美文网首页
UIMenuController使用

UIMenuController使用

作者: 风继续吹0 | 来源:发表于2016-08-25 14:03 被阅读23次

    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 UITableViewCellUILabel 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 via touchesBegan: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.

    相关文章

      网友评论

          本文标题:UIMenuController使用

          本文链接:https://www.haomeiwen.com/subject/ijcasttx.html