Gson - convert from Json to a typed ArrayList<T>

Question

Using the Gson library, how do I convert a JSON string to an ArrayList of a custom class JsonLog? Basically, JsonLog is an interface implemented by different kinds of logs made by my Android app--SMS logs, call logs, data logs--and this ArrayList is a collection of all of them. I keep getting an error in line 6.

public static void log(File destination, JsonLog log) {
    Collection<JsonLog> logs = null;
    if (destination.exists()) {
        Gson gson = new Gson();
        BufferedReader br = new BufferedReader(new FileReader(destination));
        logs = gson.fromJson(br, ArrayList<JsonLog>.class); // line 6
        // logs.add(log);
        // serialize "logs" again
    }
}

It seems the compiler doesn't understand I'm referring to a typed ArrayList. What do I do?

Please also refer to stackoverflow.com/questions/44589381/… for another approach — Pankaj Jaiswal
– Pankaj Jaiswal, Commented Jun 18, 2018 at 6:32

Alex · Accepted Answer · 2020-03-04 20:13:58Z

618

You may use TypeToken to load the json string into a custom object.

logs = gson.fromJson(br, new TypeToken<List<JsonLog>>(){}.getType());

Documentation:

Represents a generic type T.

Java doesn't yet provide a way to represent generic types, so this class does. Forces clients to create a subclass of this class which enables retrieval the type information even at runtime.

For example, to create a type literal for List<String>, you can create an empty anonymous inner class:

TypeToken<List<String>> list = new TypeToken<List<String>>() {};

This syntax cannot be used to create type literals that have wildcard parameters, such as Class<?> or List<? extends CharSequence>.

Kotlin:

If you need to do it in Kotlin you can do it like this:

val myType = object : TypeToken<List<JsonLong>>() {}.type
val logs = gson.fromJson<List<JsonLong>>(br, myType)

Or you can see this answer for various alternatives.

edited Mar 4, 2020 at 20:13

answered Sep 12, 2012 at 8:34

Alex

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10 Comments

Jono Over a year ago

tried this approach and i got a compile error saying TypeToken() has protected access only

Zapnologica Over a year ago

@jonney I have the same issue. Any solutions?

Zapnologica Over a year ago

if you are getting has protected access error, you left out the { } before .getType()

Adnan Ali Over a year ago

@Alex does it deserialize first to an Array then that Array is converted into List or does it directly deserialize to List ?

John Ward Over a year ago

Wholly cow! I would have figured that like...well...never. Thank you.

|

rupinderjeet · Accepted Answer · 2018-01-23 16:59:49Z

30

Your JSON sample is:

{
    "status": "ok",
    "comment": "",
    "result": {
    "id": 276,
    "firstName": "mohamed",
    "lastName": "hussien",
    "players": [
            "player 1",
            "player 2",
            "player 3",
            "player 4",
            "player 5"
    ]
}

so if you want to save arraylist of modules in your SharedPrefrences so :

1- will convert your returned arraylist for json format using this method

public static String toJson(Object jsonObject) {
    return new Gson().toJson(jsonObject);
}

2- Save it in shared prefreneces

PreferencesUtils.getInstance(context).setString("players", toJson((.....ArrayList you want to convert.....)));

3- to retrieve it at any time get JsonString from Shared preferences like that

String playersString= PreferencesUtils.getInstance(this).getString("players");

4- convert it again to array list

public static Object fromJson(String jsonString, Type type) {
    return new Gson().fromJson(jsonString, type);
}

ArrayList<String> playersList= (ArrayList<String>) fromJson(playersString,
                    new TypeToken<ArrayList<String>>() {
                    }.getType());

this solution also doable if you want to parse ArrayList of Objects Hope it's help you by using Gson Library .

edited Jan 23, 2018 at 16:59

rupinderjeet

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answered Feb 23, 2015 at 9:14

Mohamed Hussien

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1 Comment

xeruf Over a year ago

this is the correct solution, but essentially the same as the three years older answer with a lot of unneccessary information - I didn't come here to learn about sharedPreferences

FarukT · Accepted Answer · 2020-03-26 09:49:39Z

13

Why nobody wrote this simple way of converting JSON string in List ?

