首先我想知道,当我点击本地通知时,会调用哪种特定方法.我想点击通知打开一个网址.下面是代码应用程序委托.现在的问题是,即使我没有点击通知,网址也会自动打开.如果你知道的话请指导我.谢谢
- (void)application:(UIApplication *)application didReceiveLocalNotification: (UILocalNotification *)notifyAlarm
{
application.applicationIconBadgeNumber = 0;
NSLog(@"Notification tapped :) ");
[[UIApplication sharedApplication] openURL:[NSURL URLWithString:@"http://www.google.com.pk"]]; }
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- (void)application:(UIApplication *)application didReceiveLocalNotification: (UILocalNotification *)notifyAlarm
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每次通知火灾时都会调用此方法.
要在点击通知时打开网址,您必须检查应用的状态.
- (void)application:(UIApplication *)application didReceiveLocalNotification:(UILocalNotification *)notification
{
UIApplicationState appState = UIApplicationStateActive;
if ([application respondsToSelector:@selector(applicationState)])
appState = application.applicationState;
if (appState == UIApplicationStateActive)
{
// Don't open Url.
}
else
{
// Open Url.
}
}
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