从iOS应用程序打开用户的Twitter个人资料页面

问题描述：

我正在尝试从我的应用程序到本机Twitter应用程序上用户的Twitter个人资料的深层链接.我已经为twitter和以下代码添加了架构规则:

I am trying to deep link from my app to a user's twitter profile on the native twitter app. I have added schema rules for twitter and the following code:

    application.open(  URL(string:"twitter://user?screen_name=BarackObama", options[:],  completionHandler:{(success) in 
        print("Success")
    })

我可以成功打开twitter应用程序并看到控制台打印成功"，但是我看到的是我自己的twitter feed，而不是用户的twitter页面.这个网址架构仍然有效吗?

I can successfully open the twitter app and see the console print "Success" but my own twitter feed is what I see, not the user's twitter page. Is this url schema still valid?

谢谢

答

好的，有两个简单的步骤可以在Swift 4中实现:

OK, there are two easy steps to achieve this in Swift 4:

首先，您必须修改Info.plist以使用LSApplicationQueriesSchemes列出instagram和facebook.只需打开Info.plist作为源代码，然后将其粘贴:

First, you have to modify Info.plist to list instagram and facebook with LSApplicationQueriesSchemes. Simply open Info.plist as a Source Code, and paste this:

<key>LSApplicationQueriesSchemes</key>
<array>
    <string>twitter</string>
</array>

之后，您可以打开Twitter应用程序.这是Twitter的完整代码，您可以将此代码链接到操作中具有的任何按钮:

After that, you can open twitter apps. Here is a complete code for twitter you can link this code to any button you have as an Action:

@IBAction func followOnTwitter(sender: AnyObject) {
   let screenName =  "AffordIt_App"
   let appURL = NSURL(string: "twitter://user?screen_name=\(screenName)")!
   let webURL = NSURL(string: "https://twitter.com/\(screenName)")!

   let application = UIApplication.shared

   if application.canOpenURL(appURL as URL) {
        application.open(appURL as URL)
   } else {
        application.open(webURL as URL)
   }
}

从iOS应用程序打开用户的Twitter个人资料页面

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