I've a problem trying to make my page printing out the JSONObject in the order i want. In my code, I entered this:
JSONObject myObject = new JSONObject();
myObject.put("userid", "User 1");
myObject.put("amount", "24.23");
myObject.put("success", "NO");
However, when I see the display on my page, it gives:
JSON formatted string: [{"success":"NO", "userid":"User 1", "bid":24.23}]
I need it in the order of userid, amount, then success. Already tried re-ordering in the code, but to no avail. I've also tried .append....need some help here thanks!!
You cannot and should not rely on the ordering of elements within a JSON object.
From the JSON specification at https://www.json.org/
An object is an unordered set of
name/value pairs
As a consequence,
JSON libraries are free to rearrange the order of the elements as they see fit.
This is not a bug.
I agree with the other answers. You cannot rely on the ordering of JSON elements.
However if we need to have an ordered JSON, one solution might be to prepare a LinkedHashMap object with elements and convert it to JSONObject.
#Test
def void testOrdered() {
Map obj = new LinkedHashMap()
obj.put("a", "foo1")
obj.put("b", new Integer(100))
obj.put("c", new Double(1000.21))
obj.put("d", new Boolean(true))
obj.put("e", "foo2")
obj.put("f", "foo3")
obj.put("g", "foo4")
obj.put("h", "foo5")
obj.put("x", null)
JSONObject json = (JSONObject) obj
logger.info("Ordered Json : %s", json.toString())
String expectedJsonString = """{"a":"foo1","b":100,"c":1000.21,"d":true,"e":"foo2","f":"foo3","g":"foo4","h":"foo5"}"""
assertEquals(expectedJsonString, json.toString())
JSONAssert.assertEquals(JSONSerializer.toJSON(expectedJsonString), json)
}
Normally the order is not preserved as below.
#Test
def void testUnordered() {
Map obj = new HashMap()
obj.put("a", "foo1")
obj.put("b", new Integer(100))
obj.put("c", new Double(1000.21))
obj.put("d", new Boolean(true))
obj.put("e", "foo2")
obj.put("f", "foo3")
obj.put("g", "foo4")
obj.put("h", "foo5")
obj.put("x", null)
JSONObject json = (JSONObject) obj
logger.info("Unordered Json : %s", json.toString(3, 3))
String unexpectedJsonString = """{"a":"foo1","b":100,"c":1000.21,"d":true,"e":"foo2","f":"foo3","g":"foo4","h":"foo5"}"""
// string representation of json objects are different
assertFalse(unexpectedJsonString.equals(json.toString()))
// json objects are equal
JSONAssert.assertEquals(JSONSerializer.toJSON(unexpectedJsonString), json)
}
You may check my post too: http://www.flyingtomoon.com/2011/04/preserving-order-in-json.html
u can retain the order, if u use JsonObject that belongs to com.google.gson :D
JsonObject responseObj = new JsonObject();
responseObj.addProperty("userid", "User 1");
responseObj.addProperty("amount", "24.23");
responseObj.addProperty("success", "NO");
Usage of this JsonObject doesn't even bother using Map<>
CHEERS!!!
Real answer can be found in specification, json is unordered.
However as a human reader I ordered my elements in order of importance. Not only is it a more logic way, it happened to be easier to read. Maybe the author of the specification never had to read JSON, I do.. So, Here comes a fix:
/**
* I got really tired of JSON rearranging added properties.
* Specification states:
* "An object is an unordered set of name/value pairs"
* StackOverflow states:
* As a consequence, JSON libraries are free to rearrange the order of the elements as they see fit.
* I state:
* My implementation will freely arrange added properties, IN SEQUENCE ORDER!
* Why did I do it? Cause of readability of created JSON document!
