2017-07-30 87 views
-2
let params = "name=Thobio Joseph&googleid=24356567890uyy4546&[email protected]&avatar=https://media.licdn.com/mpr/mpr/shrinknp_200_200/AAEAAQAAAAAAAAv0AAAAJDZjZGJjMTFjLWNiNzAtNGYzNy1iMDE4LTA2MzBmNzUwZGExNQ.jpg" 

func postMethodUploadDataToServerLoginPage() { 

Alamofire.request(loginUrl, method:.post,parameters:params.data(using: String.Encoding.utf8),encoding:URLEncoding.default).responseJSON { response in 

switch response.result { 

case .success: 
       print(response) 

case .failure(let error): 
       failure(0,"Error") 

} 

} 
+1

你能解释一下你正在尝试做的,这个问题你可能会面对的? – Nisarg

回答

0

首先你PARAMS是错误的,你需要传递一个字典[key:value],然后你需要将其转换成JSON,并把你的请求主体,也如果您正在使用Alamofire.request你不”吨需要通过urlString,只有一个请求,这个代码尝试

func postMethodUploadDataToServerLoginPage() { 

let paramToSend = ["name":"Thobio Joseph","googleid":"24356567890uyy4546","email":"[email protected]","avatar":"https://media.licdn.com/mpr/mpr/shrinknp_200_200/AAEAAQAAAAAAAAv0AAAAJDZjZGJjMTFjLWNiNzAtNGYzNy1iMDE4LTA2MzBmNzUwZGExNQ.jpg"] 

let request = NSMutableURLRequest(url: URL(string: loginUrl)!) 
request.httpMethod = "POST" 
request.setValue("application/json", forHTTPHeaderField: "Content-Type") 
request.httpBody = try! JSONSerialization.data(withJSONObject: parameters) 

Alamofire.request(request).responseJSON { (response) in 

switch response.result { 

case .success: 
       print(response) 

case .failure(let error): 
       failure(0,"Error") 

} 

} 

希望这有助于

+0

如果php post方法如$ _POST ['name'],$ _ post ['email']会不会需要json –

+0

@ThobioJoseph你试过我的答案?,必须工作,试试并让我知道 –

+0

yes我尝试我得到它非常感谢你:) –

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