傳遞參數在迅速

Question

我做什麼用AFNetworking： -

1.I想用AFHTTPSessionManager

2.I調用服務想通過只在JSON格式參數使用POST方法 和我也試過通過參數JSON但我收到錯誤的請求

3，採用AFNetworking 3.0

簡稱鏈路Convert Dictionary to JSON in Swift

我的代碼： -

 dict = ["email_id":user! , "password": pass!] 

    do { 
       jsonData = try NSJSONSerialization.dataWithJSONObject(dict, options: NSJSONWritingOptions.PrettyPrinted) 
       // here "jsonData" is the dictionary encoded in JSON data 
       theJSONText = String(data: jsonData!, 
              encoding: NSASCIIStringEncoding) 
       print("json string = \(theJSONText)") 


      } catch let error as NSError { 
       print(error) 
      } 



    manager.POST(urlString, parameters: theJSONText as String!, success: 
       { 
        requestOperation, response in 

        let result = NSString(data: response as! NSData, encoding: NSUTF8StringEncoding)! 

        print(result) 
       }, 
         failure: 
       { 
        requestOperation, error in 
        //let result = NSString(data: error as! NSData, encoding: NSUTF8StringEncoding)! 

        print(error) 

      }) 

我想DO： - 我想將參數作爲JSON格式傳遞

 Printing description of theJSONText: 
"{\n \"email_id\" : \"bbh\",\n \"password\" : \"vh\"\n}" 

轉換成JSON字符串也經過我不是JSON格式越來越

請幫我做這個..

Answer 1

刪除下面從代碼行，

theJSONText = String(data: jsonData!, 
             encoding: NSASCIIStringEncoding) 
      print("json string = \(theJSONText)") 

設置選項0而不是在json序列化中的NSJSONWritingOptions.PrettyPrinted

並通過jsonData（作爲數據）作爲參數而不是theJSONText。

更新：

嘗試這樣，

let urlString = "http://example.com/file.php" 
let dictionary = ["key1": [1,2,3], "key2": [2,4,6]] 

var error: NSError? 
let data = NSJSONSerialization.dataWithJSONObject(dictionary, options: NSJSONWritingOptions.allZeros, error: &error) 
let jsonString = NSString(data: data!, encoding: NSUTF8StringEncoding) 
let parameters = ["data" : jsonString!] 

let manager = AFHTTPSessionManager() 
manager.responseSerializer = AFHTTPResponseSerializer() 
manager.POST(urlString, parameters: parameters, success: 
{ 
    requestOperation, response in 

    let result = NSString(data: response as! NSData, encoding: NSUTF8StringEncoding)! 

    println(result) 
}, 
failure: 
{ 
    requestOperation, error in 
}) 

在這個例子中的響應

是字符串，如果響應JSON格式，那麼你應該在JSON使用nsjsonserialization，而不是字符串

序列化

希望這會有所幫助:)