*/
private static class OrderedJSONObjectFactory {
private static Logger log = Logger.getLogger(OrderedJSONObjectFactory.class.getName());
private static boolean setupDone = false;
private static Field JSONObjectMapField = null;
private static void setupFieldAccessor() {
if( !setupDone ) {
setupDone = true;
try {
JSONObjectMapField = JSONObject.class.getDeclaredField("map");
JSONObjectMapField.setAccessible(true);
} catch (NoSuchFieldException ignored) {
log.warning("JSONObject implementation has changed, returning unmodified instance");
}
}
}
private static JSONObject create() {
setupFieldAccessor();
JSONObject result = new JSONObject();
try {
if (JSONObjectMapField != null) {
JSONObjectMapField.set(result, new LinkedHashMap<>());
}
}catch (IllegalAccessException ignored) {}
return result;
}
}
from lemiorhan example
i can solve with just change some line of lemiorhan's code
use:
JSONObject json = new JSONObject(obj);
instead of this:
JSONObject json = (JSONObject) obj
so in my test code is :
Map item_sub2 = new LinkedHashMap();
item_sub2.put("name", "flare");
item_sub2.put("val1", "val1");
item_sub2.put("val2", "val2");
item_sub2.put("size",102);
JSONArray itemarray2 = new JSONArray();
itemarray2.add(item_sub2);
itemarray2.add(item_sub2);//just for test
itemarray2.add(item_sub2);//just for test
Map item_sub1 = new LinkedHashMap();
item_sub1.put("name", "flare");
item_sub1.put("val1", "val1");
item_sub1.put("val2", "val2");
item_sub1.put("children",itemarray2);
JSONArray itemarray = new JSONArray();
itemarray.add(item_sub1);
itemarray.add(item_sub1);//just for test
itemarray.add(item_sub1);//just for test
Map item_root = new LinkedHashMap();
item_root.put("name", "flare");
item_root.put("children",itemarray);
JSONObject json = new JSONObject(item_root);
System.out.println(json.toJSONString());
JavaScript objects, and JSON, have no way to set the order for the keys. You might get it right in Java (I don't know how Java objects work, really) but if it's going to a web client or another consumer of the JSON, there is no guarantee as to the order of keys.
Download "json simple 1.1 jar" from this https://code.google.com/p/json-simple/downloads/detail?name=json_simple-1.1.jar&can=2&q=
And add the jar file to your lib folder
using JSONValue you can convert LinkedHashMap to json string
For those who're using maven, please try com.github.tsohr/json
<!-- https://mvnrepository.com/artifact/com.github.tsohr/json -->
<dependency>
<groupId>com.github.tsohr</groupId>
<artifactId>json</artifactId>
<version>0.0.1</version>
</dependency>
It's forked from JSON-java but switch its map implementation with LinkedHashMap which #lemiorhan noted above.
As all are telling you, JSON does not maintain "sequence" but array does, maybe this could convince you:
Ordered JSONObject
For Java code, Create a POJO class for your object instead of a JSONObject.
and use JSONEncapsulator for your POJO class.
that way order of elements depends on the order of getter setters in your POJO class.
for eg. POJO class will be like
Class myObj{
String userID;
String amount;
String success;
// getter setters in any order that you want
and where you need to send your json object in response
JSONContentEncapsulator<myObj> JSONObject = new JSONEncapsulator<myObj>("myObject");
JSONObject.setObject(myObj);
return Response.status(Status.OK).entity(JSONObject).build();
The response of this line will be
{myObject : {//attributes order same as getter setter order.}}
The main intention here is to send an ordered JSON object as response. We don't need javax.json.JsonObject to achieve that. We could create the ordered json as a string.
First create a LinkedHashMap with all key value pairs in required order. Then generate the json in string as shown below.
Its much easier with Java 8.
public Response getJSONResponse() {
Map<String, String> linkedHashMap = new LinkedHashMap<>();
linkedHashMap.put("A", "1");
linkedHashMap.put("B", "2");
linkedHashMap.put("C", "3");
String jsonStr = linkedHashMap.entrySet().stream()
.map(x -> "\"" + x.getKey() + "\":\"" + x.getValue() + "\"")
.collect(Collectors.joining(",", "{", "}"));
return Response.ok(jsonStr).build();
}
The response return by this function would be following:
{"A":"1","B":"2","C":"3"}
Underscore-java uses linkedhashmap to store key/value for json. I am the maintainer of the project.
Map<String, Object> myObject = new LinkedHashMap<>();
myObject.put("userid", "User 1");
myObject.put("amount", "24.23");
myObject.put("success", "NO");
System.out.println(U.toJson(myObject));
I found a "neat" reflection tweak on "the interwebs" that I like to share.
(origin: https://towardsdatascience.com/create-an-ordered-jsonobject-in-java-fb9629247d76)
It is about to change underlying collection in org.json.JSONObject to an un-ordering one (LinkedHashMap) by reflection API.
I tested succesfully:
import java.lang.reflect.Field;
import java.util.LinkedHashMap;
import org.json.JSONObject;
private static void makeJSONObjLinear(JSONObject jsonObject) {
try {
Field changeMap = jsonObject.getClass().getDeclaredField("map");
changeMap.setAccessible(true);
changeMap.set(jsonObject, new LinkedHashMap<>());
changeMap.setAccessible(false);
} catch (IllegalAccessException | NoSuchFieldException e) {
e.printStackTrace();
}
}
[...]
JSONObject requestBody = new JSONObject();
makeJSONObjLinear(requestBody);
requestBody.put("username", login);
requestBody.put("password", password);
[...]
