创建一个弹出按钮接口 [英] Creating a pop up button interface
问题描述
在苹果的股票消息应用程序,轻按拍照键显示弹出式菜单按钮,允许用户拍摄照片/视频,或者选择一个现有的。我将如何实现此相同的按键设计?程序是相同的两个iPhone和放大器; iPad的?
In Apple's stock 'Messages' app, tapping the camera button reveals popup buttons allowing the user to take a photo/video or choose an existing one. How would I implemented this same button design? Is the procedure the same for both iPhone & iPad?
推荐答案
这就是所谓的一个 UIActionSheet
。您可以使用像这样
It's called a UIActionSheet
. You use it like this
UIActionSheet *action = [[UIActionSheet alloc] initWithTitle:@"Foo" delegate:self cancelButtonTitle:@"Cancel" destructiveButtonTitle:nil otherButtonTitles:@"foo1", @"foo2", nil];
[action showInView:self.view];
(改变FOOS到任何)。为了检测被点击哪个按钮,落实 UIActionSheetDelegate
的 actionSheet:clickedButtonAtIndex:
委托方法。例如:
- (void)actionSheet:(UIActionSheet *)actionSheet clickedButtonAtIndex:(NSInteger)buttonIndex{
NSString *title = [actionSheet buttonTitleAtIndex:buttonIndex];
if ([title isEqualToString:@"foo1"]) {
// do stuff...
}
}
是的,这部作品同时在iPhone和iPad(如@bobnoble指出,iPad版采用了酥料饼的观点,而不是动作片,但动作片在两个工作)。
And yes, this works on both the iPhone and iPad (as @bobnoble pointed out, the iPad version uses a popover view, not an action sheet, but action sheets work on both).
这篇关于创建一个弹出按钮接口的文章就介绍到这了,希望我们推荐的答案对大家有所帮助,也希望大家多多支持IT屋!