List<Object> list = Arrays.asList(new GsonBuilder().create().fromJson(jsonString, Object[].class));

answered Mar 26, 2020 at 9:49

FarukT

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1 Comment

Peter Schorn Over a year ago

Because the poster wants the list to be a specific type. It's literally right there in the title of the question. Also, you can just use new Gson() instead of new GsonBuilder().create().

murgupluoglu · Accepted Answer · 2019-10-28 08:30:04Z

12

Kotlin

data class Player(val name : String, val surname: String)

val json = [
  {
    "name": "name 1",
    "surname": "surname 1"
  },
  {
    "name": "name 2",
    "surname": "surname 2"
  },
  {
    "name": "name 3",
    "surname": "surname 3"
  }
]

val typeToken = object : TypeToken<List<Player>>() {}.type
val playerArray = Gson().fromJson<List<Player>>(json, typeToken)

OR

val playerArray = Gson().fromJson(json, Array<Player>::class.java)

edited Oct 28, 2019 at 8:30

answered Oct 28, 2019 at 8:24

murgupluoglu

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Comments

Ali Asadi · Accepted Answer · 2018-12-12 12:59:19Z

9

If you want to use Arrays, it's pretty simple.

logs = gson.fromJson(br, JsonLog[].class); // line 6

Provide the JsonLog as an array JsonLog[].class

answered Dec 12, 2018 at 12:59

Ali Asadi

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Comments

Eraldo · Accepted Answer · 2020-11-25 23:08:11Z

8

If you want convert from Json to a typed ArrayList , it's wrong to specify the type of the object contained in the list. The correct syntax is as follows:

 Gson gson = new Gson(); 
 List<MyClass> myList = gson.fromJson(inputString, ArrayList.class);

answered Nov 25, 2020 at 23:08

Eraldo

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Comments

Shihab Uddin · Accepted Answer · 2022-05-19 04:37:00Z

7

Let's say, You have a string like this.

"[{"id":2550,"cityName":"Langkawi","hotelName":"favehotel Cenang Beach - Langkawi","hotelId":"H1266070"},
{"id":2551,"cityName":"Kuala Lumpur","hotelName":"Metro Hotel Bukit Bintang","hotelId":"H835758"}]"

Then you can covert it to ArrayList via Gson like

var hotels = Gson().fromJson(historyItem.hotels, Array<HotelInfo>::class.java).toList()

Your HotelInfo class should like this.

import com.squareup.moshi.Json

data class HotelInfo(

    @Json(name="cityName")
    val cityName: String? = null,

    @Json(name="id")
    val id: Int? = null,

    @Json(name="hotelId")
    val hotelId: String? = null,

    @Json(name="hotelName")
    val hotelName: String? = null
)

edited May 19, 2022 at 4:37

answered Apr 13, 2020 at 5:29

Shihab Uddin

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Comments

arun kongara · Accepted Answer · 2018-11-20 14:02:46Z

0

I am not sure about gson but this is how you do it with Jon.sample hope there must be similar way using gson

{ "Players": [ "player 1", "player 2", "player 3", "player 4", "player 5" ] }

===============================================

import java.io.FileReader;
import java.util.List;

import org.json.simple.JSONObject;
import org.json.simple.parser.JSONParser;

public class JosnFileDemo {

    public static void main(String[] args) throws Exception
    {
        String jsonfile ="fileloaction/fileName.json";


                FileReader reader = null;
                JSONObject jsb = null;
                try {
                    reader = new FileReader(jsonfile);
                    JSONParser jsonParser = new JSONParser();
                     jsb = (JSONObject) jsonParser.parse(reader);
                } catch (Exception e) {
                    throw new Exception(e);
                } finally {
                    if (reader != null)
                        reader.close();
                }
                List<String> Players=(List<String>) jsb.get("Players");
                for (String player : Players) {
                    System.out.println(player);
                }           
    }
}

edited Nov 20, 2018 at 14:02

answered Nov 20, 2018 at 3:28

arun kongara

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1 Comment

arun kongara Over a year ago

@ Sagar Zala :what's example you are expecting ..?

Safal Bhatia · Accepted Answer · 2022-03-14 12:37:00Z

0

In Kotlin For sending : If you send Arraylist

Gson().toJson(arraylist)

For receiving: If you receive ArrayList

var arraylist = Gson().fromJson(argument, object : TypeToken<ArrayList<LatLng>>() {}.type)

For sending : If you send ModelClass( e.g. LatLngModel.class)

var latlngmodel = LatlngModel()
latlngmodel.lat = 32.0154
latlngmodel.lng = 70.1254

Gson().toJson(latlngModel)

For receiving: If you receive ModelClass

var arraylist = Gson().fromJson(argument,LatLngModel::class.java )

answered Mar 14, 2022 at 12:37

Safal Bhatia

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