// returned '{"username": "billy_778", "password": "********"}' == unordered
// instead of '{"password": "********", "username": "billy_778"}' == ordered (by key)
Just add the order with this tag
#JsonPropertyOrder({ "property1", "property2"})
Not sure if I am late to the party but I found this nice example that overrides the JSONObject constructor and makes sure that the JSON data are output in the same way as they are added. Behind the scenes JSONObject uses the MAP and MAP does not guarantee the order hence we need to override it to make sure we are receiving our JSON as per our order.
If you add this to your JSONObject then the resulting JSON would be in the same order as you have created it.
import java.io.IOException;
import java.lang.reflect.Field;
import java.util.LinkedHashMap;
import org.json.JSONObject;
import lombok.extern.java.Log;
#Log
public class JSONOrder {
public static void main(String[] args) throws IOException {
JSONObject jsontest = new JSONObject();
try {
Field changeMap = jsonEvent.getClass().getDeclaredField("map");
changeMap.setAccessible(true);
changeMap.set(jsonEvent, new LinkedHashMap<>());
changeMap.setAccessible(false);
} catch (IllegalAccessException | NoSuchFieldException e) {
log.info(e.getMessage());
}
jsontest.put("one", "I should be first");
jsonEvent.put("two", "I should be second");
jsonEvent.put("third", "I should be third");
System.out.println(jsonEvent);
}
}
Just use LinkedHashMap to keep de order and transform it to json with jackson
import com.fasterxml.jackson.databind.ObjectMapper;
import java.util.LinkedHashMap;
LinkedHashMap<String, Object> obj = new LinkedHashMap<String, Object>();
stats.put("aaa", "aaa");
stats.put("bbb", "bbb");
stats.put("ccc", "ccc");
ObjectMapper mapper = new ObjectMapper();
String json = mapper.writerWithDefaultPrettyPrinter().writeValueAsString(obj);
System.out.println(json);
maven dependency
<dependency>
<groupId>com.fasterxml.jackson.core</groupId>
<artifactId>jackson-databind</artifactId>
<version>2.9.10.7</version>
</dependency>
I just want the order for android unit tests that are somehow randomly changing overtime with this cool org.json.JSONObject, even thou it looks like it uses linked map but probably depends on api you compile it with or something, so it has different impl. with different android api probably.
I would suggest something like this:
object Json {
#SuppressLint("DiscouragedPrivateApi")
fun Object() = org.json.JSONObject().apply {
runCatching {
val nameValuePairs: Field = javaClass.getDeclaredField("nameValuePairs")
nameValuePairs.isAccessible = true
nameValuePairs.set(this, LinkedHashMap<String, Any?>())
}.onFailure { it.printStackTrace() }
}
}
Usage:
val jsonObject = Json.Object()
...
This is just some possibility I use it little differently so I modified it to post here. Sure gson or other lib is another option.
Suggestions that specification is bla bla are so shortsighted here, why you guys even post it, who cares about 15 years old json spec, everyone wants it ordered anyway.
This question already has answers here:
How to parse JSON in Java
(36 answers)
Closed 5 years ago.
I have a Java String that contains a json object and I do not know how to get this json object from it?
My string is like this:
String myString = "[1,\"{\\\"Status\\\":0,\\\"InstanceNumber\\\":9}\"]";
How can i get the json object from this String?
I would recommend simple plain org.json library. Pass the string in JSONArray and then get the JSONObject. For example something like below :
String myString = "[1,\"{\\\"Status\\\":0,\\\"InstanceNumber\\\":9}\"]";
JSONArray js = new JSONArray(myString);
System.out.println(js);
JSONObject obj = new JSONObject(js.getString(1));
System.out.println(obj);
Output :
[1,"{\"Status\":0,\"InstanceNumber\":9}"]
{"Status":0,"InstanceNumber":9}
Downloadable jar: http://mvnrepository.com/artifact/org.json/json
For sure that you need to use a library lie Jackson or Gson.
I work mostly with gson when I don't have complicated stuff.
So here the output of what you are asking for. I suppose that you don't have the type that you want to convert to (for that I am taking Object).
Here is the code:
import com.google.gson.Gson;
public class Json {
public static void main(String[] args) {
Gson g = new Gson();
String myString = "[1,\"{\\\"Status\\\":0,\\\"InstanceNumber\\\":9}\"]";
Object p = g.fromJson(myString, Object.class);
System.out.println(p.toString());
}
}
And here is the output :
run:
[1.0, {"Status":0,"InstanceNumber":9}]
BUILD SUCCESSFUL (total time: 0 seconds)
You may wanting to manipulate the output object as you wish (I just printed it out).
NOTE: Don't forget to add gson jar to you classpath.
