ios - 无法对推送通知自定义按钮执行操作

Question

我正在开发一个涉及设备到设备推送通知的 iOS 应用程序。在前台和后台状态下，我能够接收通知并能够在各自的自定义按钮(接受和拒绝)中执行各自的操作。在上述两个状态下一切正常。但是在终止/终止状态下，虽然我能够收到通知，但我无法对单击自定义按钮(接受和拒绝)执行操作。你们能帮我这个吗？

//Notification action button function
func setActionCategories(){
    let acceptAction = UNNotificationAction(
        identifier: NAString().notificationAcceptIdentifier(),
        title: NAString().accept().capitalized,
        options: [.init(rawValue: 0)])

    let rejectAction = UNNotificationAction(
        identifier: NAString().notificationRejectIdentifier(),
        title: NAString().reject().capitalized,
        options: [.init(rawValue: 0)])

    let actionCategory = UNNotificationCategory(
        identifier: NAString().notificationActionCategory(),
        actions: [acceptAction,rejectAction],
        intentIdentifiers: [],
        options: [.customDismissAction])

    UNUserNotificationCenter.current().setNotificationCategories(
        [actionCategory])
}

func userNotificationCenter(_ center: UNUserNotificationCenter, didReceive response: UNNotificationResponse, withCompletionHandler completionHandler: @escaping () -> Void) {
    let userInfo = response.notification.request.content.userInfo

    //Here we are performing Action on Notification Buttons & We created this buttons in  "setActionCategories" function.
    if response.notification.request.content.categoryIdentifier == NAString().notificationActionCategory() {

        //Created Firebase reference to get currently invited visitor by E-Intercom
        var gateNotificationRef : DatabaseReference?
        gateNotificationRef = GlobalUserData.shared.getUserDataReference().child(Constants.FIREBASE_CHILD_GATE_NOTIFICATION).child(userUID).child(guestType!).child(guestUID!)

        //Performing accept & reject on click of recently invited visitor by E-Intercom from Notification view.
        switch response.actionIdentifier {

        //If Accept button will pressed
        case NAString().notificationAcceptIdentifier():
            gateNotificationRef?.child(NAString().status()).setValue(NAString().accepted())
            }
            break

        //If Reject button will pressed
        case NAString().notificationRejectIdentifier():                                                         gateNotificationRef?.child(NAString().status()).setValue(NAString().rejected())
            break

        default:
            break
        }
    }
    UIApplication.shared.applicationIconBadgeNumber = 0
    completionHandler()
}

Answer 1

您好 Ashish 能否提供一些代码以便我们更好地帮助您。那里应该有一个完成处理程序，您可以在其中添加一个 Action 函数。然后您可以执行您需要按钮执行的任何操作。