使用 JSON 响应从 Java 调用 c# Web 服务
本文关键字:Web 服务 调用 Java JSON 响应 使用 | 更新日期: 2023-09-27 18:32:43
在我们的编程环境中,我们同时拥有Java和C#开发人员。我有一个用 C# 创建的 Web 服务,Java 开发人员正在尝试使用它。我一直在编写 Java 来使用此 Web 服务,当我得到一个 json 结果时,它的格式是错误的。
这是我在 c# 方面所拥有的:
[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;
}
那么这是Java端:
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());
}
我得到的回应是这种形式...
{"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法来发送响应,使 JSON 如下所示:
{"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","###"]}
好的,
我想我在这里看到了你的问题。您希望将数据序列化为
{"1":["Sample1","Sample2","Sample3","BD","CN","Name","###"],"2":["Sample1","Sample2","Sample3","BD","CN","Name","###"],"3":["Sample1","Sample2","Sample3","BD","CN","Name","###"] ... etc
然而,您正在序列化的数据结构是单个链表,这就是将其序列化为单个长列表的原因。您需要做的是更改数据结构。Dictionary
将是完美的,因为它很容易序列化为 JSON。
[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;
}