[英]Open a users twitter profile page from iOS app
我正試圖從我的應用程序深入鏈接到本機Twitter應用程序上的用戶的Twitter個人資料。 我已經為twitter添加了架構規則和以下代碼:
application.open( URL(string:"twitter://user?screen_name=BarackObama", options[:], completionHandler:{(success) in
print("Success")
})
我可以成功打開Twitter應用程序並看到控制台打印“成功”,但我自己的推文是我看到的,而不是用戶的推特頁面。 此網址架構是否仍然有效?
謝謝
好的,在Swift 4中有兩個簡單的步驟來實現這一點:
首先,您必須修改Info.plist以使用LSApplicationQueriesSchemes列出instagram和facebook。 只需打開Info.plist作為源代碼,然后粘貼:
<key>LSApplicationQueriesSchemes</key>
<array>
<string>twitter</string>
</array>
之后,您可以打開Twitter應用程序。 這是一個完整的twitter代碼,您可以將此代碼鏈接到您作為操作的任何按鈕:
@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)
}
}
將此用於Twitter配置文件共享,Swift 4:
let screenName = "NJMINISTRIESINC"
let appURL = URL(string: "twitter://user?screen_name=\(screenName)")!
let webURL = URL(string: "https://twitter.com/\(screenName)")!
if UIApplication.shared.canOpenURL(appURL as URL) {
if #available(iOS 10.0, *) {
UIApplication.shared.open(appURL)
} else {
UIApplication.shared.openURL(appURL)
}
} else {
//redirect to safari because the user doesn't have Instagram
if #available(iOS 10.0, *) {
UIApplication.shared.open(webURL)
} else {
UIApplication.shared.openURL(webURL)
}
}
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