我得到的JSON对象如下：
{
"id": "1",
"name": "Hw",
"price": {
"value": "10"
},
{
"items": [{
"id": "1"
}]
}
}
我想将其表示为平面地图，但我想将项目数组表示为列表。
我的输出应该如下：
{
"id": "1",
"name":"Hw",
"price":"10",
"items": ["1"]
}
有人可以建议我如何实现这一目标吗？我试过这种方法：
How to deserialize JSON into flat, Map-like structure?
以上尝试过的链接输出：
{
"id": "1",
"name":"Hw",
"price.value":"10",
"items[0].id": "1"
}
但它将数组值表示为array[0]，array[1]，我不需要它。我需要这个数组作为列表。
答案 0 :(得分：0)
您提供的JSON无效。我认为是：
{
"id": "1",
"name": "Hw",
"price": {
"value": "10"
},
"items": [{
"id": "1"
}]
}
对于您所要求的内容，不可能有通用的解决方案。但对于这个特殊的JSON，这将（使用json-simple）：
@SuppressWarnings("unchecked")
public Map<String, String> transform(String inputJSON) throws ParseException {
Map<String, String> result = new LinkedHashMap<>();
JSONObject inputJSONObj = (JSONObject) new JSONParser().parse(inputJSON);
String id = inputJSONObj.getOrDefault("id", "").toString();
String name = inputJSONObj.getOrDefault("name", "").toString();
String price = ((JSONObject) inputJSONObj.getOrDefault("price", new JSONObject())).getOrDefault("value", "")
.toString();
JSONArray itemsArray = (JSONArray) inputJSONObj.getOrDefault("items", new JSONArray());
int n = itemsArray.size();
String[] itemIDs = new String[n];
for (int i = 0; i < n; i++) {
JSONObject itemObj = (JSONObject) itemsArray.get(i);
String itemId = itemObj.getOrDefault("id", "").toString();
itemIDs[i] = itemId;
}
result.put("id", id);
result.put("name", name);
result.put("price", price);
result.put("items", Arrays.toString(itemIDs));
return result;
}
答案 1 :(得分：0)
使用Gson的方法。这完全符合您的要求“将其表示为平面地图，但我希望将项目数组表示为列表”
public class ParseJson1 {
public static void main (String[] args){
Type type = new TypeToken<HashMap<String, Object>>() {
}.getType();
Gson gson = new Gson();
String json = "{\n" +
" \"id\": \"1\",\n" +
" \"name\": \"Hw\", \n" +
" \"price\": {\n" +
" \"value\": \"10\"\n" +
" },\n" +
" \"items\": [{\n" +
" \"id\": \"1\"\n" +
" }]\n" +
" }\n";
HashMap<String, Object> map = gson.fromJson(json, type);
Object val = null;
for(String key : map.keySet()){
val = map.get(key);
if(val instanceof List){
for(Object s : (List)val){
System.out.println(key + ":" + s);
}
} else
System.out.println(key + ":" + map.get(key));
}
}
}
答案 2 :(得分：0)
您必须转换Map集合Map<String, String>中的String，这将帮助您将Map数组转换为JSON格式。
JSONObject jsonObject = new JSONObject();
Map<String, String> mapObject = new HashMap<String, String>();
mapObject.put("id", "1");
mapObject.put("name", "VBage");
mapObject.put("mobile", "654321");
jsonObject.put("myJSON", mapObject);
System.out.println(jsonObject.toString());
答案 3 :(得分：-1)
首先，JSON似乎没有正确的格式。你是说这个吗？
public string insertUsers(string Name, string Password)
{
string status;
String ConStr = ConfigurationManager.ConnectionStrings["ConStr"].ConnectionString;
SqlConnection sqlCon = new SqlConnection(ConStr); // to make a connection with DB
SqlCommand sqlCom = new SqlCommand("InsertUsers", sqlCon); // now in order to perform action such as insert SP, we must create command object which needs command name and conncetion only
sqlCom.CommandType = CommandType.StoredProcedure; // you must tell the system that insertInfo is a storedprocedure
SqlParameter sqlParamName = new SqlParameter("@UserName", Name);
SqlParameter sqlParamPwd= new SqlParameter("@Password", Password);
sqlCom.Parameters.Add(sqlParamName);
sqlCom.Parameters.Add(sqlParamPwd);
try
{
sqlCon.Open();
sqlCom.ExecuteNonQuery(); // executenonquery is used for INSERT, UPDATE, DELETE
//sqlCom.ExecuteScalar(); // used to pick or read a single value from procedure
// Response.Write("Done");
sqlCon.Close();
status= "Success";
}
catch (Exception ex)
{
//response.Write(ex.Message);
status = ex.Message;
}
return status;
}
另外，由于你附加了（How to deserialize JSON into flat, Map-like structure?）的链接，我假设你想以相同的方式展平JSON，结果应该是
<?xml version="1.0" encoding="utf-8"?><soap:Envelope xmlns:soap="http://www.w3.org/2003/05/soap-envelope" xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xmlns:xsd="http://www.w3.org/2001/XMLSchema"><soap:Body><soap:Fault><soap:Code><soap:Value>soap:Receiver</soap:Value></soap:Code><soap:Reason><soap:Text xml:lang="en">System.Web.Services.Protocols.SoapException: Server was unable to process request. ---> System.Xml.XmlException: Data at the root level is invalid. Line 1, position 1.
at System.Xml.XmlTextReaderImpl.Throw(Exception e)
at System.Xml.XmlTextReaderImpl.Throw(String res, String arg)
at System.Xml.XmlTextReaderImpl.ParseRootLevelWhitespace()
at System.Xml.XmlTextReaderImpl.ParseDocumentContent()
at System.Xml.XmlTextReaderImpl.Read()
at System.Xml.XmlTextReader.Read()
at System.Web.Services.Protocols.SoapServerProtocol.SoapEnvelopeReader.Read()
at System.Xml.XmlReader.MoveToContent()
at System.Web.Services.Protocols.SoapServerProtocol.SoapEnvelopeReader.MoveToContent()
at System.Web.Services.Protocols.SoapServerProtocolHelper.GetRequestElement()
at System.Web.Services.Protocols.Soap12ServerProtocolHelper.RouteRequest()
at System.Web.Services.Protocols.SoapServerProtocol.RouteRequest(SoapServerMessage message)
at System.Web.Services.Protocols.SoapServerProtocol.Initialize()
at System.Web.Services.Protocols.ServerProtocolFactory.Create(Type type, HttpContext context, HttpRequest request, HttpResponse response, Boolean& abortProcessing)
--- End of inner exception stack trace ---</soap:Text></soap:Reason><soap:Detail /></soap:Fault></soap:Body></soap:Envelope>
另外，如果你只是想让项目返回一个id列表（即＆＃34; items＆＃34;：[＆＃34; 1＆＃34;]），那么获得JSON更合乎逻辑
{
"id": "1",
"name": "Hw",
"price": {
"value": "10"
},
"items": [{
"id": "1"
}]
}
您附加的链接（How to deserialize JSON into flat, Map-like structure?）提供了一个没有任何自定义的常规解决方案。它不应该知道＆＃34; id＆＃34;是要在项目上追加的值。
因此，我的第一个建议是将JSON 更改为{
id=1,
name=Hw,
price.value=10,
items[0]=1,
}
如果由于任何原因无法更改JSON，那么您需要执行一些自定义，如下所示：
{
"id": "1",
"name": "Hw",
"price": {
"value": "10"
},
"items": [ "1" ] // instead of "items": [{"id": "1"}]
}
尝试了解所需的格式，然后研究上述代码。它应该给你答案。