Bas*_*sel 2 iphone cocoa-touch uialertview grand-central-dispatch ios
我想在一段时间后显示一个UIAlertView(比如在应用程序中做一些事情后的5分钟).我已经在应用程序关闭或在后台通知用户.但我希望在应用程序运行时显示UIAlertView.
我尝试dispatch_async如下,但警报永远弹出:
[NSThread sleepForTimeInterval:minutes];
dispatch_async(dispatch_get_main_queue(),
^{
UIAlertView * alert = [[UIAlertView alloc] initWithTitle:@"title!" message:@"message!" delegate:self cancelButtonTitle:@"Cancel" otherButtonTitles:nil];
[alert show];
[alert release];
}
);
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另外,我读到线程在30到60分钟后死亡.我想能够在超过60分钟后显示警报.
Jac*_*kin 12
为什么不使用NSTimer
,为什么在这种情况下你需要使用GCD?
[NSTimer scheduledTimerWithTimeInterval:5*60 target:self selector:@selector(showAlert:) userInfo:nil repeats:NO];
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然后,在同一个类中,你会有这样的事情:
- (void) showAlert:(NSTimer *) timer {
UIAlertView * alert = [[UIAlertView alloc] initWithTitle:@"title!"
message:@"message!"
delegate:self
cancelButtonTitle:@"Cancel"
otherButtonTitles:nil];
[alert show];
[alert release];
}
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另外,正如@PeyloW所说,你也可以使用performSelector:withObject:afterDelay:
:
UIAlertView * alert = [[UIAlertView alloc] initWithTitle:@"title!"
message:@"message!"
delegate:self
cancelButtonTitle:@"Cancel"
otherButtonTitles:nil];
[alert performSelector:@selector(show) withObject:nil afterDelay:5*60];
[alert release];
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编辑你现在也可以使用GCD的dispatch_after
API:
double delayInSeconds = 5;
dispatch_time_t popTime = dispatch_time(DISPATCH_TIME_NOW, delayInSeconds * NSEC_PER_SEC);
dispatch_after(popTime, dispatch_get_main_queue(), ^(void){
UIAlertView *alertView = [[UIAlertView alloc] initWithTitle:@"title!"
message:@"message"
delegate:self
cancelButtonTitle:@"Cancel"
otherButtonTitles:nil];
[alertView show];
[alertView release]; //Obviously you should not call this if you're using ARC
});
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