Converting JSONarray to ArrayList

Question

I am downloading a JSON string and converting it to JSONArray. Im putting it into a listview and need to be able to delete from that listview later, and since JSONArray has no .remove method (Thanks Obama), I am trying to convert it to an arraylist.

here is my JSON (the array.toString()):

[
    {
         "thumb_url":"tb1370913834.jpg",
         "event_id":"15","count":"44",
         "event_tagline":"this is a tagline",
         "event_name":"5th birthday",
         "event_end":"1370919600",
         "event_start":"1370876400"
    }
]

I need to get it into an array and be able to call the strings by their respective keys. Appreciate any help!

When you say keys, doesn't that imply a Dictionary (Map in Android) of some sort? An array will be index-based. Take a look at http://developer.android.com/reference/java/util/Map.html for how to create and use. — brianestey, Jun 11 '13 at 06:02
Can a map be used to populate a listview with a custom baseadapter? Id rather not use indexes in case the JSON ever changes order. — TheGeekNess, Jun 11 '13 at 16:26
ListView will get objects from the adapter by index, so in any case you will need to maintain the list in some order. If you want to maintain the order of the keys (and know what the keys are at compile time), you can hard-code an array of those keys in the order you like and use that for your ordering when fetching from the Map. — brianestey, Jun 12 '13 at 02:02

score 201 · Accepted Answer · edited Dec 05 '16 at 20:27

201

ArrayList<String> listdata = new ArrayList<String>();     
JSONArray jArray = (JSONArray)jsonObject; 
if (jArray != null) { 
   for (int i=0;i<jArray.length();i++){ 
    listdata.add(jArray.getString(i));
   } 
}

edited Dec 05 '16 at 20:27

Vasilii Suricov

871
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answered Jun 11 '13 at 05:55

Sagar Maiyad

12,655
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5

And `listdata.add(jArray.optJSONObject(i));` if your listdata is a JSONObject arrayList. `ArrayList listdata = new ArrayList(); ` – C-- Jul 19 '14 at 12:55
2

Good snippet. Just in case if anyone wants: there is a helper class that converts JSONObject/JSONArray to a standard Map/List on the github https://gist.github.com/codebutler/2339666 – inexcii Aug 27 '14 at 01:54
2

Why not use an `ArrayList – natanavra Nov 04 '15 at 15:21
How may i handle the Json Array is empty. please reply me bro. – MohanRaj S Nov 28 '16 at 08:01
4

Is there any other way to do this without loop? – K.Sopheak Jan 26 '17 at 08:56
1

@K.Sopheak, `Streams` (Java 8 & up) can do. – Dut A. Nov 20 '21 at 17:57

score 82 · Answer 2 · edited Apr 02 '18 at 09:17

I've done it using Gson (by Google).

Add the following line to your module's build.gradle:

dependencies {
  // ...
  // Note that `compile` will be deprecated. Use `implementation` instead.
  // See https://stackoverflow.com/a/44409111 for more info
  implementation 'com.google.code.gson:gson:2.8.2'
}

JSON string:

private String jsonString = "[\n" +
            "        {\n" +
            "                \"id\": \"c200\",\n" +
            "                \"name\": \"Ravi Tamada\",\n" +
            "                \"email\": \"ravi@gmail.com\",\n" +
            "                \"address\": \"xx-xx-xxxx,x - street, x - country\",\n" +
            "                \"gender\" : \"male\",\n" +
            "                \"phone\": {\n" +
            "                    \"mobile\": \"+91 0000000000\",\n" +
            "                    \"home\": \"00 000000\",\n" +
            "                    \"office\": \"00 000000\"\n" +
            "                }\n" +
            "        },\n" +
            "        {\n" +
            "                \"id\": \"c201\",\n" +
            "                \"name\": \"Johnny Depp\",\n" +
            "                \"email\": \"johnny_depp@gmail.com\",\n" +
            "                \"address\": \"xx-xx-xxxx,x - street, x - country\",\n" +
            "                \"gender\" : \"male\",\n" +
            "                \"phone\": {\n" +
            "                    \"mobile\": \"+91 0000000000\",\n" +
            "                    \"home\": \"00 000000\",\n" +
            "                    \"office\": \"00 000000\"\n" +
            "                }\n" +
            "        },\n" +
            "        {\n" +
            "                \"id\": \"c202\",\n" +
            "                \"name\": \"Leonardo Dicaprio\",\n" +
            "                \"email\": \"leonardo_dicaprio@gmail.com\",\n" +
            "                \"address\": \"xx-xx-xxxx,x - street, x - country\",\n" +
            "                \"gender\" : \"male\",\n" +
            "                \"phone\": {\n" +
            "                    \"mobile\": \"+91 0000000000\",\n" +
            "                    \"home\": \"00 000000\",\n" +
            "                    \"office\": \"00 000000\"\n" +
            "                }\n" +
            "        },\n" +
            "        {\n" +
            "                \"id\": \"c203\",\n" +
            "                \"name\": \"John Wayne\",\n" +
            "                \"email\": \"john_wayne@gmail.com\",\n" +
            "                \"address\": \"xx-xx-xxxx,x - street, x - country\",\n" +
            "                \"gender\" : \"male\",\n" +
            "                \"phone\": {\n" +
            "                    \"mobile\": \"+91 0000000000\",\n" +
            "                    \"home\": \"00 000000\",\n" +
            "                    \"office\": \"00 000000\"\n" +
            "                }\n" +
            "        },\n" +
            "        {\n" +
            "                \"id\": \"c204\",\n" +
            "                \"name\": \"Angelina Jolie\",\n" +
            "                \"email\": \"angelina_jolie@gmail.com\",\n" +
            "                \"address\": \"xx-xx-xxxx,x - street, x - country\",\n" +
            "                \"gender\" : \"female\",\n" +
            "                \"phone\": {\n" +
            "                    \"mobile\": \"+91 0000000000\",\n" +
            "                    \"home\": \"00 000000\",\n" +
            "                    \"office\": \"00 000000\"\n" +
            "                }\n" +
            "        },\n" +
            "        {\n" +
            "                \"id\": \"c205\",\n" +
            "                \"name\": \"Dido\",\n" +
            "                \"email\": \"dido@gmail.com\",\n" +
            "                \"address\": \"xx-xx-xxxx,x - street, x - country\",\n" +
            "                \"gender\" : \"female\",\n" +
            "                \"phone\": {\n" +
            "                    \"mobile\": \"+91 0000000000\",\n" +
            "                    \"home\": \"00 000000\",\n" +
            "                    \"office\": \"00 000000\"\n" +
            "                }\n" +
            "        },\n" +
            "        {\n" +
            "                \"id\": \"c206\",\n" +
            "                \"name\": \"Adele\",\n" +
            "                \"email\": \"adele@gmail.com\",\n" +
            "                \"address\": \"xx-xx-xxxx,x - street, x - country\",\n" +
            "                \"gender\" : \"female\",\n" +
            "                \"phone\": {\n" +
            "                    \"mobile\": \"+91 0000000000\",\n" +
            "                    \"home\": \"00 000000\",\n" +
            "                    \"office\": \"00 000000\"\n" +
            "                }\n" +
            "        },\n" +
            "        {\n" +
            "                \"id\": \"c207\",\n" +
            "                \"name\": \"Hugh Jackman\",\n" +
            "                \"email\": \"hugh_jackman@gmail.com\",\n" +
            "                \"address\": \"xx-xx-xxxx,x - street, x - country\",\n" +
            "                \"gender\" : \"male\",\n" +
            "                \"phone\": {\n" +
            "                    \"mobile\": \"+91 0000000000\",\n" +
            "                    \"home\": \"00 000000\",\n" +
            "                    \"office\": \"00 000000\"\n" +
            "                }\n" +
            "        },\n" +
            "        {\n" +
            "                \"id\": \"c208\",\n" +
            "                \"name\": \"Will Smith\",\n" +
            "                \"email\": \"will_smith@gmail.com\",\n" +
            "                \"address\": \"xx-xx-xxxx,x - street, x - country\",\n" +
            "                \"gender\" : \"male\",\n" +
            "                \"phone\": {\n" +
            "                    \"mobile\": \"+91 0000000000\",\n" +
            "                    \"home\": \"00 000000\",\n" +
            "                    \"office\": \"00 000000\"\n" +
            "                }\n" +
            "        },\n" +
            "        {\n" +
            "                \"id\": \"c209\",\n" +
            "                \"name\": \"Clint Eastwood\",\n" +
            "                \"email\": \"clint_eastwood@gmail.com\",\n" +
            "                \"address\": \"xx-xx-xxxx,x - street, x - country\",\n" +
            "                \"gender\" : \"male\",\n" +
            "                \"phone\": {\n" +
            "                    \"mobile\": \"+91 0000000000\",\n" +
            "                    \"home\": \"00 000000\",\n" +
            "                    \"office\": \"00 000000\"\n" +
            "                }\n" +
            "        },\n" +
            "        {\n" +
            "                \"id\": \"c2010\",\n" +
            "                \"name\": \"Barack Obama\",\n" +
            "                \"email\": \"barack_obama@gmail.com\",\n" +
            "                \"address\": \"xx-xx-xxxx,x - street, x - country\",\n" +
            "                \"gender\" : \"male\",\n" +
            "                \"phone\": {\n" +
            "                    \"mobile\": \"+91 0000000000\",\n" +
            "                    \"home\": \"00 000000\",\n" +
            "                    \"office\": \"00 000000\"\n" +
            "                }\n" +
            "        },\n" +
            "        {\n" +
            "                \"id\": \"c2011\",\n" +
            "                \"name\": \"Kate Winslet\",\n" +
            "                \"email\": \"kate_winslet@gmail.com\",\n" +
            "                \"address\": \"xx-xx-xxxx,x - street, x - country\",\n" +
            "                \"gender\" : \"female\",\n" +
            "                \"phone\": {\n" +
            "                    \"mobile\": \"+91 0000000000\",\n" +
            "                    \"home\": \"00 000000\",\n" +
            "                    \"office\": \"00 000000\"\n" +
            "                }\n" +
            "        },\n" +
            "        {\n" +
            "                \"id\": \"c2012\",\n" +
            "                \"name\": \"Eminem\",\n" +
            "                \"email\": \"eminem@gmail.com\",\n" +
            "                \"address\": \"xx-xx-xxxx,x - street, x - country\",\n" +
            "                \"gender\" : \"male\",\n" +
            "                \"phone\": {\n" +
            "                    \"mobile\": \"+91 0000000000\",\n" +
            "                    \"home\": \"00 000000\",\n" +
            "                    \"office\": \"00 000000\"\n" +
            "                }\n" +
            "        }\n" +
            "    ]";

