如何为 UITableViewCell 显示自定义 UIMenuItem? [英] How to show a custom UIMenuItem for a UITableViewCell?

问题描述

我希望在我长按 UITableViewCell 时弹出的 UIMenuController 显示自定义 UIMenuItems.

I want the UIMenuController that pops up when I long-press a UITableViewCell to show custom UIMenuItems.

我在 viewDidLoad 中设置了自定义项

I set up the custom item in viewDidLoad

UIMenuItem *testMenuItem = [[UIMenuItem alloc] initWithTitle:@"Test" action:@selector(test:)];
[[UIMenuController sharedMenuController] setMenuItems: @[testMenuItem]];

然后我设置了所有正确的委托方法.

And then I set all the right delegate methods.

- (BOOL)tableView:(UITableView *)tableView shouldShowMenuForRowAtIndexPath:(NSIndexPath *)indexPath {
    return YES;
}

-(BOOL)tableView:(UITableView *)tableView canPerformAction:(SEL)action forRowAtIndexPath:(NSIndexPath *)indexPath withSender:(id)sender {
    return (action == @selector(copy:) || action == @selector(test:));
}

- (BOOL)tableView:(UITableView *)tableView performAction:(SEL)action forRowAtIndexPath:(NSIndexPath *)indexPath withSender:(id)sender {
    if (action == @selector(copy:)) {
         // do stuff
    }

    return YES;
}

但它所做的只是显示复制"项目，因为我只允许它和我的自定义项目.但是，自定义项目不会显示.

But all it does, is show the "Copy" item, since I only allow it and my custom item. The custom item, however, won't show up.

我意识到，我可以向单元格本身添加一个手势识别器，但这违背了 UIMenuController 共享实例的目的，不是吗?

I realize, I could add a gesture recognizer to the cell itself, but that kind of defeats the purpose of the shared instance of UIMenuController, doesn't it?

推荐答案

据我了解主要有两个问题:

As far as I understand there are two main problems:

1) 您希望 tableView canPerformAction: 支持自定义选择器，而文档说它只支持两个 UIResponderStandardEditActions(复制和/或粘贴)；

1) you expect tableView canPerformAction: to support custom selectors while the documentation says it supports only two of UIResponderStandardEditActions (copy and/or paste);

2) 不需要 ||action == @selector(test:) 当您通过初始化 menuItems 属性添加自定义菜单选项时.对于这个项目选择器，检查将是自动的.

2) there's no need for the part || action == @selector(test:) as you are adding the custom menu options by initializing menuItems property. For this items selectors the check will be automatical.

您可以做些什么来显示和工作的自定义菜单项:

What you can do to get the custom menu item displayed and work is:

1) 使用

a)

UIMenuItem *testMenuItem = [[UIMenuItem alloc] initWithTitle:@"Test" action:@selector(test:)];
[[UIMenuController sharedMenuController] setMenuItems: @[testMenuItem]];
[[UIMenuController sharedMenuController] update];

b)

- (BOOL)tableView:(UITableView *)tableView shouldShowMenuForRowAtIndexPath:(NSIndexPath *)indexPath {
    return YES;
}

-(BOOL)tableView:(UITableView *)tableView canPerformAction:(SEL)action forRowAtIndexPath:(NSIndexPath *)indexPath withSender:(id)sender {
    return (action == @selector(copy:));
}

- (void)tableView:(UITableView *)tableView performAction:(SEL)action forRowAtIndexPath:(NSIndexPath *)indexPath withSender:(id)sender {
    // required
}

2) 使用

-(BOOL) canPerformAction:(SEL)action withSender:(id)sender {
    return (action == @selector(copy:) || action == @selector(test:));
}

-(BOOL)canBecomeFirstResponder {
    return YES;
}

/// this methods will be called for the cell menu items
-(void) test: (id) sender {

}

-(void) copy:(id)sender {

}
///////////////////////////////////////////////////////

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