如何使用branch.io访问我分配给URL的参数?

时间:2017-03-22 10:21:38

标签: ios swift branch.io

我只需创建一个这样的网址:

Branch.getInstance().getShortURL(withParams: ["a": [1, 2, 3], "b": ["c": 34, "d": "Malwina is OK:-)"]]) { url, error in

    let controller = UIActivityViewController(activityItems: [URL(string: url!)!], applicationActivities: nil)
    self.present(controller, animated: true, completion: nil)
}

我得到这样的东西:

https://fieldserviceios.app.link/Hjp9gesGIB

从该链接我尝试打开应用程序并捕获该参数:

func application(_ application: UIApplication, continue userActivity: NSUserActivity, restorationHandler: @escaping ([Any]?) -> Void) -> Bool {

    let a = Branch.getInstance().getLatestReferringParams() //no my params here
}

1 个答案:

答案 0 :(得分:1)

您创建链接的方式很好,您正在生成有效的链接。您可以通过附加"?debug = true"来检查任何Branch链接的内容。链接并在浏览器的地址栏中输入。

对于您的链接:https://fieldserviceios.app.link/Hjp9gesGIB?debug=true我看到了:

{ 
    "$identity_id": "372863071192286427", 
    "$one_time_use": false, 
    "a": [1,2,3], 
    "b": { 
        "c": 34, 
        "d": "Malwina jest OK:-)" 
    }, 
    "~creation_source": 3, 
    "~id": "373038618178427139" 
}

我将介绍如何从我对其他Stack Overflow问题的响应中的链接中读取参数,例如:How do I access metadata when branch link was clicked and opened my app...?,但请说明您提供的具体代码: