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));
- }
- }
- }
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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);
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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:Google In-App billing, IllegalArgumentException: Service Intent must be explicit, after upgrading to Android L Dev Preview - Stack Overflow
- public static Intent getExplicitIntent(Context context, Intent implicitIntent) {
- // Retrieve all services that can match the given intent
- PackageManager pm = context.getPackageManager();
- List 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;
- }
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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);