Android 警告( Android 出现的警告(Service Intent must be explicit)解决办法详解
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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action mIntent.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);
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