PushDemo

1.本地推送自定义Action实现快速操作界面，简单实现例如微信推送消息，在接收到推送消息的时候，下拉有按钮，点击按钮后弹出文本输入框

2.在锁屏界面接收到推送消息的时候左滑显示快速操作消息按钮

代码实现：

//1.创建消息上面要显示的按钮
UIMutableUserNotificationAction * action = [[UIMutableUserNotificationAction alloc]init];
action.identifier = @"action";
action.title = @"Accept";
action.behavior = UIUserNotificationActionBehaviorTextInput;
//点击启动程序
action.activationMode = UIUserNotificationActivationModeForeground;

UIMutableUserNotificationAction * action2 = [[UIMutableUserNotificationAction alloc]init];
action2.identifier = @"action2";
action2.title = @"refuse";
//当点击的时候不启动程序，在后台处理
action2.activationMode = UIUserNotificationActivationModeBackground;
action2.authenticationRequired = YES; //需要解锁才能处理  如果action2.activationMode = UIUserNotificationActivationModeForeground  那么这个属性没用
action2.destructive = YES;

//2.创建动作类别集合
UIMutableUserNotificationCategory * categorys = [[UIMutableUserNotificationCategory alloc]init];
categorys.identifier = @"alert"; //这组动作的唯一标识
[categorys setActions:@[action,action2] forContext:(UIUserNotificationActionContextMinimal)];

//3.ios8之后必须先进行注册
//询问用户是否允许发送通知
UIUserNotificationSettings * uns = [UIUserNotificationSettings settingsForTypes:UIUserNotificationTypeAlert|UIUserNotificationTypeBadge|UIUserNotificationTypeSound categories:[NSSet setWithObjects:categorys, nil]];
//注册通知
[[UIApplication sharedApplication] registerUserNotificationSettings:uns];