How to deserialize json string into object

Question

{
   "LocalLocationId [id=1]":{
      "type":"folderlocation",
      "id":{
         "type":"locallocationid",
         "id":1
      },
      "parentId":{
         "type":"locallocationid",
         "id":0
      },
      "name":"Test",
      "accessibleToUser":true,
      "defaultLocation":false,
      "timezoneId":"Asia/Calcutta",
      "children":[]
   },
   "LocalLocationId [id=0]":{
      "type":"folderlocation",
      "id":{
         "type":"locallocationid",
         "id":0
      },
      "parentId":null,
      "name":"Locations",
      "accessibleToUser":false,
      "defaultLocation":false,
      "timezoneId":"Asia/Calcutta",
      "children":[{
         "type":"locallocationid",
         "id":1
      }]
   },
   "allAllowedChildren":[{
      "type":"locallocationid",
      "id":1
   }]
}

How to deserialize above string into java object.

Class im using is

public class Tree {

    @SerializedName("allAllowedChildren")
    private List<Id> allAllowedChildren;

    @SerializedName("LocalLocationId")
    private Map<String, LocalLocationId> localLocationId;

    public class LocalLocationId {
        @SerializedName("type")
        private String type;

        @SerializedName("name")
        private String name;

        @SerializedName("accessibleToUser")
        private boolean accessibleToUser;

        @SerializedName("defaultLocation")
        private boolean defaultLocation;

        @SerializedName("timezoneId")
        private String timezoneId;

        @SerializedName("id")
        private Id id;

        @SerializedName("parentId")
        private Id parentId;

        @SerializedName("children")
        private List<Id> children;

        public String getType() {
            return type;
        }
        public String getName() {
            return name;
        }
        public boolean isAccessibleToUser() {
            return accessibleToUser;
        }
        public boolean isDefaultLocation() {
            return defaultLocation;
        }
        public String getTimezoneId() {
            return timezoneId;
        }
        public Id getId() {
            return id;
        }
        public Id getParentId() {
            return parentId;
        }
        public List<Id> getChildren() {
            return children;
        }
    }

    public class Id {
        private String type;
        private Integer id;

        public String getType() {
            return type;
        }
        public Integer getId() {
            return id;
        }
    }

    public List<Id> getAllAllowedChildren() {
        return allAllowedChildren;
    }
    public Map<String, LocalLocationId> getLocalLocationId() {
        return localLocationId;
    }
}

Deserialize above string into java object is impossible to do so with the class that you are using right now. — Ryan Fung, Sep 30 '15 at 07:51
@RyanFung care to elaborate? I'm guessing OP doesn't know why that is the case. — Andy Turner, Sep 30 '15 at 07:54
@dotvav., after `Tree tree = gson.fromJson(locationTree, Tree.class);` i get LocalLocationId as `null` — Kedar Javalkar, Sep 30 '15 at 08:11
@KedarJavalkar Yes sorry for my quick reply. Your class object will not be stored or hard to stored. A "tree like" structure will be more suited. — Ryan Fung, Sep 30 '15 at 08:59
I hope, this link will solve ur problem http://programmerbruce.blogspot.com/2011/06/gson-v-jackson.html#TOC-Nested-Classes-including-Inner-Clas — Noor Nawaz, Sep 30 '15 at 10:28

vvs · Accepted Answer · 2015-09-30T11:18:50.100

@Kedar

I'll assume you are in control of how the JSON input string is created. I think the JSON string is not formatted correctly for default GSON deserialization of Map types.

I have modified the input string for your consideration and this results in a non null LocalLocationId

{
   "LocalLocationId":[
   [
     "1",
       {
          "type":"folderlocation",
          "id":{
             "type":"locallocationid",
             "id":1
          },
          "parentId":{
             "type":"locallocationid",
             "id":0
          },
          "name":"Test",
          "accessibleToUser":true,
          "defaultLocation":false,
          "timezoneId":"Asia/Calcutta",
          "children":[]
       }
   ],
   [
     "2",
       {
          "type":"folderlocation",
          "id":{
             "type":"locallocationid",
             "id":0
          },
          "parentId":null,
          "name":"Locations",
          "accessibleToUser":false,
          "defaultLocation":false,
          "timezoneId":"Asia/Calcutta",
          "children":[{
             "type":"locallocationid",
             "id":1
          }]
       }
   ]
   ],
   "allAllowedChildren":[{
      "type":"locallocationid",
      "id":1
   }]
}

Please comment if my assumption about the input string is incorrect.

