How to create array in json object

Question

I am using something like this-

String Number1=to_1;
        String Number2=to_2;
        String[] arrayNumbers = new String[] {(char)34+Number1+(char)34,(char)34+Number2+(char)34};
        System.out.println(Arrays.toString(arrayNumbers));
        JSONObject jsonObj = new JSONObject();
        jsonObj.put("to",Arrays.toString(arrayNumbers));
        jsonObj.put("type",type);
        jsonObj.put("callback",callbackUrl);
        JSONArray array = new JSONArray();
        JSONObject Array_item = new JSONObject();
        jsonObj.put(type, array);
        Array_item.put("caption",captionName);
        array.add(Array_item);

        System.out.println(jsonObj.toString());

Expected-

{
  "to":["91890xx", "91890xx"],
 "type": "document", "document" : {"caption" : "doc"},
"callback":"{{callback}}"
}

Actual- {"document":[{"caption":"hello"}],"callback":"{{callback}}","to":"[\"91890xxx\", \"91890xx\"]","type":"document"}

I don't know any more logic to remove the out double quotes of to number where as its considering that to a string where as both the numbers should be array format as mentioned in Expected.

Why don't you use just jsonObj.put("to",new JSONArray().put("").put(""))? — Ehcnalb
– Ehcnalb, Commented Mar 11, 2019 at 9:50

Allen Kerr · Accepted Answer · 2019-03-14 12:23:21Z

First, I show you the correct codes:

import org.json.JSONArray;
import org.json.JSONException;
import org.json.JSONObject;

public class Test {
  /**
   * <pre>
   *  {
   *    "to":["91890xx", "91890xx"],
   *    "type": "document",
   *    "document" : {"caption" : "doc"},
   *    "callback":"{{callback}}"
   *  }
   * </pre>
   *
   * @param args
   * @throws JSONException
   */
   public static void main(String[] args) throws JSONException {
    String number1 = "91890";
    String number2 = "91890";
    String[] numbers = new String[]{number1, number2};

    JSONArray toNode = new JSONArray();
    for (String number : numbers) {
      toNode.put(number);
    }

    JSONObject jsonObj = new JSONObject();
    jsonObj.put("to", toNode);
    jsonObj.put("type", "document");
    jsonObj.put("document", new JSONObject().put("caption", "doc"));
    jsonObj.put("callback", "{{callback}}");

    System.out.println(jsonObj.toString());
  }
}

Result:

{"document":{"caption":"doc"},"callback":"{{callback}}","to":["91890","91890"],"type":"document"}

If you want create a josn array node, you show use JSONArray, and use JSONArray#put(*) method to add elements.

Put string into JSONArray or JSONObject, you don't need wrap the string with quote("). Besides, you should write \" instead of (char) 34 which is a little obscure in Java.

The following case is used to reply comments.

import org.json.JSONArray;
import org.json.JSONException;
import org.json.JSONObject;

import java.math.BigDecimal;
import java.net.URI;
import java.net.URL;
import java.util.*;

public class Test1 {

  public static void main(String[] args) throws JSONException {
    Map<String, Object> map = new HashMap<>();
    map.put("key1", "value1");
    map.put("key2", new Date());
    map.put("key3", 1);
    map.put("key4", null);
    map.put("key5", Collections.singletonMap("key5-key1", "value"));
    map.put("key6", Arrays.asList(1, 2, 3, 4));
    map.put("key7", BigDecimal.TEN);
    map.put("key8", new String[]{"a", "b", "c"});
    map.put("key9", TestEnum.A);
    map.put("key10", new TestEnum[]{TestEnum.A, TestEnum.B, TestEnum.C});

