Java: Parsing JSON with unknown keys to Map

Question

In Java, I need to consume JSON (example below), with a series of arbitrary keys, and produce Map<String, String>. I'd like to use a standard, long term supported JSON library for the parsing. My research, however, shows that these libraries are setup for deserialization to Java classes, where you know the fields in advance. I need just to build Maps.

It's actually one step more complicated than that, because the arbitrary keys aren't the top level of JSON; they only occur as a sub-object for prefs. The rest is known and can fit in a pre-defined class.

{
    "al" : { "type": "admin", "prefs" : { "arbitrary_key_a":"arbitary_value_a", "arbitrary_key_b":"arbitary_value_b"}},
    "bert" : {"type": "user", "prefs" : { "arbitrary_key_x":"arbitary_value_x", "arbitrary_key_y":"arbitary_value_y"}},
    ...
}

In Java, I want to be able to take that String, and do something like:

people.get("al").get("prefs"); // Returns Map<String, String>

How can I do this? I'd like to use a standard well-supported parser, avoid exceptions, and keep things simple.

UPDATE

@kumensa has pointed out that this is harder than it looks. Being able to do:

people.get("al").getPrefs(); // Returns Map<String, String>
people.get("al").getType();  // Returns String

is just as good.

That should parse the JSON to something like:

public class Person {
    public String type;
    public HashMap<String, String> prefs;
}
// JSON parsed to:
HashMap<String, Person>

It looks like your people json property should be an array. — Michał Krzywański
– Michał Krzywański, Commented May 10, 2019 at 7:25
But what about people.get("al").get("type");? It cannot return a Map. You'd need an arbitrary tree node and checking content types by yourself. Now if it was more about people.get("al").getPrefs() that returns a Map, it's doable. — kumesana
– kumesana, Commented May 10, 2019 at 7:27
@kumensa people.get("al").getPrefs() would be fully acceptable — SRobertJames
– SRobertJames, Commented May 10, 2019 at 7:32

Abrikot · Accepted Answer · 2019-05-10 08:19:42Z

1

Having your Person class and using Gson, you can simply do:

final Map<String, Person> result = new Gson().fromJson(json, new TypeToken<Map<String, Person>>() {}.getType());

Then, retrieving prefs is achieved with people.get("al").getPrefs();.

But be careful: your json string is not valid. It shouldn't start with "people:".

answered May 10, 2019 at 8:19

Abrikot

1,0051 gold badge10 silver badges22 bronze badges

Sign up to request clarification or add additional context in comments.

Comments

Oleg Cherednik · Accepted Answer · 2019-05-10 07:34:13Z

0

public static <T> Map<String, T> readMap(String json) {
    if (StringUtils.isEmpty(json))
        return Collections.emptyMap();

    ObjectReader reader = new ObjectMapper().readerFor(Map.class);
    MappingIterator<Map<String, T>> it = reader.readValues(json);

    if (it.hasNextValue()) {
        Map<String, T> res = it.next();
        return res.isEmpty() ? Collections.emptyMap() : res;
    }

    return Collections.emptyMap();
}

All you need to do next, it that check the type of the Object. If it is Map, then you have an object. Otherwise, this is a simple value.

edited May 10, 2019 at 7:34

answered May 10, 2019 at 7:28

Oleg Cherednik

18.3k4 gold badges25 silver badges38 bronze badges

Comments

azhar · Accepted Answer · 2019-05-10 08:08:29Z

0

You can use Jackson lib to achieve this. Put the following in pom.xml.

<dependency>
    <groupId>com.fasterxml.jackson.core</groupId>
    <artifactId>jackson-databind</artifactId>
    <version>2.9.8</version>
</dependency>

Refer the following snippet that demonstrates the same.

ObjectMapper mapper = new ObjectMapper();
HashMap<String, Object> people = mapper.readValue(jsonString, new TypeReference<HashMap>(){});

Now, it is deserialized as a Map;

Full example:

import java.io.IOException;
import java.util.HashMap;

import com.fasterxml.jackson.core.JsonParseException;
import com.fasterxml.jackson.core.type.TypeReference;
import com.fasterxml.jackson.databind.JsonMappingException;
import com.fasterxml.jackson.databind.ObjectMapper;

public class testMain {

    public static void main(String[] args) throws JsonParseException, JsonMappingException, IOException {

        String json = "{\"address\":\"3, 43, Cashier Layout, Tavarekere Main Road, 1st Stage, BTM Layout, Ambika Medical, 560029\",\"addressparts\":{\"apartment\":\"Cashier Layout\",\"area\":\"BTM Layout\",\"floor\":\"3\",\"house\":\"43\",\"landmark\":\"Ambika Medical\",\"pincode\":\"560029\",\"street\":\"Tavarekere Main Road\",\"subarea\":\"1st Stage\"}}";

        ObjectMapper mapper = new ObjectMapper();
        HashMap<String, Object> people = mapper.readValue(json, new TypeReference<HashMap>(){});

        System.out.println(((HashMap<String, String>)people.get("addressparts")).get("apartment"));
    }

}

Output: Cashier Layout

edited May 10, 2019 at 8:08

answered May 10, 2019 at 7:25

azhar

16712 bronze badges

2 Comments

kumesana Over a year ago

Ah, yes. The famous get() method of String. (No you can't do that)

azhar Over a year ago

Sorry, my bad I missed that.

Collectives™ on Stack Overflow

Java: Parsing JSON with unknown keys to Map

UPDATE

3 Answers 3

Comments

Comments

2 Comments

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

UPDATE

3 Answers 3

Comments

Comments

2 Comments

Your Answer

Sign up or log in

Post as a guest

Related