The org.json library is easy to use.
Just remember (while casting or using methods like getJSONObject and getJSONArray) that in JSON notation
[ … ]represents an array, so library will parse it toJSONArray{ … }represents an object, so library will parse it toJSONObject
Example code below:
import org.json.*;
String jsonString = ... ; //assign your JSON String here
JSONObject obj = new JSONObject(jsonString);
String pageName = obj.getJSONObject("pageInfo").getString("pageName");
JSONArray arr = obj.getJSONArray("posts"); // notice that `"posts": [...]`
for (int i = 0; i < arr.length(); i++)
{
String post_id = arr.getJSONObject(i).getString("post_id");
......
}
You may find more examples from: Parse JSON in Java
Downloadable jar: http://mvnrepository.com/artifact/org.json/json
Answer from user1931858 on Stack OverflowVideos
The org.json library is easy to use.
Just remember (while casting or using methods like getJSONObject and getJSONArray) that in JSON notation
[ … ]represents an array, so library will parse it toJSONArray{ … }represents an object, so library will parse it toJSONObject
Example code below:
import org.json.*;
String jsonString = ... ; //assign your JSON String here
JSONObject obj = new JSONObject(jsonString);
String pageName = obj.getJSONObject("pageInfo").getString("pageName");
JSONArray arr = obj.getJSONArray("posts"); // notice that `"posts": [...]`
for (int i = 0; i < arr.length(); i++)
{
String post_id = arr.getJSONObject(i).getString("post_id");
......
}
You may find more examples from: Parse JSON in Java
Downloadable jar: http://mvnrepository.com/artifact/org.json/json
For the sake of the example lets assume you have a class Person with just a name.
private class Person {
public String name;
public Person(String name) {
this.name = name;
}
}
Jackson (Maven)
My personal favourite and probably the most widely used.
ObjectMapper mapper = new ObjectMapper();
// De-serialize to an object
Person user = mapper.readValue("{\"name\": \"John\"}", Person.class);
System.out.println(user.name); //John
// Read a single attribute
JsonNode nameNode = mapper.readTree("{\"name\": \"John\"}");
System.out.println(nameNode.get("name").asText());
Google GSON (Maven)
Gson g = new Gson();
// De-serialize to an object
Person person = g.fromJson("{\"name\": \"John\"}", Person.class);
System.out.println(person.name); //John
// Read a single attribute
JsonObject jsonObject = new JsonParser().parseString("{\"name\": \"John\"}").getAsJsonObject();
System.out.println(jsonObject.get("name").getAsString()); //John
Org.JSON (Maven)
This suggestion is listed here simply because it appears to be quite popular due to stackoverflow reference to it. I would not recommend using it as it is more a proof-of-concept project than an actual library.
JSONObject obj = new JSONObject("{\"name\": \"John\"}");
System.out.println(obj.getString("name")); //John
Based on the javadoc for Gson 2.8.6
No need to instantiate this class, use the static methods instead.
and following are the alternatives to be used.
// jsonString is of type java.lang.String
JsonObject jsonObject = JsonParser.parseString(jsonString).getAsJsonObject();
// reader is of type java.io.Reader
JsonObject jsonObject = JsonParser.parseReader(reader).getAsJsonObject();
// jsonReader is of type com.google.gson.stream.JsonReader
JsonObject jsonObject = JsonParser.parseReader(jsonReader).getAsJsonObject();
Example
import static org.junit.Assert.assertTrue;
import com.google.gson.JsonObject;
import com.google.gson.JsonParser;
public class Test {
public static void main(String[] args) {
String jsonString = "{ \"name\":\"John\"}";
JsonObject jsonObjectAlt = JsonParser.parseString(jsonString).getAsJsonObject();
// Shows deprecated warning for new JsonParser() and parse(jsonString)
JsonObject jsonObject = new JsonParser().parse(jsonString).getAsJsonObject();
assertTrue(jsonObjectAlt.equals(jsonObject));
}
}
new GsonBuilder().setPrettyPrinting().create().toJson(new JSONParser(jsonString).parse());
If you have the JsonParser then you can use jsonParser.readValueAsTree().toString().
However, this likely requires that the JSON being parsed is indeed valid JSON.
I had a situation where I was using a custom deserializer, but I wanted the default deserializer to do most of the work, and then using the SAME json do some additional custom work. However, after the default deserializer does its work, the JsonParser object current location was beyond the json text I needed. So I had the same problem as you: how to get access to the underlying json string.
You can use JsonParser.getCurrentLocation.getSourceRef() to get access to the underlying json source. Use JsonParser.getCurrentLocation().getCharOffset() to find the current location in the json source.
Here's the solution I used:
public class WalkStepDeserializer extends StdDeserializer<WalkStep> implements
ResolvableDeserializer {
// constructor, logger, and ResolvableDeserializer methods not shown
@Override
public MyObj deserialize(JsonParser jp, DeserializationContext ctxt) throws IOException,
JsonProcessingException {
MyObj myObj = null;
JsonLocation startLocation = jp.getCurrentLocation();
long charOffsetStart = startLocation.getCharOffset();
try {
myObj = (MyObj) defaultDeserializer.deserialize(jp, ctxt);
} catch (UnrecognizedPropertyException e) {
logger.info(e.getMessage());
}
JsonLocation endLocation = jp.getCurrentLocation();
long charOffsetEnd = endLocation.getCharOffset();
String jsonSubString = endLocation.getSourceRef().toString().substring((int)charOffsetStart - 1, (int)charOffsetEnd);
logger.info(strWalkStep);
// Special logic - use JsonLocation.getSourceRef() to get and use the entire Json
// string for further processing
return myObj;
}
}
And info about using a default deserializer in a custom deserializer is at How do I call the default deserializer from a custom deserializer in Jackson