Question

我有一个非常简单的SwiftUI应用程序，该应用程序在菜单栏中运行，应该从后台的重复计时器内部定期打开一个应用程序窗口（整个应用程序中只有一个窗口/视图）。

我实际上如何从代码中打开应用程序窗口？

这是一个简化的AppDelegate.swift示例，显示了我要执行的操作：

import Cocoa
import SwiftUI

@NSApplicationMain
class AppDelegate: NSObject, NSApplicationDelegate {  
    var window: NSWindow!

    func applicationDidFinishLaunching(_ aNotification: Notification) {
        // Create the SwiftUI view that provides the window contents.
        let contentView = ContentView()

        // Create the window and set the content view. 
        window = NSWindow(
            contentRect: NSRect(x: 0, y: 0, width: 480, height: 300),
            styleMask: [.titled, .closable, .miniaturizable, .resizable, .fullSizeContentView],
            backing: .buffered, defer: false)
        window.center()
        window.setFrameAutosaveName("Main Window")
        window.contentView = NSHostingView(rootView: contentView)
        window.makeKeyAndOrderFront(nil)

        var loop = 0
        Timer.scheduledTimer(withTimeInterval: 1.0, repeats: true) { timer in
            loop+=1
            if (loop % 10 == 0) {
                // TODO: How to close the window?
            } else {
                // TODO: How to reopen the window?
            }
        }
    }
}

Answer 1

一种方法是隐藏/取消隐藏应用程序本身

func applicationDidFinishLaunching(_ aNotification: Notification) {
    //setup of window etc ...

    Timer.scheduledTimer(withTimeInterval: 3, repeats: true, block: { _ in
        if self.window.isVisible {
            NSApp.hide(self)
        } else {
            NSApp.unhide(self)
        }
    })
}

如何在macOS（不是iOS）上的SwiftUI中以编程方式从后台打开应用程序窗口？

1 个答案: