【发布时间】：2017-01-15 08:12:12
【问题描述】：
我需要创建一个如下所示的 json 响应。我尝试了一些代码，但无法获得我需要的东西。在 java 代码中需要帮助来创建嵌套数组以根据类别以及下面 json 中的类别详细信息对食物进行分组
{
"menu": {
"items": [{
"id": 1,
"code": "hot1_sub1_mnu",
"name": "Mutton",
"status": "1",
"sub_items": [{
"id": "01",
"name": "Mutton Pepper Fry",
"price": "100"
}, {
"id": "02",
"name": "Mutton Curry",
"price": "100"
}]
},
{
"id": "2",
"code": "hot1_sub2_mnu",
"name": "Sea Food",
"status": "1",
"sub_items": [{
"id": "01",
"name": "Fish Fry",
"price": "150"
}]
},
{
"id": "3",
"code": "hot1_sub3_mnu",
"name": "Noodles",
"status": "1",
"sub_items": [{
"id": "01",
"name": "Chicken Noodles",
"price": "70"
}, {
"id": "02",
"name": "Egg Noodles",
"price": "60"
}]
}
]
}
}
到目前为止，我尝试过的将在一个数组中给出响应。
@Path("/items")
public class HotelsMenu {
@GET
@Path("/getitems")
@Produces(MediaType.APPLICATION_JSON)
public String doLogin(@QueryParam("hotelcode") String hotelcode) {
JSONObject response = new JSONObject();
JSONArray hotelDetails = checkCredentials(hotelcode);
try {
response.put("hotels", hotelDetails);
response.put("status", (hotelDetails == new JSONArray()) ? "false" : "true");
} catch (JSONException e) {
// TODO Auto-generated catch block
e.printStackTrace();
}
return response.toString();
}
private JSONArray checkCredentials(String hotelcode) {
System.out.println("Inside checkCredentials");
JSONArray result = new JSONArray();
try {
result = DBConnection.checkItems(hotelcode);
} catch (Exception e) {
// TODO Auto-generated catch block
e.printStackTrace();
}
return result;
}
public static JSONArray checkItems(String hotelcode) throws Exception {
int id;
String code = hotelcode + "_mnu";
String name = null;
String name1 = null;
String status;
String price;
Connection dbConn = null;
Connection dbConn1 = null;
JSONArray hotels = new JSONArray();
JSONArray menu = new JSONArray();
try {
try {
dbConn = DBConnection.createConnection();
} catch (Exception e) {
// TODO Auto-generated catch block
e.printStackTrace();
}
Statement stmt = dbConn.createStatement();
String query = "SELECT * FROM " + code + " where Status='1'";
System.out.println(query);
ResultSet rs1 = stmt.executeQuery(query);
System.out.println("hai");
while (rs1.next()) {
JSONObject hotel = new JSONObject();
id = rs1.getInt("Id");
hotel.put("id", id);
code = rs1.getString("Code");
System.out.println(code);
hotel.put("code", code);
name = rs1.getString("Name");
hotel.put("name", name);
status = rs1.getString("Status");
hotel.put("status", status);
hotels.put(hotel);
System.out.println("Hotel1:" + hotels);
try {
dbConn1 = DBConnection.createConnection();
} catch (Exception e) {
// TODO Auto-generated catch block
e.printStackTrace();
}
Statement stmt1 = dbConn1.createStatement();
String query1 = "SELECT * FROM " + code + " where Status='1' ";
System.out.println(query1);
ResultSet rs2 = stmt1.executeQuery(query1);
while (rs2.next()) {
JSONArray hotel1 = new JSONArray();
JSONObject hotelmenu = new JSONObject();
id = rs2.getInt("Id");
hotelmenu.put("id", id);
name1 = rs2.getString("Name");
hotelmenu.put("name", name1);
price = rs2.getString("Price");
hotelmenu.put("price", price);
hotel1.put(hotelmenu);
hotels.put(hotel1);
System.out.println(hotels);
}
}
} catch (SQLException sqle) {
throw sqle;
} catch (Exception e) {
// TODO Auto-generated catch block
if (dbConn != null) {
dbConn.close();
}
throw e;
} finally {
if (dbConn != null) {
dbConn.close();
}
}
return hotels;
}
}
【问题讨论】：
-
您想要的 JSON 结构是无效的 JSON。请发布有效的 JSON，否则很难理解您真正想要什么。我的建议：不要在同一个结构中混合数组和对象。设计您的对象模型，并将其映射到 JSON。在 Java 中，您不会使用包含混合类型值的 List。对 JSON 做同样的事情。