ios Instagram api POST 使用 Alamofire 的关系端点 - 错误
ios Instagram api POST Relationship Endpoints using Alamofire - Error
我在我的应用程序中使用 swift
我正在尝试使用 Alamofire post 与范围 ("follower_list+relationships") 建立关系 (Follow)
注意:我正在使用沙盒用户进行测试
使用此代码
let userid = "testAccount"
let urlString = "https://api.instagram.com/v1/users/\(userid)/relationship?access_token=\(INSTAGRAM_IDS.INSTAGRAM_Token)&action=follow"
Alamofire.request(urlString).responseJSON { response in switch response.result {
case .success(let JSON):
let response = JSON as! NSDictionary
let userModel = response
print("SUCCESS")
print(userModel)
case .failure(let error):
print("ERROR")
print(error)
}
}
也试过这个
let userid = "testAccount"
let urlString = "https://api.instagram.com/v1/users/\(userid)/relationship?access_token=\(INSTAGRAM_IDS.INSTAGRAM_Token)"
let parameters: Parameters = ["action": "follow"]
Alamofire.request(urlString, method: .post, parameters: parameters, encoding: URLEncoding.default, headers: nil).responseJSON { response in switch response.result {
case .success(let JSON):
let response = JSON as! NSDictionary
let userModel = response
print("SUCCESS")
print(userModel)
case .failure(let error):
print("ERROR")
print(error)
}
}
但仍然收到这条消息 ??
ERROR
responseSerializationFailed(Alamofire.AFError.ResponseSerializationFailureReason.jsonSerializationFailed(Error Domain=NSCocoaErrorDomain Code=3840 "Invalid value around character 0." UserInfo={NSDebugDescription=Invalid value around character 0.}))
有什么想法吗??
如果您尝试 post json 我怀疑您需要使用 encoding: JSONEncoding.default
更改编码
The JSONEncoding type creates a JSON representation of the parameters object, which is set as the HTTP body of the request. The Content-Type HTTP header field of an encoded request is set to application/json.
上的使用
我发现了问题....我一直在使用帐户用户名而不是用户 ID
有区别...我不好
您可以从(搜索)端点获取用户 ID api 或者使用像这样的网站
https://smashballoon.com/instagram-feed/find-instagram-user-id/
你输入用户名你会得到ID
我在我的应用程序中使用 swift
我正在尝试使用 Alamofire post 与范围 ("follower_list+relationships") 建立关系 (Follow)
注意:我正在使用沙盒用户进行测试
使用此代码
let userid = "testAccount"
let urlString = "https://api.instagram.com/v1/users/\(userid)/relationship?access_token=\(INSTAGRAM_IDS.INSTAGRAM_Token)&action=follow"
Alamofire.request(urlString).responseJSON { response in switch response.result {
case .success(let JSON):
let response = JSON as! NSDictionary
let userModel = response
print("SUCCESS")
print(userModel)
case .failure(let error):
print("ERROR")
print(error)
}
}
也试过这个
let userid = "testAccount"
let urlString = "https://api.instagram.com/v1/users/\(userid)/relationship?access_token=\(INSTAGRAM_IDS.INSTAGRAM_Token)"
let parameters: Parameters = ["action": "follow"]
Alamofire.request(urlString, method: .post, parameters: parameters, encoding: URLEncoding.default, headers: nil).responseJSON { response in switch response.result {
case .success(let JSON):
let response = JSON as! NSDictionary
let userModel = response
print("SUCCESS")
print(userModel)
case .failure(let error):
print("ERROR")
print(error)
}
}
但仍然收到这条消息 ??
ERROR responseSerializationFailed(Alamofire.AFError.ResponseSerializationFailureReason.jsonSerializationFailed(Error Domain=NSCocoaErrorDomain Code=3840 "Invalid value around character 0." UserInfo={NSDebugDescription=Invalid value around character 0.}))
有什么想法吗??
如果您尝试 post json 我怀疑您需要使用 encoding: JSONEncoding.default
上的使用The JSONEncoding type creates a JSON representation of the parameters object, which is set as the HTTP body of the request. The Content-Type HTTP header field of an encoded request is set to application/json.
我发现了问题....我一直在使用帐户用户名而不是用户 ID 有区别...我不好
您可以从(搜索)端点获取用户 ID api 或者使用像这样的网站 https://smashballoon.com/instagram-feed/find-instagram-user-id/
你输入用户名你会得到ID