从OSX中的菜单栏状态项中删除突出显示？

I have create a menubar 'button' which runs a function when it is clicked. Below is my code for the menubar button.

我创建了一个菜单栏“按钮”，它在点击时运行一个功能。下面是我的菜单栏按钮的代码。

let statusItem = NSStatusBar.systemStatusBar().statusItemWithLength(-1)

func applicationDidFinishLaunching(aNotification: NSNotification) {

    if let button = statusItem.button {
        button.title = "This is in the menu bar."
        button.action = Selector("myFunction")
    }
}

The problem is, this is going to be made into an image, not text, so I need it not to be highlighted blue when the user clicks it, which is what happens at the moment.

问题是，这将被制作成一个图像，而不是文本，因此我需要在用户点击它时不要将其突出显示为蓝色，这就是目前发生的情况。

Is there anything I can do to stop it being highlighted when the user clicks it?

当用户点击它时，我能做些什么来阻止它突出显示？

Thanks!

谢谢！

1 个解决方案

#1

I guess you could set the button type to MomentaryChangeButton, a type of button that does not get highlighted when clicked.

我猜你可以将按钮类型设置为MomentaryChangeButton，这是一种在点击时不会突出显示的按钮。

button.setButtonType(NSButtonType.MomentaryChangeButton)

Short syntax:

语法短：

button.setButtonType(.MomentaryChangeButton)

In case you also want to get rid of the focus ring:

如果你也想摆脱焦点圈：

button.focusRingType = NSFocusRingType.None

Short syntax:

语法短：

button.focusRingType = .None

Note: I wrote "I guess" because I didn't verify that it works in your specific case of status bar button.

注意：我写了“我猜”，因为我没有验证它是否适用于您的状态栏按钮的特定情况。