Dog*_*fee 2 uimenucontroller ios
这就是我设置弹出窗口的方式
UIMenuItem *menuItem = [[UIMenuItem alloc] initWithTitle:@"Delete Patient"
action:@selector(customAction:)];
[[UIMenuController sharedMenuController] setMenuItems:@[menuItem]];
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然后添加require方法
- (BOOL)canBecomeFirstResponder {
return YES;
}
- (BOOL)canPerformAction:(SEL)action withSender:(id)sender {
NSLog(@"canPerformAction");
// The selector(s) should match your UIMenuItem selector
if (action == @selector(customAction:)) {
return YES;
}
return NO;
}
- (void) customAction:(id) sender
{
for (Treatment *t in self.ptToDelete.patientRx) {
[self.managedObjectContext deleteObject:t];
}
[self.managedObjectContext deleteObject:self.ptToDelete];
NSError *error = nil;
if (![self.managedObjectContext save:&error]) {
NSLog(@"Error! %@", error);
}
}
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这适用于iOS6,但现在不适用.以下方法未被调用,应该在我点击并按住时调用
- (BOOL)canPerformAction:(SEL)action withSender:(id)sender
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我发现我需要在CollectionViewCell类中包含以下内容.但是,ios6中不需要这样做.希望这可以节省几个小时的人.
- (BOOL)canPerformAction:(SEL)action withSender:(id)sender
{
// The selector/s should match your UIMenuItem selector
if (action == @selector(customAction:)) {
return YES;
}
return NO;
}
- (void) customAction:(id)sender
{
// do stuff
}
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