我一直在寻找答案,但我似乎找不到答案。我想要做的是让我的应用程序在后台运行应用程序时发送本地通知,当用户打开通知时,它会将他们带到网站。我已经完成了所有设置,但它不断打开应用程序,而不是去网站。
我的问题是,这甚至可以吗?如果是这样,请问下面我的代码出错了?谢谢你的帮助。
CODE:
- (void)applicationDidEnterBackground:(UIApplication *)应用程序 { //使用此方法释放共享资源,保存用户数据,使计时器无效,并存储足够的应用程序状态信息,以便将应用程序恢复到当前状态,以防以后终止。 //如果您的应用程序支持后台执行,则在用户退出时调用此方法而不是applicationWillTerminate:
NSDate *date = [NSDate date];
NSDate *futureDate = [date dateByAddingTimeInterval:3];
notifyAlarm.fireDate = futureDate;
notifyAlarm.timeZone = [NSTimeZone defaultTimeZone];
notifyAlarm.repeatInterval = 0;
notifyAlarm.alertBody = @"Visit Our Website for more info";
[app scheduleLocalNotification:notifyAlarm];
if ( [notifyAlarm.alertBody isEqualToString:@"Visit Our Website for more info"] ) {
[[UIApplication sharedApplication] openURL:[NSURL URLWithString:@"https://www.bluemoonstudios.com.au"]];
}
答案 0 :(得分:7)
在您的业务逻辑中
-(void)scheduleNotificationForDate:(NSDate *)fireDate{
UILocalNotification *notification = [[UILocalNotification alloc] init];
notification.fireDate = fireDate;
notification.alertAction = @"View";
notification.alertBody = @"New Message Received";
notification.userInfo = @{@"SiteURLKey": @"http://www.google.com"};
[[UIApplication sharedApplication] scheduleLocalNotification:notification];
}
你的AppDelegate中的
- (BOOL)application:(UIApplication *)application didFinishLaunchingWithOptions:(NSDictionary *)launchOptions
{
UILocalNotification *notification = [launchOptions objectForKey:UIApplicationLaunchOptionsLocalNotificationKey];
if(notification != nil){
NSDictionary *userInfo = notification.userInfo;
NSURL *siteURL = [NSURL URLWithString:[userInfo objectForKey:@"SiteURLKey"]];
[[UIApplication sharedApplication] openURL:siteURL];
}
}
-(void)application:(UIApplication *)application didReceiveLocalNotification:(UILocalNotification *)notification{
NSDictionary *userInfo = notification.userInfo;
NSURL *siteURL = [NSURL URLWithString:[userInfo objectForKey:@"SiteURLKey"]];
[[UIApplication sharedApplication] openURL:siteURL];
}