You can use any Json mapping framework to deserialise the String into Java object. Below example shows how to do it with Jackson:
String myString = "[1,\"{\\\"Status\\\":0,\\\"InstanceNumber\\\":9}\"]";
ObjectMapper mapper = new ObjectMapper();
List<Object> value = mapper.readValue(myString, new TypeReference<List<Object>>() {});
Map<String, Object> map = mapper.readValue(value.get(1).toString(), new TypeReference<Map<String, Object>>() {});
System.out.println(map);
Here's the documentation.
I have one Model Object. In which, i have multiple values. I want to store this Values in SQLite. But data is large, so i want to store Direct Model object
in databse. So i convert model Object to string and store it into database.
Now, Problem is that how to convert this String value to Model Object.
If you have any idea, please share that with Me.
For example,
Person p = new Person();
p.setname("xyz");
p.setage("18");`
String person=p.toString();
Now How to get this "person" string back to Person "p" model object.
This is my code.
ContentValues values = new ContentValues();
String favorite_id = UUID.randomUUID().toString();
values.put(EMuseumLocalData.KEY_FAVORITE_EXHIBITS_ID, favorite_id);
values.put(EMuseumLocalData.KEY_EXHIBIT_SUBCATEGORY_ITEM_ID, Integer.parseInt(categoryByCustomerList.get(position).getSubCategoryItemID()));
try {
Gson gson = new Gson();
String personString = gson.toJson(getAllCategory.get(position).toString());
values.put(EMuseumLocalData.KEY_EXHIBIT_SUBCATEGORY_ITEM_DATA, personString);
Gson gson1 = new Gson();
CategoryByCustomer categoryByCustomer = gson1.fromJson(personString, categoryByCustomer.getName());
} catch (JSONException e) {
e.printStackTrace();
}
You should use GSON or similar libs for this.
Store to DB
For example If you use GSON
Person p = new Person();
p.setname("xyz");
p.setage("18");
Gson gson = new Gson();
String personString = gson.toJson(p);
Now store this personString to DB.
Read from DB
Get back this object from database, read string from DB and convert it to object like below
String personStringFromDB = READ_LOGIC_OF_DB;
Gson gson = new Gson();
Person p = gson.fromJson(personStringFromDB, Person.class);
For more information, read GSON - Gson Example
Consider using a json string representation of the Model Object. There are many java libraries like Jackson, Gson etc., available to help you with serialization/deserialization part.
Here's a sample code to do this in Jackson
//For conversion of Person object(person) to json String:
String personJsonString = new com.fasterxml.jackson.databind.ObjectMapper().writeValueAsString(person);
//For conversion of json String back to Person object(person)
Person person = new com.fasterxml.jackson.databind.ObjectMapper().readValue(personJsonString, Person.class);
You can make Model Object serializable. You need to store the serialized object in SQLite. When you need it, you just get that serialized object from SOLite and deserialize it.
I have JSON string which is in a standalone Java project:
{"MsgType":"AB","TID":"1","ItemID":"34532136","TransactTime":1389260033223}
I want to extract MsgType from this which is AB
Which is the best way to do it?
You can use Gson library for this.
String json="{MsgType:AB,TID:1,ItemID:34532136,TransactTime:1389260033223}";
Map jsonJavaRootObject = new Gson().fromJson(json, Map.class);
System.out.println(jsonJavaRootObject.get("MsgType"));
where the jsonJavaRootObject will contain a map of keyvalues like this
{MsgType=AB, TID=1.0, ItemID=3.4532136E7, TransactTime=1.389260033223E12}
I use JSONObject under android but from Oracle docs I see its also available under javax.json package:
http://docs.oracle.com/javaee/7/api/javax/json/package-summary.html
If you want Gson then your code should look like below (sorry not compiled/tested):
/*
{"MsgType":"AB","TID":"1","ItemID":"34532136","TransactTime":1389260033223}
*/
Gson gson = new Gson();
static class Data{
String MsgType;
String TID;
String ItemID;
int TransactTime;
}
Data data = gson.fromJson(yourJsonString, Data.class);
I have an example with json-simple:
JSONParser parser = new JSONParser();
JSONObject jsonObject = (JSONObject) parser.parse(yourString);
String msgType = (String) jsonObject.get("MsgType");
Check this link for a complete example.
JsonPath
For parsing simple JSON use JsonPath
JsonPath is to JSON what XPATH is to XML, a simple way to extract parts of a given document.
Example code
String json = "{\"MsgType\":\"AB\",\"TID\":\"1\",\"ItemID\":\"34532136\",\"TransactTime\":1389260033223}";
String author = JsonPath.read(json, "$.MsgType");
System.out.println(author);
Result
AB
Dependency
'com.jayway.jsonpath:json-path:0.9.1'