How to get the underlying String from a JsonParser (Jackson Json)

Question

Looking through the documentation and source code I don\'t see a clear way to do this. Curious if I\'m missing something.

Say I receive an InputStream from a server resp

Answer 1

Building my own Deserialiser in which I wanted to deserialise a specific field as text i.s.o. a proper DTO, this is the solution I came up with.

I wrote my own JsonToStringDeserializer like this:

import com.fasterxml.jackson.core.JsonParser;
import com.fasterxml.jackson.core.TreeNode;
import com.fasterxml.jackson.databind.DeserializationContext;
import com.fasterxml.jackson.databind.JsonDeserializer;
import lombok.NoArgsConstructor;
import org.apache.commons.lang3.StringEscapeUtils;

import java.io.IOException;

/**
 * Deserialiser to deserialise any Json content to a String.
 */
@NoArgsConstructor
public class JsonToStringDeserializer extends JsonDeserializer {


    /**
     * Deserialise a Json attribute that is a fully fledged Json object, into a {@link String}.
     * @param jsonParser Parsed used for reading JSON content
     * @param context Context that can be used to access information about this deserialization activity.
     * @return The deserialized value as a {@link String}.
     * @throws IOException
     */
    @Override
    public String deserialize(JsonParser jsonParser, DeserializationContext context) throws IOException {

        final TreeNode node = jsonParser.getCodec().readTree(jsonParser);

        final String unescapedString = StringEscapeUtils.unescapeJava(node.toString());
        return unescapedString.substring(1, unescapedString.length()-1);
    }
}

Annotate the field you want to deserialize like this:

@JsonDeserialize(using = JsonToStringDeserializer.class)

I initially followed advice that said to use a TreeNode like this:

    final TreeNode treeNode = jsonParser.getCodec().readTree(jsonParser);
    return treeNode.toString();

But then you get a Json String that contains escape characters.