Swift 调用此完成处理程序的正确语法是什么?
我在使用下面的代码时出错 这是我的密码:Swift 调用此完成处理程序的正确语法是什么?,swift,branch.io,Swift,Branch.io,我在使用下面的代码时出错 这是我的密码: let params = [ "referringUsername": "vihar", "referringUserId": "78457" ] Branch.getInstance().getShortURL(withParams: params, andChannel: "SMS", andFeature: "Referral", andCallback: { (url: String!, error: NS
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!”(又称“隐式包装()>”)
替换
Branch.getInstance().getShortURL(withParams: params, andChannel: "SMS",
andFeature: "Referral", andCallback: { (url: String!, error: NSError!) -> Void in
与
替换
Branch.getInstance().getShortURL(withParams: params, andChannel: "SMS",
andFeature: "Referral", andCallback: { (url: String!, error: NSError!) -> Void in
与
使用此代码:
Branch.getInstance().getShortURL(带参数:params和通道:“SMS”,
andFeature:“引用”,andCallback:{(url,错误)在
分享
如报告中所述
如有任何疑问,请写信至integration@branch.io.使用以下代码:
Branch.getInstance().getShortURL(带参数:params和通道:“SMS”,
andFeature:“引用”,andCallback:{(url,错误)在
分享
如报告中所述
如有任何疑问,请写信至integration@branch.io.@Sh_Khan你能解释一下吗?@Sh_Khan你能解释一下吗?