Android 出現的警告(Service Intent must be explicit)解決辦法詳解
Android 出現的警告(Service Intent must be explicit)解決辦法詳解
有些時候我們使用Service的時需要采用隱私啟動的方式,但是Android 5.0一出來后,其中有個特性就是Service Intent must be explitict,也就是說從Lollipop開始,service服務必須采用顯示方式啟動。
而android源碼是這樣寫的(源碼位置: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,源碼里是這樣寫的,那么這里有兩種解決方法:
1、設置Action和packageName:
參考代碼如下:
Intent mIntent = new Intent(); mIntent.setAction("XXX.XXX.XXX");//你定義的service的action mIntent.setPackage(getPackageName());//這里你需要設置你應用的包名 context.startService(mIntent);
此方式是google官方推薦使用的解決方法。
在此附上地址供大家參考:http://developer.android.com/goo ... tml#billing-service,有興趣的可以去看看。
2、將隱式啟動轉換為顯示啟動:--參考地址: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; }
調用方式如下:
Intent mIntent = new Intent(); mIntent.setAction("XXX.XXX.XXX"); Intent eintent = new Intent(getExplicitIntent(mContext,mIntent)); context.startService(eintent);
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