ContactModel.java:

public class ContactModel {
     public String id;
     public String name;
     public String email;
}

Code for converting a JSON string to ArrayList<Model>:

Note: You have to import java.lang.reflect.Type;:

// Top of file
import java.lang.reflect.Type;

// ...

private void parseJSON() {
    Gson gson = new Gson();
    Type type = new TypeToken<List<ContactModel>>(){}.getType();
    List<ContactModel> contactList = gson.fromJson(jsonString, type);
    for (ContactModel contact : contactList){
        Log.i("Contact Details", contact.id + "-" + contact.name + "-" + contact.email);
    }
}

Hope this will help you.

THREE LINES...DONE! I think this should have been the accepted answer since (a) it's only three lines of code that work beautifully, and (b) few of the answers actually perform a typed translation from JSONArray to List. Thank you! — John Ward, Jan 05 '17 at 23:27
It must be a JSONArray, the JSONElement seems not work. so, use the `getAsJsonArray()` method appearently. Thank you! — Jiaheng Tao, Mar 13 '19 at 10:09
Exactly what I needed, thanks! Note: You have to import these: java.lang.reflect.Type; com.google.gson.reflect.TypeToken; — Chandrani H, Jan 14 '20 at 06:11
This is chaos when having date and Time objects like LocalDate and LocalDateTime — David Kariuki, Nov 18 '22 at 07:51

