Question

我正在构建状态栏应用，并希望根据用户是向左还是向右单击来调用不同的操作。这是我到目前为止所做的：

var statusItem = NSStatusBar.system().statusItem(withLength: -1)
statusItem.action = #selector(AppDelegate.doSomeAction(sender:))

let leftClick = NSEventMask.leftMouseDown
let rightClick = NSEventMask.rightMouseDown

statusItem.button?.sendAction(on: leftClick)
statusItem.button?.sendAction(on: rightClick)

func doSomeAction(sender: NSStatusItem) {
    print("hello world")
}

我的功能没有调用，我找不到原因。我感谢任何帮助！

Answer 1

你试过了吗？

button.sendAction(on: [.leftMouseUp, .rightMouseUp])

然后查看doSomeAction()功能中按下了哪个鼠标按钮？

所以它看起来像......

let statusItem = NSStatusBar.system().statusItem(withLength: NSSquareStatusItemLength)

func applicationDidFinishLaunching(_ aNotification: Notification) {

    if let button = statusItem.button {
        button.action = #selector(self.doSomeAction(sender:))
        button.sendAction(on: [.leftMouseUp, .rightMouseUp])
    }

}

func doSomeAction(sender: NSStatusItem) {

    let event = NSApp.currentEvent!

    if event.type == NSEventType.rightMouseUp {
        // Right button click
    } else {
        // Left button click
    }

}

https://github.com/craigfrancis/datetime/blob/master/xcode/DateTime/AppDelegate.swift

Answer 2

已更新：SWIFT 4

我已经更新了（克雷格弗朗西斯）回答

func doSomeAction(sender: NSStatusItem) {

    let event = NSApp.currentEvent!

    if event.type == NSEvent.EventType.rightMouseUp{
        // Right button click
    } else {
        // Left button click
    }

检测NSStatusItem（Swift）上的左右键单击事件

2 个答案: