收藏
回答

ios使用微信登录sdk可以拉起微信但是闪一下就退了,无法正确获取授权页面该怎么办?

环境描述:系统IOS 12.4 ,微信版本 7.0.8, SDK 1.8.6,

错误日志:

Error:set token fail, errCode:4, errLog:wx token[] or contextId[(null)] is nil

状态描述:可以看到有跳转微信的过程,但是不能正确获取微信授权界面。但是又能拿到相应的返回值,很奇怪,有人遇到过嘛?

相关代码:

    

//三个重写的API

func application(_ app: UIApplication, open url: URL, options: [UIApplication.OpenURLOptionsKey : Any] = [:]) -> Bool {

        returnWXApi.handleOpen(url, delegate: self)

    

    

//    func application(_ application: UIApplication, open url: URL, sourceApplication: String?, annotation: Any) -> Bool {

//        return WXApi.handleOpen(url, delegate: self)

//    }

    

func application(_application: UIApplication, handleOpen url: URL) -> Bool{

        returnWXApi.handleOpen(url, delegate: self)

    

    

func application(_ application: UIApplication, continueuserActivity: NSUserActivity, restorationHandler: @escaping ([UIUserActivityRestoring]?) -> Void) -> Bool {

        logger.info("调用useractivity"

        returnWXApi.handleOpenUniversalLink(userActivity, delegate: self)

    


//回调 

func onResp(_ resp: BaseResp) {

        logger.info(String(format: "返回的resp是:%p", resp))        

        //微信登录

        ifresp is SendAuthResp {

            logger.info("微信登录接口"

//            switch(resp.errCode){

//            case 0:

//                let response = resp as! SendAuthResp

//                NotificationCenter.default.post(name: NSNotification.Name(rawValue: "WXLoginSuccessNotification"), object: response.code)

//                break

//            case -4: //用户拒绝

//                NotificationCenter.default.post(name: NSNotification.Name(rawValue: "WXLoginDenyNotification"), object: nil)

//                break

//            case -2: //用户取消

//                NotificationCenter.default.post(name: NSNotification.Name(rawValue: "WXLoginCancelNotification"), object: nil)

//                break

//            default:

//                break

//            }

            ifresp.errCode == 0 && resp.type == 0 {//授权成功

                let response = resp as! SendAuthResp

                NotificationCenter.default.post(name: NSNotification.Name(rawValue: "WXLoginSuccessNotification"), object: response.code)

            // end of internal if

            return

        // end of external if

    // end of onResp


// send

@IBAction func wechatLogin(_ sender: LGButton) {

    sender.isLoading = true

    ifWXApi.isWXAppInstalled() {

        let app = UIApplication.shared

        ifapp.canOpenURL(URL.init(string: WEIXIN_SCHEME)!) {

            let req = SendAuthReq.init()

            req.scope = "snsapi_message,snsapi_userinfo,snsapi_friend,snsapi_contact"

            req.state = "\(arc4random()%100)"

            logger.info("准备开启微信登录!"

            let deadlineTime = DispatchTime.now() + .seconds(1)

            DispatchQueue.main.asyncAfter(deadline: deadlineTime) {

                //停止转圈圈

               WXApi.send(req,  completion: { (success) in

               sender.isLoading = false

               logger.info("成功获取微信app的授权,开始向微信服务器发起请求!"

               })}

        } else

            if#available(iOS 10.0, *) {

                UIApplication.shared.open(URL.init(string: "http://weixin.qq.com/r/qUQVDfDEVK0rrbRu9xG7")!, options: [:], completionHandler: nil)

            } else

                // Fallback on earlier versions

                UIApplication.shared.openURL(URL.init(string: "http://weixin.qq.com/r/qUQVDfDEVK0rrbRu9xG7")!)

            

        

   else

        self.sendmsgToWxServer()

        sender.isLoading = false

    

    



// 登录成功之后

@objc func WXLoginSuccess(notification:Notification) {

        logger.info("WXLoginSuccess开始执行"

        let code = notification.object as! String

        let requestUrl = "https://api.weixin.qq.com/sns/oauth2/access_token?appid=\(WX_APPID)&secret=\(WX_APPSecret)&code=\(code)&grant_type=authorization_code"

        

        

        DispatchQueue.global().async {

            let requestURL: URL = URL.init(string: requestUrl)!

            let data = try? Data.init(contentsOf: requestURL, options: Data.ReadingOptions())

            

            //TODO: 从微信获取到的用户信息,然后再回传到app的后台服务器

            DispatchQueue.main.async {

                let jsonResult = try! JSONSerialization.jsonObject(with: data!, options: JSONSerialization.ReadingOptions.mutableContainers) as! Dictionary<String,Any>

                logger.info("微信app传回来的数据: \(jsonResult)"

                let openid: String = jsonResult["openid"] as! String

                let access_token: String = jsonResult["access_token"] as! String

                let unionid: String = jsonResult["unionid"] as! String

                

                self.openid = openid

                

                //向app服务器保存用户的openid和access_token

                let parameters: Parameters = ["openId": openid, "accessToken": access_token]

                AF.request(SAVE_USER_INFO_URL_STR, parameters: parameters).validate()

                

                let homeVcUrlStr = String(format: "%@%@", arguments: [HOMEPAGE_LOGINED_URL_STR, unionid])

                

                logger.info(String(format:"homeStr: %@, openid: %@", arguments:[homeVcUrlStr, openid]))           

     

                

                let mainNavigationController = self.presentingViewController as! LFMainNavigationController

                let homePageController = mainNavigationController.viewControllers[0] as! LFHomeViewController

                

                homePageController.openid = self.openid

                homePageController.homePageUrlStr = homeVcUrlStr

                

                self.dismiss(animated: true, completion: {

                }) // end of dismiss

                

                //暂时不需要从微信服务器获取用户的信息

//                self.getUserInfo(openid: openid, access_token: access_token)

            

        

    


有人能帮忙看看嘛? 或者有人遇到相似的问题嘛 ? 求交流 !!!

网页回复不及时可加微信:dzjMichael    ( 加时注明来源  不甚感激!!!


————————————————————————————

2020 - 01 -07更新

问题已经解决,发现是UL配置的问题,建议各位老铁多多检查自己的配置吧,从scheme到ul,然后再试自己的代理函数,基本上照着文档来一般不会有问题。


最后一次编辑于  2020-01-07
回答关注问题邀请回答
收藏

2 个回答

  • 深浅色
    深浅色
    2019-11-13

    这个问题我也遇到了,你的问题解决了吗?我的是每次拉起微信之后微信弹窗提示bundleid验证不通过,但是可以获取到信息。

    2019-11-13
    有用
    回复 1
    • 随便起个名
      随便起个名
      2019-11-27
      请问你解决了吗,怎么解决的
      2019-11-27
      回复
  • paulwley
    paulwley
    2019-10-29

    相应的  universal link 也都配置完成,Safari中输入域名可以看到相关跳转, 以及info.list中相关的值也都已设置  但还是有问题

    2019-10-29
    有用
    回复
登录 后发表内容
问题标签