Vasilii Suricov · Answer 3 · 2016-12-05T21:04:00.323

8

I have fast solution. Just create a file ArrayUtil.java

import java.util.ArrayList;
import java.util.Collection;
import org.json.JSONArray;
import org.json.JSONException;

public class ArrayUtil
{
    public static ArrayList<Object> convert(JSONArray jArr)
    {
        ArrayList<Object> list = new ArrayList<Object>();
        try {
            for (int i=0, l=jArr.length(); i<l; i++){
                 list.add(jArr.get(i));
            }
        } catch (JSONException e) {}

        return list;
    }

    public static JSONArray convert(Collection<Object> list)
    {
        return new JSONArray(list);
    }

}

Usage:

ArrayList<Object> list = ArrayUtil.convert(jArray);

or

JSONArray jArr = ArrayUtil.convert(list);

edited Dec 05 '16 at 21:04

answered Dec 05 '16 at 20:49

Vasilii Suricov

871
12
18

Wha I do if need to return a `ArrayList` – Feb 01 '17 at 17:40
I think it's not possible to cast ArrayList – Vasilii Suricov Feb 02 '17 at 19:03
1

Kudos to this answer. U solved my day! I spent 2 hrs trying to fix this. Actually my problem was I wanted to put the converted `JSONArray` to a list so I could put the `List` in a `HashMap`, so this util worked for me pretty well. Thanks @Vasilii Suricov – Jose Mhlanga Jul 26 '20 at 10:30

score 7 · Answer 4 · answered Jun 11 '13 at 06:03

try this way Simply loop through that, building your own array. This code assumes it's an array of strings, it shouldn't be hard to modify to suit your particular array structure.

JSONArray jsonArray = new JSONArray(jsonArrayString);
List<String> list = new ArrayList<String>();
for (int i=0; i<jsonArray.length(); i++) {
    list.add( jsonArray.getString(i) );

score 7 · Answer 5 · answered Jun 11 '13 at 06:07

7

Instead of converting the JSON string to an ArrayList or even a Map, you can just create a JSONObject itself. This object has the ability to get string values by key, as you want and also to remove objects.

To create a JSONObject from a properly formatted JSON string, you simply call the appropriate constructor.

JSONObject json = new JSONObject(jsonString);

answered Jun 11 '13 at 06:07

brianestey

8,202
5
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1

The problem I run into with this solution in particular, is that when dealing with JSONObject's and JSONArray's is they throw JSONException's. Sometimes it's handy to pass the contents of a JSONArray to a function that shouldn't have to be aware of JSON. – Aaron Dougherty Apr 19 '15 at 22:21
2

I agree. I would use the JSONObject as part of parsing the JSON text into a usable model object. Any JSONException that is raised would indicate a problem with the source JSON, meaning you can't parse it into a JSONArray or JSONObject, let alone a different model object. – brianestey Apr 20 '15 at 22:32

Anant Simran Singh · Answer 6 · 2021-10-08T03:51:05.640

7

In Java 8,

IntStream.range(0,jsonArray.length()).mapToObj(i->jsonArray.getString(i)).collect(Collectors.toList())

edited Oct 08 '21 at 03:51

answered Sep 18 '17 at 10:28

Anant Simran Singh

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2

The only problem with this is that JSONArray.getString(...) throws an exception which has to be handled inside of the mapping, so you end up with `List listOfStrings = IntStream.range(0, array.length()).mapToObj(i -> { try { return array.getString(i); } catch (JSONException e) { throw new AssertionFailedError("JSONArray is not a list of Strings! " + e.getMessage()); } }).collect(Collectors.toList());` which is not that elegant any more. So I'll go with a foreach :) – LazR Mar 04 '19 at 13:29
This is the best answer here (imho) but would be elegant if there was a way to not need to access the jsonArray reference more than once. – djangofan Apr 22 '19 at 21:32

Sorter · Answer 7 · 2015-04-03T06:24:39.027

4

 JSONArray array = new JSONArray(json);
 List<JSONObject> list = new ArrayList();
 for (int i = 0; i < array.length();list.add(array.getJSONObject(i++)));

edited Apr 03 '15 at 06:24

answered Dec 02 '14 at 10:17

Sorter

9,704
6
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score 2 · Answer 8 · answered Jun 11 '13 at 06:28

To make it handy, use POJO.

try like this..

