How to parse a Java List of already parsed JSON into a Big JSON?

问题

I use Jackson to serialize/deserialize JSON.

I have a List<String> in which all elements inside are already serialized in JSON format. I would like to generate a big JSON from that List.

In other word, I have:

List<String> a = new ArrayList<>();
a[0] = JSON_0
a[1] = JSON_1
...
a[N] = JSON_N

And I would like to render:

[
   {JSON_0},
   {JSON_1},
   ...
   {JSON_N}
]

What is the best way to do so using Jackson?

回答1:

Probably the simpler solution would be to create ArrayNode and use addRawValue method:

import com.fasterxml.jackson.databind.ObjectMapper;
import com.fasterxml.jackson.databind.node.ArrayNode;
import com.fasterxml.jackson.databind.util.RawValue;

public class JsonApp {

    public static void main(String[] args) throws Exception {
        ObjectMapper mapper = new ObjectMapper();

        ArrayNode nodes = mapper.getNodeFactory().arrayNode();
        nodes.addRawValue(new RawValue("{}"));
        nodes.addRawValue(new RawValue("true"));
        nodes.addRawValue(new RawValue("{\"id\":1}"));

        System.out.println(mapper.writeValueAsString(nodes));
    }
}

Above code prints:

[{},true,{"id":1}]

You can also, create a POJO with list and use @JsonRawValue annotation. But if you can not have extra root object you need to implement custom serialiser for it. Example with POJO and custom serialiser:

import com.fasterxml.jackson.core.JsonGenerator;
import com.fasterxml.jackson.databind.JsonSerializer;
import com.fasterxml.jackson.databind.ObjectMapper;
import com.fasterxml.jackson.databind.SerializerProvider;
import com.fasterxml.jackson.databind.annotation.JsonSerialize;

import java.io.IOException;
import java.util.ArrayList;
import java.util.List;

public class JsonApp {

    public static void main(String[] args) throws Exception {
        ObjectMapper mapper = new ObjectMapper();

        List<String> jsons = new ArrayList<>();
        jsons.add("{}");
        jsons.add("true");
        jsons.add("{\"id\":1}");

        RawJsons root = new RawJsons();
        root.setJsons(jsons);
        System.out.println(mapper.writeValueAsString(root));
    }
}

@JsonSerialize(using = RawJsonSerializer.class)
class RawJsons {

    private List<String> jsons;

    public List<String> getJsons() {
        return jsons;
    }

    public void setJsons(List<String> jsons) {
        this.jsons = jsons;
    }
}

class RawJsonSerializer extends JsonSerializer<RawJsons> {

    @Override
    public void serialize(RawJsons value, JsonGenerator gen, SerializerProvider serializers) throws IOException {
        gen.writeStartArray();
        if (value != null && value.getJsons() != null) {
            for (String json : value.getJsons()) {
                gen.writeRawValue(json);
            }
        }
        gen.writeEndArray();
    }
}

If you need to have SerializationFeature.INDENT_OUTPUT feature enabled for all items in array, you need to deserialise all inner objects and serialise them again.

回答2:

the simple fact of having the character '[' we are marking that it is an array so what I recommend to put the list into a JSON array.

I would need a little more information to help you, since it doesn't make much sense to use a JSON String, since a JSON is composed of Key / Value, it is best to make a bean / object with the attribute. Example:

class Object {
    private String attribute = value;
}

{attribute: value}

来源：https://stackoverflow.com/questions/57731892/how-to-parse-a-java-list-of-already-parsed-json-into-a-big-json

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java

json

serialization

jackson

json-serialization