使用JSON響應從Java調用c＃webservice

Question

在我們的編程環境中，我們同時擁有Java和C＃開發人員。我有一個在C＃中創建的Web開發人員正在嘗試使用的Web服務。我一直在編寫Java來使用這個Web服務，而當我得到一個JSON結果時，它的格式是錯誤的。

以下是我對C＃的一面：

[WebMethod] 
public static LinkedList<string> GetInfo(string InfoID, string Username, string Password) 
{ 
    LinkedList<string> Result = new LinkedList<string>(); 
    try 
    { 
     // Do some stuff I can't show you to get the information... 
     foreach (Result from data operations) 
     { 
      Result.AddLast(sample1); 
      Result.AddLast(sample2); 
      Result.AddLast(sample3); 
      Result.AddLast(BD)); 
      Result.AddLast(CN); 
      Result.AddLast(Name); 
      Result.AddLast("###"); 
     } 
    }catch(Exception exc) 
    { 
     Result.AddLast(exc.ToString()); 
     return Result; 
    }    
    return Result; 
} 

那麼這就是Java方面：

try { 
    String uri = "http://example.com/service.asmx/GetInfo"; 

    URL url = new URL(uri); 
    HttpURLConnection connection = (HttpURLConnection) url.openConnection(); 

    // Setup Connection Properties 
    connection.setRequestMethod("POST"); 
    connection.setDoInput(true); 
    connection.setDoOutput(true); 
    connection.setRequestProperty("Content-Type", "application/json"); 
    connection.setRequestProperty("charset", "utf-8"); 
    connection.setRequestProperty("Accept", "application/json");    
    connection.setChunkedStreamingMode(0); 
    connection.connect(); 

    // Create the JSON Going out 
    byte[] parameters = "{'InfoID':'123456789','Username':'usernametoken','Password':'passwordtoken'}".getBytes("UTF-8"); 


    // Start doing stuff     
    DataOutputStream os = new DataOutputStream(connection.getOutputStream()); 
    os.write(parameters); 
    os.close();   
    InputStream response;     

    // Check for error , if none store response 
    if(connection.getResponseCode() == 200){response = connection.getInputStream();} 
    else{response = connection.getErrorStream();} 

    InputStreamReader isr = new InputStreamReader(response); 
    StringBuilder sb = new StringBuilder(); 
    BufferedReader br = new BufferedReader(isr); 
    String read = br.readLine(); 

    while(read != null){ 
     sb.append(read); 
     read = br.readLine(); 
    } 
    // Print the String  
    System.out.println(sb.toString()); 

    // Creat JSON off of String 
    JSONObject token = new JSONObject(sb.toString()); 

    // print JSON 
    System.out.println("Tokener: " + token.toString()); 
    response.close(); 

} catch(IOException exc) { 
    System.out.println("There was an error creating the HTTP Call: " + exc.toString()); 
} 

而我得到的迴應是這種形式...

{"d":["Sample1","Sample2","Sample3","BD","CN","Name","###","Sample1","Sample2","Sample3","BD","CN","Name","###","Sample1","Sample2","Sample3","BD","CN","Name","###"]} 

我想知道是否有更好的方式發送響應，使JSON看起來像這樣：

{"1":["Sample1","Sample2","Sample3","BD","CN","Name","###"],"2":["Sample1","Sample2","Sample3","BD","CN","Name","###"],"3":["Sample1","Sample2","Sample3","BD","CN","Name","###"],"4":["Sample1","Sample2","Sample3","BD","CN","Name","###"]} 

Answer 1

好的我想我在這裏看到你的問題。你希望自己的數據被序列化爲

{"1":["Sample1","Sample2","Sample3","BD","CN","Name","###"],"2":["Sample1","Sample2","Sample3","BD","CN","Name","###"],"3":["Sample1","Sample2","Sample3","BD","CN","Name","###"] ... etc 

然而，你正在序列化的數據結構是單鏈表，這就是爲什麼它被序列化爲一個長長的名單。你需要做的是改變數據結構。 A Dictionary將是完美的，因爲它很容易作爲JSON序列化。

[ScriptMethod(ResponseFormat = ResponseFormat.Json)] 
[WebMethod] 
public static Dictionary<int,LinkedList<string>> GetInfo(string InfoID, string Username, string Password) 
{ 
    var Result = new Dictionary<int,LinkedList<string>>(); 
    try 
    { 
     // Do some stuff I can't show you to get the information... 

     foreach (Result from data operations) 
     { 
      var newList = new LinkedList<string>();  
      newList.AddLast(sample1); 
      newList.AddLast(sample2); 
      newList.AddLast(sample3); 
      newList.AddLast(BD)); 
      newList.AddLast(CN); 
      newList.AddLast(Name); 
      newList.AddLast("###"); 
      int number = something //the number before the list 
      Result.add(number, newList); 
     } 
    }catch(Exception exc) 
    { 
     . 
     . 
     . 
    }    
    return Result; 
}