我正在使用FBSDKLoginKit和FBSDKShareKit进行登录,并共享了FBSDKShareOpenGraphContent之后。
我在android中有类似的行为,并且工作正常。
我能够迅速在Facebook中成功登录,但是当我调用FBSDKShareDialog.show(来自:self,with:content,委托:nil)时,可以从详细视图控制器,应用程序中共享我的FBSDKShareOpenGraphContent,以显示登录脸书页面再次在浏览器中! 我希望直接共享FBSDKShareOpenGraphContent,而无需再次登录!
如果FBSDKAccessToken.currentAccessTokenIsActive()返回true,为什么会出现这种情况? 如何在不同的ViewController之间保持相同的会话? 我读了十亿篇有关该机器人无法解决我问题的文章。
请注意,如果我键入用户名和密码并第二次重新登录,则FBSDKShareOpenGraphContent将正确共享! 但是这是糟糕的用户体验,我希望用户只能登录1次!
请帮助 这是我的代码:
//APPDELEGATE
func application(_ application: UIApplication, didFinishLaunchingWithOptions launchOptions: [UIApplicationLaunchOptionsKey: Any]?) -> Bool {
// Override point for customization after application launch.
FBSDKApplicationDelegate.sharedInstance().application(application, didFinishLaunchingWithOptions: launchOptions)
return true
}
private func application(application: UIApplication, didFinishLaunchingWithOptions launchOptions: [NSObject: AnyObject]?) -> Bool {
// Override point for customization after application launch.
return FBSDKApplicationDelegate.sharedInstance().application(application, didFinishLaunchingWithOptions: launchOptions)
}
private func application(application: UIApplication, openURL url: URL, sourceApplication: String?, annotation: Any?) -> Bool {
return FBSDKApplicationDelegate.sharedInstance().application(
application,
open: url as URL?,
sourceApplication: sourceApplication,
annotation: annotation)
}
func applicationDidBecomeActive(_ application: UIApplication) {
// Restart any tasks that were paused (or not yet started) while the application was inactive. If the application was previously in the background, optionally refresh the user interface.
FBSDKAppEvents.activateApp()
}
----
//登录VIEWCONTROLLER @IBAction
@IBAction func fbLoginButtonTapped(_ sender: UIButton) {
let fbLoginManager : FBSDKLoginManager = FBSDKLoginManager()
fbLoginManager.logIn(withReadPermissions: ["public_profile"], from: self) { (result, error) -> Void in
if (error == nil) {
let fbloginresult : FBSDKLoginManagerLoginResult = result!
// if user cancel the login
if (result?.isCancelled)! {
print("LOGIN FB CANCELLED")
return
}
if(fbloginresult.grantedPermissions.contains("public_profile")) {
print("LOGIN FB SUCCESS")
/* DO MY STUFF */
}
} else {
print("LOGIN FB ERROR")
}
}
}
//Page Detail ViewController with share button @IBAction
@IBAction func facebookButtonTapped(_ sender: UIButton) {
if (FBSDKAccessToken.currentAccessTokenIsActive() == true) {
//THIS RETURNS TRUE AS USER IS ALWAYS LOGGED
let properties = ["og:type": "article",
"og:title": "\(self.post?.title ?? "")",
"og:image": "\(self.post?.image ?? "")",
"og:description": "\(self.post?.content ?? "")"]
let object: FBSDKShareOpenGraphObject = FBSDKShareOpenGraphObject(properties: properties)
let action: FBSDKShareOpenGraphAction = FBSDKShareOpenGraphAction(type: "news.reads", object: object, key: "article")
let content: FBSDKShareOpenGraphContent = FBSDKShareOpenGraphContent()
content.action = action
content.previewPropertyName = "article"
FBSDKShareDialog.show(from: self, with: content, delegate: nil)
}
}
用于Facebook集成的Info.plist
<!-- FBSDK INTEGRATION -->
<key>CFBundleURLTypes</key>
<array>
<dict>
<key>CFBundleURLSchemes</key>
<array>
<string>fb************</string>
</array>
</dict>
</array>
<key>FacebookAppID</key>
<string>**************</string>
<key>FacebookDisplayName</key>
<string>*********</string>
<key>LSApplicationQueriesSchemes</key>
<array>
<string>fbapi</string>
<string>fb-messenger-api</string>
<string>fbauth2</string>
<string>fbshareextension</string>
</array>
First Login page shown after login button tapped:
Login page again on share button tapped, after first login success
答案 0 :(得分:1)
我用以下代码解决了这个问题:
let shareDialog = FBSDKShareDialog()
shareDialog.delegate = self
shareDialog.shareContent = content
if (UIApplication.shared.canOpenURL( URL(string: "fbauth2:/")!) ) {
shareDialog.mode = .native
} else {
shareDialog.mode = .web
}
shareDialog.show()
以前我用过
shareDialog.mode = .auto
但是,如果这无法选择适当的操作或由于某种原因而失败,则似乎重新实例化了Web浏览器,而没有保持活动会话。
使用.native和.web模式,我可以处理是否安装本机应用程序的情况,并保持活动会话状态。