Swift : 通过 Twitter 分享文字

Question

所以基本上我正在制作一个活动应用程序。一切进展顺利，但只是将活动分享到推特。

我已经在互联网上进行了搜索，但我得到的只是使用我不想要的 twitter 的本机应用程序。我想用浏览器发微博

这个方法我已经实现了FB分享

任何想法都会对我有很大帮助。

let content = FBSDKShareLinkContent()

    content.contentURL=NSURL(string: "http://facebook.com")
    content.imageURL = NSURL(string: "http://facebook.com")

    content.contentTitle = "Shou 3emlin test app "
    content.contentDescription = "testing testing testing"

    let shareDialog = FBSDKShareDialog()
    shareDialog.fromViewController = self
    shareDialog.mode=FBSDKShareDialogMode.Browser
    shareDialog.shareContent = content


    if !shareDialog.canShow() {
        shareDialog.mode=FBSDKShareDialogMode.Native
        shareDialog.shareContent = content
    }

    if shareDialog.canShow() {
        shareDialog.show()
    }

Answer 1

看看Fabric.io。此 SDK 允许您直接从您的应用撰写推文。

let composer = TWTRComposer()

composer.setText("just setting up my Fabric")
composer.setImage(UIImage(named: "fabric"))

// Called from a UIViewController
composer.showFromViewController(self) { result in
    if (result == TWTRComposerResult.Cancelled) {
         print("Tweet composition cancelled")
    }
    else {
       print("Sending tweet!")
    }
}

Answer 2

将其放在按钮的操作方法中或要使用浏览器发送文本的方法中 Swift 3.0:

let tweetText = "your text"
let tweetUrl = "http://whosebug.com/"

let shareString = "https://twitter.com/intent/tweet?text=\(tweetText)&url=\(tweetUrl)"

// encode a space to %20 for example
let escapedShareString = shareString.addingPercentEncoding(withAllowedCharacters: CharacterSet.urlQueryAllowed)!

// cast to an url
let url = URL(string: escapedShareString)

// open in safari
UIApplication.shared.openURL(url!)

结果:

Answer 3

let tweetText = "hy" 
let tweetUrl = "http://rimmi/" 

let shareString = "https://twitter.com/intent/tweet?text=\(tweetText)&url=\(tweetUrl)"

// encode a space to %20 for example 
let escapedShareString = shareString.addingPercentEncoding(withAllowedCharacters: CharacterSet.urlQueryAllowed)!

// cast to an url 
let url = URL(string: escapedShareString)

// open in safari 
UIApplication.shared.openURL(url!)

Answer 4

@ronatory 的解决方案非常有效。如果用户设备上已经安装了 Twitter 应用程序，它还会打开它。

对于 swift 5+，请使用 UIApplication.shared.open(url!) 而不是 UIApplication.shared.openURL(url!)，因为它已被弃用。

Swift : 通过 Twitter 分享文字

Swift : Share text through Twitter

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