当苹果在 iOS 3.0 中增加了剪切、复制和粘贴功能时，它同时为开发者提供了 UIMenuController 组件用来定制该弹出菜单，但不幸的是，最开始的实现要很麻烦。
- 附加在菜单的视图的
canBecomeFirstResponser必须返回 YES，这意味着必须子类化。例如最常用的显示元素UITableViewCell和UILabel默认返回的是NO
-
UILongPressGestureRecognizer直到 iOS 3.2 才提供, which means that the long press to initiate the menu display had to be implemented viatouchesBegan:withEvent:, touchesMoved:withEvent:, andtouchesEnded:withEvent:. Every custom long press recognizer might use a different delay constant, which could easily confuse users who are used to another app's implementation.
而最新的 iOS 使用两种基本方法解决了这个问题，一个是表格单元格，另外一个是定制菜单选项。
指定情景: UITableViewCell on iOS 5
如果你只是想在 UITableViewCell 中使用系统提供的复制粘贴功能（大部分情况是这样），iOS 5.0 有更简单的方法：
- (BOOL)tableView:(UITableView *)tableView shouldShowMenuForRowAtIndexPath:(NSIndexPath *)indexPath {
return YES;
}
- (BOOL)tableView:(UITableView *)tableView canPerformAction:(SEL)action forRowAtIndexPath:(NSIndexPath *)indexPath withSender:(id)sender {
if (action == @selector(copy:)) {
returnYES;
}
returnNO;
}
- (void)tableView:(UITableView *)tableView performAction:(SEL)action forRowAtIndexPath:(NSIndexPath *)indexPath withSender:(id)sender {
if (action == @selector(copy:)) {
[UIPasteboard generalPasteboard].string = [data objectAtIndex:indexPath.row];
}
}
该菜单调用 tableView:canPerformAction:forRowAtIndexPath:withSender
以确认是否该显示系统菜单选项并调用
tableView:performAction:forRowAtIndexPath:withSender:当用户选择某个选项时.
为了让菜单显示，目标视图必须在 responder 链中，很多 UIKit 视图默认并无法成为一个 responder ，因此你需要之类这些视图重载 canBecomeFirstResponder 方法范围 YES
在下面例子中，我们使用定制类 TSTableViewCell 并实现了长按选择器
- (void)longPress:(UILongPressGestureRecognizer *)recognizer {
if (recognizer.state == UIGestureRecognizerStateBegan) {
TSTableViewCell *cell = (TSTableViewCell *)recognizer.view;
[cell becomeFirstResponder];
UIMenuItem *flag = [[UIMenuItem alloc] initWithTitle:@"Flag"action:@selector(flag:)];
UIMenuItem *approve = [[UIMenuItem alloc] initWithTitle:@"Approve"action:@selector(approve:)];
UIMenuItem *deny = [[UIMenuItem alloc] initWithTitle:@"Deny"action:@selector(deny:)];
UIMenuController *menu = [UIMenuController sharedMenuController];
[menu setMenuItems:[NSArray arrayWithObjects:flag, approve, deny, nil]];
[menu setTargetRect:cell.frame inView:cell.superview];
[menu setMenuVisible:YES animated:YES];
}
}
- (void)flag:(id)sender {
NSLog(@"Cell was flagged");
}
- (void)approve:(id)sender {
NSLog(@"Cell was approved");
}
- (void)deny:(id)sender {
NSLog(@"Cell was denied");
}
There is only one small gotcha with UIMenuItem: if the specified action is not implemented by your view controller, that item will not appear in the menu.
