Gson: Directly convert String to JsonObject (no POJO)

Question

Can't seem to figure this out. I'm attempting JSON tree manipulation in GSON, but I have a case where I do not know or have a POJO to convert a string into, prior to converting to JsonObject. Is there a way to go directly from a String to JsonObject?

I've tried the following (Scala syntax):

val gson = (new GsonBuilder).create

val a: JsonObject = gson.toJsonTree("""{ "a": "A", "b": true }""").getAsJsonObject
val b: JsonObject = gson.fromJson("""{ "a": "A", "b": true }""", classOf[JsonObject])

but a fails, the JSON is escaped and parsed as a JsonString only, and b returns an empty JsonObject.

Any ideas?

Beware of gson validation pitfalls: https://stackoverflow.com/questions/43233898/how-to-check-if-json-is-valid-in-java-using-gson/47890960#47890960 — Vadzim, Dec 19 '17 at 16:20

score 559 · Accepted Answer · edited Nov 23 '21 at 12:02

559

use JsonParser; for example:

JsonObject o = JsonParser.parseString("{\"a\": \"A\"}").getAsJsonObject();

edited Nov 23 '21 at 12:02

khateeb

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answered Dec 24 '10 at 15:27

Dallan Quass

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ugh should have a static 1 liner convenience method – Blundell Dec 06 '13 at 12:36
no static example, but plenty of one liner examples below – Yaniv May 04 '14 at 11:10
44

the cast to JsonObject is unnecessary, better use `new JsonParser().parse(..).getAsJsonObject();` – Chriss Jul 18 '14 at 11:46
I think [...].getAsJsonObject() is the right implementation. Casting seems like an overkill to me and not too clean. – FeleMed Oct 23 '14 at 18:30
Note that it can throw 3 exceptions per the [docs](http://google-gson.googlecode.com/svn/trunk/gson/docs/javadocs/com/google/gson/JsonParser.html#parse(com.google.gson.stream.JsonReader)) – Kevin Meredith Dec 05 '14 at 22:12
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I guess JsonParser is an abstract class – Jatin Sehgal Mar 30 '15 at 11:08
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@KevinMeredith you link is broken ,use [this](https://static.javadoc.io/com.google.code.gson/gson/2.6.2/com/google/gson/JsonParser.html#parse-java.lang.String-) please – Ninja Feb 07 '17 at 06:54
Dallan Quass - I have a class with many static methods which can make different api calls. Should I have one instance of JsonParser for the class or should I create one in each static method ? – MasterJoe May 17 '19 at 18:01
@MasterJoe2 I'm hoping others will answer your question, or maybe you want to post it as a new question? I haven't been around java for awhile. – Dallan Quass May 23 '19 at 02:34
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Note that this method is now deprecated. Use `JsonParser.parseString(str).getAsJsonObject()`. – Michael Röhrig Nov 29 '19 at 13:45
In my case, i needed it as string, so i used this: `String data = new Gson().toJson(JsonParser.parseString("{type: someValue}"));` – Neoraptor Mar 18 '20 at 04:27
This seems like it could be a performance hit if you have to parse the string when you already have knowledge that it's a `JsonPrimitve`. But I don't see another way. – Sridhar Sarnobat Aug 25 '22 at 00:26

score 149 · Answer 2 · edited Nov 30 '18 at 22:58

149

Try to use getAsJsonObject() instead of a straight cast used in the accepted answer:

JsonObject o = new JsonParser().parse("{\"a\": \"A\"}").getAsJsonObject();

edited Nov 30 '18 at 22:58

Ram Ghadiyaram

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answered Jun 02 '11 at 18:35

maverick

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For some reason it wraps with `members` parent key. Here is a sample { "members" : { "key1" : "13756963814f2c594822982c0307fb81", "key2" : true, "key3" : 123456789 } } – Hossain Khan Nov 15 '13 at 20:01
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Use the latest gson library, like 2.2.4. The version like 2.2.2 adds members tag for some reason. – Rubin Yoo Feb 13 '15 at 23:12
5

JsonParser().parse() is deprecated in newer versions of Gson. Use `JsonObject jsonObj = JsonParser.parseString(str).getAsJsonObject()`or `Gson gson = new Gson(); JsonElement element = gson.fromJson (jsonStr, JsonElement.class); JsonObject jsonObj = element.getAsJsonObject();` – Jimmy Garpehäll Dec 06 '19 at 09:26

Purushotham · Answer 3 · 2014-03-02T13:08:22.367

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String jsonStr = "{\"a\": \"A\"}";

Gson gson = new Gson();
JsonElement element = gson.fromJson (jsonStr, JsonElement.class);
JsonObject jsonObj = element.getAsJsonObject();

edited Mar 02 '14 at 13:08

answered Jun 20 '13 at 17:53

Purushotham

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can you validate my answer with GSON way for convert List data to jsonobject by gson http://stackoverflow.com/questions/18442452/create-an-array-of-json-objects/18443395#18443395 – LOG_TAG Aug 26 '13 at 11:59
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I have validated your answer. – Purushotham Feb 07 '14 at 05:47
@knoxxs, You mean `JsonObject` class definition? It comes from Google's Gson library. You can refer the documentation [here](http://google-gson.googlecode.com/svn/trunk/gson/docs/javadocs/index.html). – Purushotham Apr 09 '15 at 10:31
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This gives me an error complaining about JsonElement not having a no-arg constructor. – clapas May 28 '15 at 14:19

