在某些情况下,开发者可能需要唤起客服的APP,这样可以快速的进入到APP客服系统进行操作。
private void doStartApplicationWithPackageName(String packagename) {
// 通过包名获取此APP详细信息,包括Activities、services、versioncode、name等等
PackageInfo packageinfo = null;
try {
packageinfo = getPackageManager().getPackageInfo(packagename, 0);
} catch (PackageManager.NameNotFoundException e) {
e.printStackTrace();
}
if (packageinfo == null) {
Toast.makeText(this, "应用未安装", Toast.LENGTH_SHORT).show();
return;
}
// 创建一个类别为CATEGORY_LAUNCHER的该包名的Intent
Intent resolveIntent = new Intent(Intent.ACTION_MAIN, null);
resolveIntent.addCategory(Intent.CATEGORY_LAUNCHER);
resolveIntent.setPackage(packageinfo.packageName);
// 通过getPackageManager()的queryIntentActivities方法遍历
List<ResolveInfo> resolveinfoList = getPackageManager()
.queryIntentActivities(resolveIntent, 0);
ResolveInfo resolveinfo = resolveinfoList.iterator().next();
if (resolveinfo != null) {
// packagename = 参数packname
String packageName = resolveinfo.activityInfo.packageName;
// 这个就是我们要找的该APP的LAUNCHER的Activity[组织形式:packagename.mainActivityname]
String className = resolveinfo.activityInfo.name;
// LAUNCHER Intent
Intent intent = new Intent(Intent.ACTION_MAIN);
intent.addCategory(Intent.CATEGORY_LAUNCHER);
// 设置ComponentName参数1:packagename参数2:MainActivity路径
ComponentName cn = new ComponentName(packageName, className);
intent.setComponent(cn);
startActivity(intent);
}
}
// 执行这句即可
doStartApplicationWithPackageName("com.easemob.helpdesk");
可以动态注册,也可以在AndroidManifest.xml中注册,只需要指定对Action即可,在此只举一个例子:
//1. 在AndroidManifest.xml注册广播
<receiver android:name=".NewMessageReceive">
<intent-filter>
<action android:name="com.easemob.helpdesk.newmessage"/>
</intent-filter>
</receiver>
// 2. 继承自BroadcastReceiver的类,根据extra获取unreadcount
public class NewMessageReceive extends BroadcastReceiver {
@Override
public void onReceive(Context context, Intent intent) {
int count = intent.getIntExtra("com.easemob.helpdesk.unreadcount", 0);
Toast.makeText(context.getApplicationContext(), "收到消息->" + count, Toast.LENGTH_SHORT).show();
}
}