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?
Try to use getAsJsonObject()
instead of a straight cast used in the accepted answer:
JsonObject o = new JsonParser().parse("{\"a\": \"A\"}").getAsJsonObject();
String jsonStr = "{\"a\": \"A\"}";
Gson gson = new Gson();
JsonElement element = gson.fromJson (jsonStr, JsonElement.class);
JsonObject jsonObj = element.getAsJsonObject();
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();
//import com.google.gson.JsonObject;
JsonObject complaint = new JsonObject();
complaint.addProperty("key", "value");
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.
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();
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());
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();
The JsonParser
constructor has been deprecated. Use the static method instead:
JsonObject asJsonObject = JsonParser.parseString(request.schema).getAsJsonObject();
Source: Stackoverflow.com