List<YourPojoObject> yourPojos = new ArrayList<YourPojoObject>();

JSONObject jsonObject = new JSONObject(jsonString);
YourPojoObject yourPojo = new YourPojoObject();
yourPojo.setId(jsonObject.getString("idName"));
...
...

yourPojos.add(yourPojo);

score 2 · Answer 9 · edited Jun 20 '19 at 06:25

2

I have fast solution. Just create a file ArrayUtil.java

ObjectMapper mapper = new ObjectMapper(); 
List<Student> list = Arrays.asList(mapper.readValue(jsonString, Student[].class));

Usage:

ArrayList<Object> list = ArrayUtil.convert(jArray);

or

JSONArray jArr = ArrayUtil.convert(list);

edited Jun 20 '19 at 06:25

adiga

34,372
9
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answered Jun 20 '19 at 06:17

Bheem Singh

607
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score 2 · Answer 10 · answered Apr 20 '20 at 11:01

if you want to extract data form JSON string array, here is my working code. change parameter as your data.

PoJo class

public class AllAppModel {
    private String appName;
    private String packageName;
    private int uid;
    private boolean isSelected;
    private boolean isSystemApp;
    private boolean isFav;
}

Extract your data using below line of code

try {
    JSONArray jsonArr = new JSONArray("Your json string array");
    List<AllAppModel> lstExtrextData = new ArrayList<>();
    for (int i = 0; i < jsonArr.length(); i++) {
        JSONObject jsonObj = jsonArr.getJSONObject(i);
        AllAppModel data = new AllAppModel();
        data.setAppName(jsonObj.getString("appName"));
        data.setPackageName(jsonObj.getString("packageName"));
        data.setUid(jsonObj.getInt("uid"));
        data.setSelected(jsonObj.getBoolean("isSelected"));
        data.setSystemApp(jsonObj.getBoolean("isSystemApp"));
        data.setFav(jsonObj.getBoolean("isFav"));
        lstExtrextData.add(data);
    }
} catch (JSONException e) {
    e.printStackTrace();
}

it will return you List of PoJo class object.

score 2 · Answer 11 · answered May 29 '20 at 17:24

Just going by the original subject of the thread:

converting jsonarray to list (used jackson jsonarray and object mapper here):

ObjectMapper mapper = new ObjectMapper();
JSONArray array = new JSONArray();
array.put("IND");
array.put("CHN");
List<String> list = mapper.readValue(array.toString(), List.class);

score 2 · Answer 12 · answered Mar 11 '21 at 13:32

If the Json object consists an array of strings, then there is a way where we need not even use the JSONArray.

if the Json is not in string format, we can convert the json into a string. Now if Json string is:

String value = "[\"value1\",\"value2\"]";

Then we can use to get the List of String:

List<String> result = new Gson().fromJson(value, List.class);

score 2 · Answer 13 · answered May 26 '21 at 09:38

2

with GSON and Kotlin you just need this:

    val arr = Gson().fromJson(jsonArrayInString, Array<T>::class.java)

answered May 26 '21 at 09:38

avisper

878
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score 1 · Answer 14 · answered Oct 11 '19 at 11:39

Using Gson

    List<Student> students = new ArrayList<>();
    JSONArray jsonArray = new JSONArray(stringJsonContainArray);
    for (int i = 0; i < jsonArray.length(); i++) {
        Student student = new Gson().fromJson(jsonArray.get(i).toString(), Student.class);
        students.add(student);
    }
    return students;

score 1 · Answer 15 · answered Jul 14 '20 at 11:54

1

Java 8 style

   JSONArray data = jsonObject.getJSONArray("some-node");

   List<JSONObject> list = StreamSupport.stream(data.spliterator(), false)
                .map(e -> (JSONObject)e)
                .collect(Collectors.toList());

answered Jul 14 '20 at 11:54

Min Hyoung Hong

1,102
9
13

Not working... Getting error - `Type mismatch: cannot convert from Object to List` – Chetan Oswal Feb 11 '22 at 15:32

score 1 · Answer 16 · answered May 21 '21 at 18:36

Single line code that I used to convert json array to ArrayList for the below listed json string stored in a variable named classData.