    Object json = buildJsonObj(map);
    System.out.println(json);
  }

  private static Object buildJsonObj(Object source) throws JSONException {
    if (source == null) {
      return null;
    }
    if (isSimpleValueType(source.getClass())) {
      return source;
    }

    if (source instanceof Map) {
      Map<Object, Object> map = (Map<Object, Object>) source;
      JSONObject jsonObject = new JSONObject();
      for (Map.Entry<Object, Object> entry : map.entrySet()) {
        Object key = entry.getKey();
        if (!(key instanceof String)) {
          throw new IllegalArgumentException("key must be string.");
        }
        jsonObject.put((String) key, buildJsonObj(entry.getValue()));
      }
      return jsonObject;
    }
    if (source instanceof Iterable) {
      Iterable<Object> iterable = (Iterable<Object>) source;
      JSONArray jsonArray = new JSONArray();
      for (Object value : iterable) {
        jsonArray.put(buildJsonObj(value));
      }
      return jsonArray;
    }
    if (source.getClass().isArray()) {
      Object[] array = (Object[]) source;
      JSONArray jsonArray = new JSONArray();
      for (Object value : array) {
        jsonArray.put(buildJsonObj(value));
      }
      return jsonArray;
    }

    throw new IllegalArgumentException("Unsupported type: " + source + ".");
  }

  private static boolean isSimpleValueType(Class<?> clazz) {
    return (Enum.class.isAssignableFrom(clazz) ||
        CharSequence.class.isAssignableFrom(clazz) ||
        Number.class.isAssignableFrom(clazz) ||
        Date.class.isAssignableFrom(clazz) ||
        URI.class == clazz || URL.class == clazz ||
        Locale.class == clazz);
  }

  public enum TestEnum {
    A, B, C
  }
}

Result:

{"key1":"value1","key2":"Thu Mar 14 20:20:49 CST 2019","key5":{"key5-key1":"value"},"key6":[1,2,3,4],"key3":1,"key9":"A","key7":10,"key8":["a","b","c"],"key10":["A","B","C"]}

I have question, What if I the same function needs to be used for one number also. Because this looks like hard coded.
You can define a function to do that. The function's param can be map and list/array. And you use JSONObject to process map, JSONArray to process list/array. Recursive algorithm should be used in those functions.

Santosh b · Accepted Answer · 2019-03-11 10:01:04Z

1

Replace line

jsonObj.put("to",Arrays.toString(arrayNumbers));

to

jsonObj.put("to", Arrays.asList(arrayNumbers));

answered Mar 11, 2019 at 10:01

Santosh b

8062 gold badges9 silver badges21 bronze badges

Comments

Bheem Singh · Accepted Answer · 2019-03-11 09:47:40Z

0

Your value is String of 'to',so can you please convert string to json after that you can convert json to array.

answered Mar 11, 2019 at 9:47

Bheem Singh

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Comments

Tran Dung · Accepted Answer · 2019-03-11 09:50:57Z

0

Replace

jsonObj.put("to",Arrays.toString(arrayNumbers));

With

jsonObj.put("to",arrayNumbers);

answered Mar 11, 2019 at 9:50

Tran Dung

11 bronze badge

Comments

ricksclick · Accepted Answer · 2019-03-11 09:51:31Z

0

let a = {"document":[{"caption":"hello"}],"callback":"{{callback}}","to":"[\"91890xxx\", \"91890xx\"]","type":"document"};

    a = JSON.parse(JSON.stringify(a));

    let b = JSON.parse(a.to);

    a.to = b;

    console.log(a);

answered Mar 11, 2019 at 9:51

ricksclick

215 bronze badges

Comments

Bathri Nathan · Accepted Answer · 2019-03-11 11:46:18Z

0

This is the exact structure of you required json.

 {
        "to": ["91890xx", "91890xx"],
        "type": "document",
        "document": {
            "caption": "doc"
        },
        "callback": "{{callback}}"
    }

To Remove the double quotes, please use replace function as shown below

String  Number1="[\"91890xxx\", \"91890xx\"]";
        Number1.replace("\\/", "");
        System.out.println("Excepted output:"+Number1);
output:

Excepted output:["91890xxx", "91890xx"]

edited Mar 11, 2019 at 11:46

answered Mar 11, 2019 at 9:50

Bathri Nathan

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