使用迅速5的Instagram登录名，获得[“ error_type”：OAuthException，“ error_message”：无效的平台应用，“代码”：400]

Question

我正在尝试使用新文档登录instagram，并收到类似[[无效的平台应用程序，“代码”：400]之类的错误>我成功地使用

https://api.instagram.com/oauth/authorize

获得了[[code，并且在获得代码之后，我致电https://api.instagram.com/oauth/access_token以获取access_token，但它给了我错误。我正在遵循本文档：https://developers.facebook.com/docs/instagram-basic-display-api/getting-started

请帮助我解决这个问题。

提前感谢。

[我正在尝试使用新文档登录instagram，并收到诸如无效平台应用程序之类的错误，“代码”：400，我正在使用https://api.instagram.com/oauth/authorize并成功获取代码，并...] >

Answer 1

let urlString = "https://api.instagram.com/oauth/access_token"

let appendedURI = "client_id=\(API.INSTAGRAM_CLIENT_ID)&client_secret=\(API.INSTAGRAM_CLIENTSERCRET)&grant_type=authorization_code&redirect_uri=\(API.INSTAGRAM_REDIRECT_URI)&code=\(code)"

    let url = URL(string: urlString)!

    let session = URLSession.shared
    var request = URLRequest(url: url)
    request.httpMethod = "Post"
    request.setValue("application/x-www-form-urlencoded", forHTTPHeaderField:"Content-Type");
    request.cachePolicy = URLRequest.CachePolicy.reloadIgnoringCacheData
    request.httpBody = appendedURI.data(using: .utf8)

    let task = session.dataTask(with: request) { (data, response, error) in
        if error == nil {
        let dataString = NSString(data: data!, encoding: String.Encoding.utf8.rawValue)
        print("result=",dataString!)


let task = session.dataTask(with: request) { (data, response, error) in
            if error == nil {
            let dataString = NSString(data: data!, encoding: String.Encoding.utf8.rawValue)
            print("result=",dataString!)
                do {

                if let accDetail = try JSONSerialization.jsonObject(with: data!, options: []) as? [String : Any] {

                    let accessToken = accDetail["access_token"] as! String

                    let userID = accDetail["user_id"] as! Int64

                    completion("success", accessToken, userID)

                }

            }catch let error {



            }
        }

    }
    task.resume()