如何解決Android 5.0以上出現的警告：Service Intent must be expli

有些時候我們使用Service的時需要採用隱私啟動的方式，但是Android 5.0一齣來後，其中有個特性就是Service Intent  must be explitict，也就是說從Lollipop開始，service服務必須採用顯示方式啟動。 
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，那麼這裡有兩種解決方法： 

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());//這裡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。 


2、將隱式啟動轉換為顯示啟動：

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;
    }

就是使用上面這段代碼解決了出錯的問題

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);