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: stackoverflow.com/questions/43233898/… — Vadzim
– Vadzim, Commented Dec 19, 2017 at 16:20

khateeb · Accepted Answer · 2021-11-23 12:02:10Z

565

use JsonParser; for example:

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

edited Nov 23, 2021 at 12:02

khateeb

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answered Dec 24, 2010 at 15:27

Dallan Quass

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12 Comments

Blundell Over a year ago

ugh should have a static 1 liner convenience method

Chriss Over a year ago

the cast to JsonObject is unnecessary, better use new JsonParser().parse(..).getAsJsonObject();

Jatin Sehgal Over a year ago

I guess JsonParser is an abstract class

Ninja Over a year ago

@KevinMeredith you link is broken ,use this please

Michael Röhrig Over a year ago

Note that this method is now deprecated. Use JsonParser.parseString(str).getAsJsonObject().

|

Ram Ghadiyaram · Accepted Answer · 2018-11-30 22:58:04Z

151

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, 2018 at 22:58

Ram Ghadiyaram

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answered Jun 2, 2011 at 18:35

maverick

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3 Comments

Hossain Khan Over a year ago

For some reason it wraps with members parent key. Here is a sample { "members" : { "key1" : "13756963814f2c594822982c0307fb81", "key2" : true, "key3" : 123456789 } }

Rubin Yoo Over a year ago

Use the latest gson library, like 2.2.4. The version like 2.2.2 adds members tag for some reason.

Jimmy Garpehäll Over a year ago

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();

Purushotham · Accepted Answer · 2014-03-02 13:08:22Z

62

String jsonStr = "{\"a\": \"A\"}"; Gson gson = new Gson(); JsonElement element = gson.fromJson (jsonStr, JsonElement.class); JsonObject jsonObj = element.getAsJsonObject();

edited Mar 2, 2014 at 13:08

answered Jun 20, 2013 at 17:53

Purushotham

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4 Comments

LOG_TAG Over a year ago

can you validate my answer with GSON way for convert List data to jsonobject by gson stackoverflow.com/questions/18442452/…

Purushotham Over a year ago

I have validated your answer.

Purushotham Over a year ago

@knoxxs, You mean JsonObject class definition? It comes from Google's Gson library. You can refer the documentation here.

clapas Over a year ago

This gives me an error complaining about JsonElement not having a no-arg constructor.

Jozef Benikovský · Accepted Answer · 2015-09-14 09:06:15Z

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();

dikkini · Accepted Answer · 2017-01-11 21:58:34Z

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 · Accepted Answer · 2021-03-02 15:25:42Z

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

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

Eduardo Chico · Accepted Answer · 2014-08-17 07:58:23Z

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 · Accepted Answer · 2014-04-29 18:40:00Z

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());

Maddy · Accepted Answer · 2015-07-04 16:41:07Z

1

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

answered Jul 4, 2015 at 16:41

Maddy

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2 Comments

Maddy Over a year ago

Above is the easiest way to convert your key-value data to gson object.

7zark7 Over a year ago

Thanks, in my case I had an unparsed JSON string which I needed to start from.

Hossein Kurd · Accepted Answer · 2021-02-04 07:25:42Z

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();

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