Jozef Benikovský · Answer 4 · 2015-09-14T09:06:15.427

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The simplest way is to use the JsonPrimitive class, which derives from JsonElement, as shown below:

JsonElement element = new JsonPrimitive(yourString);
JsonObject result = element.getAsJsonObject();

edited Sep 14 '15 at 09:06

answered Feb 28 '14 at 00:02

Jozef Benikovský

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This is the simplest answer and helped me out. Thanks! – khiner Dec 20 '14 at 22:32

score 11 · Answer 5 · edited Jan 11 '17 at 21:58

Just encountered the same problem. You can write a trivial custom deserializer for the JsonElement class:

import com.google.gson.Gson;
import com.google.gson.GsonBuilder;
import com.google.gson.JsonElement;
import com.google.gson.JsonObject;

GsonBuilder gson_builder = new GsonBuilder();
gson_builder.registerTypeAdapter(
        JsonElement.class,
        new JsonDeserializer<JsonElement>() {
            @Override
            public JsonElement deserialize(JsonElement arg0,
                    Type arg1,
                    JsonDeserializationContext arg2)
                    throws JsonParseException {

                return arg0;
            }
        } );
String str = "{ \"a\": \"A\", \"b\": true }";
Gson gson = gson_builder.create();
JsonElement element = gson.fromJson(str, JsonElement.class);
JsonObject object = element.getAsJsonObject();

Stephen Paul · Answer 6 · 2021-03-02T15:25:42.460

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The JsonParser constructor has been deprecated. Use the static method instead:

JsonObject asJsonObject = JsonParser.parseString(someString).getAsJsonObject();

edited Mar 02 '21 at 15:25

answered Oct 25 '19 at 10:30

Stephen Paul

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score 4 · Answer 7 · answered Aug 17 '14 at 07:58

I believe this is a more easy approach:

public class HibernateProxyTypeAdapter implements JsonSerializer<HibernateProxy>{

    public JsonElement serialize(HibernateProxy object_,
        Type type_,
        JsonSerializationContext context_) {
        return new GsonBuilder().create().toJsonTree(initializeAndUnproxy(object_)).getAsJsonObject();
        // that will convert enum object to its ordinal value and convert it to json element

    }

    public static <T> T initializeAndUnproxy(T entity) {
        if (entity == null) {
            throw new 
               NullPointerException("Entity passed for initialization is null");
        }

        Hibernate.initialize(entity);
        if (entity instanceof HibernateProxy) {
            entity = (T) ((HibernateProxy) entity).getHibernateLazyInitializer()
                    .getImplementation();
        }
        return entity;
    }
}

And then you will be able to call it like this:

Gson gson = new GsonBuilder()
        .registerTypeHierarchyAdapter(HibernateProxy.class, new HibernateProxyTypeAdapter())
        .create();

This way all the hibernate objects will be converted automatically.

c2willi · Answer 8 · 2014-04-29T18:40:00.393

Came across a scenario with remote sorting of data store in EXTJS 4.X where the string is sent to the server as a JSON array (of only 1 object).
Similar approach to what is presented previously for a simple string, just need conversion to JsonArray first prior to JsonObject.

String from client: [{"property":"COLUMN_NAME","direction":"ASC"}]

String jsonIn = "[{\"property\":\"COLUMN_NAME\",\"direction\":\"ASC\"}]";
JsonArray o = (JsonArray)new JsonParser().parse(jsonIn);

String sortColumn = o.get(0).getAsJsonObject().get("property").getAsString());
String sortDirection = o.get(0).getAsJsonObject().get("direction").getAsString());

score 1 · Answer 9 · answered Jul 04 '15 at 16:41

1

//import com.google.gson.JsonObject;  
JsonObject complaint = new JsonObject();
complaint.addProperty("key", "value");

answered Jul 04 '15 at 16:41

Maddy

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Above is the easiest way to convert your key-value data to gson object. – Maddy Jul 04 '15 at 16:42
1

Thanks, in my case I had an unparsed JSON string which I needed to start from. – 7zark7 Jul 04 '15 at 18:45

score 1 · Answer 10 · answered Feb 04 '21 at 07:25

com.google.gson.JsonParser#parse(java.lang.String) is now deprecated

so use com.google.gson.JsonParser#parseString, it works pretty well

Kotlin Example:

val mJsonObject = JsonParser.parseString(myStringJsonbject).asJsonObject

Java Example:

JsonObject mJsonObject = JsonParser.parseString(myStringJsonbject).getAsJsonObject();

Gson: Directly convert String to JsonObject (no POJO)

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