带操作按钮的Swift open twitter配置文件

我试图通过点击按钮打开我的Twitter个人资料，当我点击按钮时，它将以Safari打开，但我想用Twitter应用程序打开，我怎么能做到这一点，我在这里做错了什么

let twUrl = URL(string: "twitter://user?screen_name=wixvii")!
let twUrlWeb = URL(string: "https://www.twitter.com/wixvii")!
if UIApplication.shared.canOpenURL(twUrl){
   UIApplication.shared.open(twUrl, options: [:],completionHandler: nil) 
}else{
   UIApplication.shared.open(twUrlWeb, options: [:], completionHandler: nil)
}

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将此命令添加到info.plist可以解决此问题

LSApplicationQueriesSchemes

推特

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1。在Xcode中添加Info.plist源代码或文本编辑下一个字符串：

func twitter() {
 
   let lowerCaseSocialNetworkName = "twitter"
   let socialNetworkDomain = "com"
   let userName = "bandyliuk"
   let appURL = URL(string: "\(lowerCaseSocialNetworkName)://user? screen_name=\(userName)")!
   let application = UIApplication.shared
 
   if application.canOpenURL(appURL) {
      application.open(appURL)
   } else {
      let webURL = URL(string: "https://\(lowerCaseSocialNetworkName).\ (socialNetworkDomain)/\(userName)")!
   application.open(webURL)
   }
 
}

func twitter() {
 
   let lowerCaseSocialNetworkName = "twitter"
   let socialNetworkDomain = "com"
   let userName = "bandyliuk"
   let appURL = URL(string: "\(lowerCaseSocialNetworkName)://user? screen_name=\(userName)")!
   let application = UIApplication.shared
 
   if application.canOpenURL(appURL) {
      application.open(appURL)
   } else {
      let webURL = URL(string: "https://\(lowerCaseSocialNetworkName).\ (socialNetworkDomain)/\(userName)")!
   application.open(webURL)
   }
 
}