NSButton的弹出菜单实现

Question

我该怎么办呢？

我在考虑......

[NSMenu popUpContextMenu:menu withEvent:event forView:(NSButton *)sender];

Answer 1

对.

按钮动作调用

[NSMenu popUpContextMenu:menu withEvent:event forView:(NSButton *)sender];

哪里

• menu :你想要显示的菜单
• sender :你点击的按钮
• event:NSEvent你创造的新事物

创建新内容时NSEvent,请指定要在何处显示弹出菜单的位置.

Answer 2

接受答案的Swift版本

@IBAction func actionOccurred(sender: NSButton) {
    if let event = NSApplication.sharedApplication().currentEvent {
        NSMenu.popUpContextMenu(sender.menu!, withEvent: event, forView: sender)
    }
}

Answer 3

正如我所评论的那样,我发现ButtonMadness的例子并不完美.我的实现似乎更好.鼠标按下时显示菜单,整个按钮保持按下状态,可以指定菜单位置,并且菜单将被解除,而不会出现虚假显示.

说实话,在大多数情况下,NSPopupButton是更好的选择.我使用这段代码主要是因为按钮和弹出窗口有一个类的便利,因为菜单不包含弹出控件图像和标题.我从单独的笔尖加载菜单,并根据需要在应用程序的其他位置重复使用它.

请注意,添加弹出窗口和菜单的附加支持是微不足道的.

NSButton subclass:

- (void)mouseDown:(NSEvent *)theEvent {

    // if a menu is defined let the cell handle its display
    if (self.menu) {
        if ([theEvent type] == NSLeftMouseDown) {
            [[self cell] setMenu:[self menu]];
        } else {
            [[self cell] setMenu:nil];
        }
    }

    [super mouseDown:theEvent];
}

NSButtonCell subclass:

- (BOOL)trackMouse:(NSEvent *)event inRect:(NSRect)cellFrame ofView:(NSView *)controlView untilMouseUp:(BOOL)untilMouseUp
{
    // if menu defined show on left mouse
    if ([event type] == NSLeftMouseDown && [self menu]) {

        NSPoint result = [controlView convertPoint:NSMakePoint(NSMidX(cellFrame), NSMidY(cellFrame)) toView:nil];

        NSEvent *newEvent = [NSEvent mouseEventWithType: [event type]
                                               location: result
                                          modifierFlags: [event modifierFlags]
                                              timestamp: [event timestamp]
                                           windowNumber: [event windowNumber]
                                                context: [event context]
                                            eventNumber: [event eventNumber]
                                             clickCount: [event clickCount]
                                               pressure: [event pressure]];

        // need to generate a new event otherwise selection of button
        // after menu display fails
        [NSMenu popUpContextMenu:[self menu] withEvent:newEvent forView:controlView];

        return YES;
    }

    return [super trackMouse:event inRect:cellFrame ofView:controlView untilMouseUp:untilMouseUp];
}

Answer 4

最近,我试图实现它,我想,我用一个更简单的解决方案

-(IBAction)buttonClick:(id)sender {
    NSButton * b = (NSButton*)sender;
    NSPoint l = [ self.window convertBaseToScreen:b.frame.origin ];

    [ self.menu popUpMenuPositioningItem:nil atLocation:l inView:nil ];
}

更新

convertBaseToScreen从10.7开始不推荐使用,而不是以convertRectToScreen下列方式使用:

NSPoint l = [self.window convertRectToScreen:b.frame].origin;