在Swift中访问userInfo字典

Question

我正在尝试访问userInfo字典,application:didReceiveRemoteNotification()以便我可以根据收到的推送类型/是否收到任何推送来选择segue.我已经尝试过这两种方式(其他一些方式,我敢肯定,我现在还记不起来).

if let info = userInfo as? Dictionary<String,String> {
        var notificationType = info["notificationType"]
}


if let info: String = userInfo["notificationType"] as? String {
        //do stuff
}

我没有得到任何错误,我根本没有得到任何东西.如果我打印userInfo字典,它只有一个成员["aps"],其中包含显示给用户的推送消息,所以即使我可以访问它,我也可以使用它进行条件化.

我尝试按照Parse API写入userData:

let data = ["notificationType" : "coffee"]
push.setData(data)

所以一个问题是这似乎没有设置任何东西,但更大的问题是我无法获取任何userInfo数据.

你怎么访问这本字典？

编辑

一些打印声明结果:

println(userInfo["aps"]) => {alert = "You've Been Invited To A Coffee Order";}

println(userInfo) => [aps: {alert = "You've Been Invited To A Coffee Order";}, type: coffee]

println(userInfo[0]) => nil

let info = userInfo as? Dictionary<String,String>
println(info) => nil

所以似乎警报键被困在这个'aps'数组中？字典？我没有创建它,它只是自动的一部分userInfo

编辑2

这是我设置数据并发送推送的代码.我userInfo从这里发送出现的数据:

func notifyUserOfCoffeeOrder(){
    var uQuery:PFQuery = PFUser.query()
    uQuery.whereKey("username", equalTo: usernameLabel.text)

    var pushQuery:PFQuery = PFInstallation.query()
    pushQuery.whereKey("user", matchesQuery: uQuery)

    var push:PFPush = PFPush()
    push.setQuery(pushQuery)
    let data = ["type" : "coffee",
                "alert" : "You've Been Invited To A Coffee Order"]
    push.setData(data)
    push.sendPush(nil)
    println("push sent")

}

这是来自AppDelegate.swift的整个函数,我将收到的数据拉出来:

func application(application: UIApplication, didReceiveRemoteNotification userInfo: [NSObject : AnyObject]) {
     NSNotificationCenter.defaultCenter().postNotificationName("getMessage", object: nil)

    if  let info = userInfo["type"] as? String {
        self.window = UIWindow(frame: UIScreen.mainScreen().bounds)

        var storyboard = UIStoryboard(name: "Main", bundle: nil)

        var initialViewController = storyboard.instantiateViewControllerWithIdentifier("coffeeVC") as UIViewController

        self.window?.rootViewController = initialViewController
        self.window?.makeKeyAndVisible()
    }

}

Answer 1

编辑:我知道重建这个答案符合给出的评论.请注意,我目前无法访问Apples推送通知服务,因此无法在与OP相同的环境中对此进行测试

似乎APN(或parse.com)在userInfo构建它时会在这里和那里添加一点点.因此,我最好的建议是使用以下代码从您收到的双字典中打印所有内容.我还提供了代码来打开在字典中查找内容时出现的选项.

所以这是我修改过的示例代码:

func myApplicationFunc(didReceiveRemoteNotification userInfo: [NSObject : AnyObject]) {
    // NSNotificationCenter.defaultCenter().postNotificationName("getMessage", object: nil)

    // Default printout of userInfo
    println("All of userInfo:\n\( userInfo)\n")

    // Print all of userInfo
    for (key, value) in userInfo {
        println("userInfo: \(key) —> value = \(value)")
    }

    if let info = userInfo["aps"] as? Dictionary<String, AnyObject> {
        // Default printout of info = userInfo["aps"]
        println("All of info: \n\(info)\n")

        for (key, value) in info {
            println("APS: \(key) —> \(value)")
        }


        if  let myType = info["type"] as? String {
            // Printout of (userInfo["aps"])["type"]
            println("\nFrom APS-dictionary with key \"type\":  \( myType)")

            // Do your stuff?
        }
    }
}

var userInfo : [ NSObject: AnyObject] =
[
    "aps" : [
        "alert" : "Coffee is not the answer",
        "type" : "coffee"
    ],
    "anotherType" : 123
]

myApplicationFunc(didReceiveRemoteNotification: userInfo)

此代码生成以下输出:

All of userInfo:
[anotherType: 123, aps: {
    alert = "Coffee is not the answer";
    type = coffee;
}]

userInfo: anotherType —> value = 123
userInfo: aps —> value = {
    alert = "Coffee is not the answer";
    type = coffee;
}

All of info: 
[alert: Coffee is not the answer, type: coffee]

APS: alert —> Coffee is not the answer
APS: type —> coffee

From APS-dictionary with key "type":  coffee

您应该能够在您的application(...)方法中添加我的代码,并希望您将看到如何解开双字典以及实际传输给您的信息.

PS!在另一篇关于他如何setMessage扰乱他的帖子的回答中setData,这可能也很有趣,如果有一些代码会影响你的结果.我的示例代码仍应列出收到的所有内容userInfo