How to convert Object type to HashMap

Question

I'm trying to parse JSON my own way and I'm stuck on converting List to Map:

   public static void main ( final String[] args ) throws Exception
   {
      {
         String test = new String("[{\"label\": 1, \"value\": 12345}, {\"label\": 2, \"value\": 12}]");
         final StringReader stringReader = new StringReader(test);
         final Object obj = new JSONReader( stringReader ).parseData();
         List list = (List)obj;  // [{label=1, value=12345}, {label=2, value=12}] looks good
         System.out.println(list);
         for (Object o : list) {
             System.out.println(o);  // {label=1, value=12345} {label=2, value=12} also looks ok
                     if (!(o instanceof Map)) {
            throw new IllegalArgumentException("Root object must be a JSON object.");
        }
         }
      }

The following code doesn't throw exception thus the o is instance of Map. However I can't use the o.get("label") to retrieve the label value. How can I get every value from label and value field?

Kemal Kaplan · Accepted Answer · 2021-08-23 09:18:13Z

1

Actually it works with objectMapper

@Test
void t2() throws IOException {
    String test = new String("[{\"label\": 1, \"value\": 12345}, {\"label\": 2, \"value\": 12}]");
    final StringReader stringReader = new StringReader(test);
    final Object obj = new ObjectMapper().readValue(stringReader, Object.class);
    List list = (List)obj;  // [{label=1, value=12345}, {label=2, value=12}] looks good
    System.out.println(list);
    for (Object o : list) {
        System.out.println(o);  // {label=1, value=12345} {label=2, value=12} also looks ok
        if (!(o instanceof Map)) {
            throw new IllegalArgumentException("Root object must be a JSON object.");
        }
        System.out.println(((Map)o).get("label"));
    }
}

The output is;

[{label=1, value=12345}, {label=2, value=12}]
{label=1, value=12345}
1
{label=2, value=12}
2

This may be due to the JsonReader that you are using. I cannot figure out the parser with parseData method...

answered Aug 23, 2021 at 9:18

Kemal Kaplan

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1 Comment

Robert Kwiatkowski Over a year ago

Okay, I just forgot about explicit conversion (Map)o. I just got pretty tired and I'm not using java too often (almost never). Thank you!

Renis1235 · Accepted Answer · 2021-08-23 09:24:55Z

1

You should try to remove the keyword Object. That way you are not sure what you are parsing and it can happen that the element that you get is not a list.

I would go with the JSON Java Library. And then parse the value this way:

String test = "[{\"label\": 1, \"value\": 12345}, {\"label\": 2, \"value\": 12}]";
JSONArray jsonArray = new JSONArray(test);

for (Object o : jsonArray)
{
    JSONObject jsonObject = (JSONObject) o;
    System.out.println(jsonObject);
}

As you can see JSONArray and JSONObject can be used for the parsing. And they provide several functions that help on working with them.

answered Aug 23, 2021 at 9:24

Renis1235

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Comments

Andrzej Więcławski · Accepted Answer · 2021-08-23 14:37:10Z

0

My method proposition uses simple org.json library (2009) and is quite different:

private Map<Integer, Integer> convertJsonToMap(String test) throws JSONException {
    Map<Integer, Integer> map = new HashMap<>();
    JSONArray jsonArray = new JSONArray(test);
    for (int i = 0; i < jsonArray.length(); i++) {
        JSONObject jsonObject = jsonArray.getJSONObject(i);
        if (jsonObject.length() == 2) {
            map.put((Integer) jsonObject.get(jsonObject.names().getString(0)),
                    (Integer) jsonObject.get(jsonObject.names().getString(1)));
        }
    }
    return map;
}

Returns only Integers and could throw ClassCastException if any "label" or "value" is String.

answered Aug 23, 2021 at 14:37

Andrzej Więcławski

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