODES.LOLLIPOP) {    IllegalArgumentException ex = new IllegalArgumentException(      "Service Intent must be explicit: " + service);    throw ex;   } else {    Log.w(TAG, "Implicit intents with startService are not safe: " + service      + " " + Debug.getCallers(2, 3));   }  } }既然,源码里是这样写的,那么这里有两种解决方法:
1、设置Action和packageName:
参考代码如下:
Intent mIntent = new Intent();mIntent.setAction("XXX.XXX.XXX");//你定义的service的actionmIntent.setPackage(getPackageName());//这里你需要设置你应用的包名context.startService(mIntent);此方式是google官方推荐使用的解决方法。
在此附上地址供大家参考:http://developer.android.com/goo ... tml#billing-service,有兴趣的可以去看看。
2、将隐式启动转换为显示启动:--参考地址:http://stackoverflow.com/a/26318757/1446466
public static Intent getExplicitIntent(Context context, Intent implicitIntent) {  // Retrieve all services that can match the given intent  PackageManager pm = context.getPackageManager();  List<ResolveInfo> resolveInfo = pm.queryIntentServices(implicitIntent, 0);  // Make sure only one match was found  if (resolveInfo == null || resolveInfo.size() != 1) {   return null;  }  // Get component info and create ComponentName  ResolveInfo serviceInfo = resolveInfo.get(0);  String packageName = serviceInfo.serviceInfo.packageName;  String className = serviceInfo.serviceInfo.name;  ComponentName component = new ComponentName(packageName, className);  // Create a new intent. Use the old one for extras and such reuse  Intent explicitIntent = new Intent(implicitIntent);  // Set the component to be explicit  explicitIntent.setComponent(component);  return explicitIntent; }调用方式如下:
Intent mIntent = new Intent();mIntent.setAction("XXX.XXX.XXX");Intent eintent = new Intent(getExplicitIntent(mContext,mIntent));context.startService(eintent);感谢阅读,希望能帮助到大家,谢谢大家对本站的支持!
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