openURL没有打开Safari

Question

我正在尝试使用操作表打开带有链接的safari.变量设置正确并相应地显示链接,但由于某种原因,Safari将无法打开,我无法弄清楚为什么...

这是代码:

-(void)actionSheet {
    sheet = [[UIActionSheet alloc] initWithTitle:@"Options"
                                    delegate:self
                           cancelButtonTitle:@"Cancel"
                      destructiveButtonTitle:nil
                           otherButtonTitles:@"Open in Safari", nil];

    [sheet showInView:[UIApplication sharedApplication].keyWindow];
}

-(void)actionSheet:(UIActionSheet *)actionSheet clickedButtonAtIndex:(NSInteger)buttonIndex {

    if (buttonIndex != -1) {
        [[UIApplication sharedApplication] openURL:[NSURL URLWithString:self.url]];
    }
}

Answer 1

NSString *strurl = [NSString stringWithFormat:@"http://%@",strMediaIconUrl];
[[UIApplication sharedApplication] openURL:[NSURL URLWithString:strurl]];
use http:// must.