如何解决Android 5.0中出现的警告：Service Intent must be explitict

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。

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;
}

就是使用上面这段代码解决了出错的问题

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);