List<String> names = new Gson().fromJson(classData.getAsJsonObject("result").getAsJsonArray("names").toString(), new TypeToken<List<String>>(){}.getType());

{   
    "result": {
        "id": "104297",
        "class": "Chemistry",
        "total": 57,
        "names": [
            "Alex",
            "Dan",
            "Rex",
            "Tony",
            "Elizabeth"
        ],
        ...
    }
}

score 0 · Answer 17 · answered Nov 09 '16 at 14:30

public static List<JSONObject> getJSONObjectListFromJSONArray(JSONArray array) 
        throws JSONException {
  ArrayList<JSONObject> jsonObjects = new ArrayList<>();
  for (int i = 0; 
           i < (array != null ? array.length() : 0);           
           jsonObjects.add(array.getJSONObject(i++))
       );
  return jsonObjects;
}

yoAlex5 · Answer 18 · 2020-04-15T15:47:21.073

0

Generic variant

public static <T> List<T> getList(JSONArray jsonArray) throws Exception {

    List<T> list = new ArrayList<>(jsonArray.length());

    for (int i = 0; i < jsonArray.length(); i++) {

        list.add((T)jsonArray.get(i));
    }

    return list;

}

//Usage

List<String> listKeyString = getList(dataJsonObject.getJSONArray("keyString"));

edited Apr 15 '20 at 15:47

answered Aug 06 '19 at 15:20

yoAlex5

29,217
8
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score 0 · Answer 19 · answered May 21 '20 at 05:47

0

ArrayList<String> listdata = new ArrayList<String>();     
JSONArray jArray = (JSONArray)jsonObject; 
if (jArray != null) { 
 listdata.addAll(jArray);
}

@simplified

answered May 21 '20 at 05:47

naveen sunkavalli

1
1

score 0 · Answer 20 · answered Aug 07 '20 at 13:14

0

A simpler Java 8 alternative:

JSONArray data = new JSONArray(); //create data from this -> [{"thumb_url":"tb-1370913834.jpg","event_id":...}]

List<JSONObject> list = data.stream().map(o -> (JSONObject) o).collect(Collectors.toList());

answered Aug 07 '20 at 13:14

Vivek Dhayaal

1
1

Tried this. Does not work. Getting error - `Type mismatch: cannot convert from Object to List` – Chetan Oswal Feb 11 '22 at 15:34

score 0 · Answer 21 · answered Feb 04 '21 at 18:54

0

With Kotlin, you can avoid a loop by wrapping the JSONArray with a MutableList, e.g.

val artistMetadata = player.metadata.optJSONArray("artist")
val artists = MutableList<String>(artistMetadata.length()) { i -> artistMetadata.getString(i)}

answered Feb 04 '21 at 18:54

Phoenix87

1,003
11
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score 0 · Answer 22 · answered Mar 05 '21 at 01:28

0

public static List<String> convertJsonArrayToStringList(JsonArray ja){
    
    List<String> result = new ArrayList<String>();     
    for(JsonElement je: ja)
        result.add(je.getAsString());
    
    return result;
    
}

answered Mar 05 '21 at 01:28

aianitro

868
11
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score 0 · Answer 23 · answered Nov 20 '21 at 17:58

This can do (Java >= 8):

    // Functional (variable)
    Function<JSONArray, ArrayList<String>> convert = jsonArray -> (ArrayList<String>) jsonArray.stream()
        .map(item -> item.toString())
        .collect(Collectors.toList());

    // Call
    convert.apply(jsonArray);


    // Regular method (OOP)
    private Function<JSONArray, ArrayList<String>> convert() {
        return jsonArray -> (ArrayList<String>) jsonArray.stream()
            .map(item -> item.toString())
            .collect(Collectors.toList());
   }

   // Call
   convert().apply(jsonArray);

Converting JSONarray to ArrayList

23 Answers23

PoJo class

Extract your data using below line of code

Linked

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