EDIT 1: Since input cannot be modified, consider writing custom Deserializer. Below is the way to register custom deserialisation class

GsonBuilder gsonb = new GsonBuilder();
        gsonb.registerTypeAdapter(Tree.class, new TreeDeserializer());
        Gson gson = gsonb.create();

Below is the TreeDeserializer

public class TreeDeserializer implements JsonDeserializer<Tree> {

    public Tree deserialize(JsonElement json, Type typeOfT,
            JsonDeserializationContext context) throws JsonParseException {
        Tree out = new Tree();

        if (json != null) {
            JsonObject obj  = json.getAsJsonObject();
            Set<Map.Entry<String,JsonElement>> entries = obj.entrySet();
            for (Map.Entry<String, JsonElement> e: entries) {
                if (e.getKey().equals("allAllowedChildren")) {
                    Type ft = List.class;
                    System.out.println(context.deserialize(e.getValue(), ft));
                    // TODO add this back into the Tree out object
                } else {
                    // LocalLocationId
                    System.out.println(e.getKey());
                    System.out.println(context.deserialize(e.getValue(), Tree.LocalLocationId.class));

                    // TODO add this back into the Tree out object
                }
            }
        } 
        return out;
    }

}

Here is the console output from the Sysouts.

LocalLocationId [id=1]
org.test.StackOverflowAnswers.Tree$LocalLocationId@464bee09
LocalLocationId [id=0]
org.test.StackOverflowAnswers.Tree$LocalLocationId@f6c48ac
[{type=locallocationid, id=1.0}]
org.test.StackOverflowAnswers.Tree@589838eb

I have left TODOs in the deserialiser where you'll need to write custom code to inject the values from deserialisation into the Tree class just created. Hope this helps. Can't provide full implementation, but I think this would be a partial solution

The json string is received from a third party api., which is in the exact format i have mentioned — Kedar Javalkar, Sep 30 '15 at 09:14
Ok then I can see the only option would be to write your own code to perform parts of the deserialisation — vvs, Sep 30 '15 at 11:06
i have done something similar to parse the string into `Tree` class.. i will definitely give it a try to your method.. cheers.... — Kedar Javalkar, Sep 30 '15 at 17:25

score 0 · Answer 2 · answered Sep 30 '15 at 08:06

User JSONParser which is faster one.

Below is sample. there can be a btter example if you google. Hope this helps.

JSONParser parser=new JSONParser();
System.out.println("=======decode=======");
String s="[0,{\"1\":{\"2\":{\"3\":{\"4\":[5,{\"6\":7}]}}}}]";  
Object obj=parser.parse(s);  
JSONArray array=(JSONArray)obj;  
System.out.println("======the 2nd element of array======");  
System.out.println(array.get(1));  
System.out.println();                  
JSONObject obj2=(JSONObject)array.get(1);  
System.out.println("======field \"1\"==========");  
System.out.println(obj2.get("1"));                      
s="{}";  
obj=parser.parse(s);  
System.out.println(obj);                  
s="[5,]";  
obj=parser.parse(s);  
System.out.println(obj);                  
s="[5,,2]";  
obj=parser.parse(s);  
System.out.println(obj);

score 0 · Answer 3 · answered Sep 30 '15 at 08:23

0

You can use Gson ..

String json = "Your json string "
Tree treeObj= new Gson().fromJson(json, Tree .class);

answered Sep 30 '15 at 08:23

Manisha Srivastava

757
6
14

running above results into `null` for localLocationId map – Kedar Javalkar Sep 30 '15 at 08:43
I could see you need to modify your json string ,its not formatted correctly. – Manisha Srivastava Sep 30 '15 at 09:11

score -1 · Answer 4 · answered Dec 15 '20 at 13:38

-1

You can use ObjectMapper from Jackson -

Tree deserializedTree = new ObjectMapper().readValue(jsonStringOfTree, Tree.class);

answered Dec 15 '20 at 13:38

Amit Dusane

1,324
10
12

How to deserialize json string into object

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