调用此完成处理程序的正确语法是什么

Question

我在使用以下代码时出错。

这是我的代码：

let params = [ "referringUsername": "vihar",
"referringUserId": "78457" ]
Branch.getInstance().getShortURL(withParams: params, andChannel: "SMS", andFeature: "Referral", andCallback: { (url: String!, error: NSError!) -> Void in
if (error == nil) {
let vc = UIActivityViewController(activityItems: [url], applicationActivities: [])
self.present(vc, animated: true, completion: nil)
} else {
print("Genrate Error")
}
})

我得到这个错误：

无法将类型为"(String！，NSError！(->Void"的值转换为预期的参数类型"callbackWithUrl！"(又名"隐含展开可选<(可选<字符串>，可选<错误>(->((>'(

Answer 1

更换

Branch.getInstance().getShortURL(withParams: params, andChannel: "SMS", 
andFeature: "Referral", andCallback: { (url: String!, error: NSError!) -> Void in

带有

Branch.getInstance().getShortURL(withParams: params, andChannel: "SMS",
andFeature: "Referral", andCallback: { (url, error) in

Answer 2

使用此代码：

Branch.getInstance((.getShortURL(带参数：params，频道："SMS"，andFeature:"Referral"，andCallback:｛(url，error(在共享

如文件中所述

如有任何进一步的问题，请写信至[e-mail].