UIMenuController使用

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：

1. 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
2. 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：

01 - (BOOL)tableView:(UITableView *)tableView shouldShowMenuForRowAtIndexPath:(NSIndexPath *)indexPath {
02     return YES;
03 }
04
05 - (BOOL)tableView:(UITableView *)tableView canPerformAction:(SEL)action forRowAtIndexPath:(NSIndexPath *)indexPath withSender:(id)sender {
06     if (action == @selector(copy:)) {
07         returnYES;
08     }
09
10     returnNO;
11 }
12
13 - (void)tableView:(UITableView *)tableView performAction:(SEL)action forRowAtIndexPath:(NSIndexPath *)indexPath withSender:(id)sender {
14     if (action == @selector(copy:)) {
15         [UIPasteboard generalPasteboard].string = [data objectAtIndex:indexPath.row];
16     }
17 }

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.

定制菜单项

如果你想使用定制菜单项，下面代码比较隐晦，但非常灵活。你需要检测是否用户长按并显示菜单，而最简单的方法就是在表格单元格中使用 UILongPressGestureRecognizer

1 UILongPressGestureRecognizer *recognizer = [[UILongPressGestureRecognizer alloc] initWithTarget:self action:@selector(longPress:)];
2 [cell addGestureRecognizer:recognizer];

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

01 - (void)longPress:(UILongPressGestureRecognizer *)recognizer {
02     if (recognizer.state == UIGestureRecognizerStateBegan) {
03         TSTableViewCell *cell = (TSTableViewCell *)recognizer.view;
04         [cell becomeFirstResponder];
05
06         UIMenuItem *flag = [[UIMenuItem alloc] initWithTitle:@"Flag"action:@selector(flag:)];
07         UIMenuItem *approve = [[UIMenuItem alloc] initWithTitle:@"Approve"action:@selector(approve:)];
08         UIMenuItem *deny = [[UIMenuItem alloc] initWithTitle:@"Deny"action:@selector(deny:)];
09
10         UIMenuController *menu = [UIMenuController sharedMenuController];
11         [menu setMenuItems:[NSArray arrayWithObjects:flag, approve, deny, nil]];
12         [menu setTargetRect:cell.frame inView:cell.superview];
13         [menu setMenuVisible:YES animated:YES];
14     }
15 }
16
17 - (void)flag:(id)sender {
18     NSLog(@"Cell was flagged");
19 }
20
21 - (void)approve:(id)sender {
22     NSLog(@"Cell was approved");
23 }
24
25 - (void)deny:(id)sender {
26     NSLog(@"Cell was denied");
27 }

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.