Gson - 使用数组或字符串字段解析 json

Question

I'm trying to parse some json with Gson and I have following problem:

This is my json:

[{
  "label": "Check Digit",
  "value": "W"
},
{
  "label": "Equipment",
  "value": [
    "With Standard Liftgate",
    "(-) Version Packages"
  ]
}]

This is my java class:

public class Decode {

    private String label;
    private List<String> value = new ArrayList<>();

    public String getLabel() {
        return label;
    }

    public void setLabel(String label) {
        this.label = label;
    }

    public List<String> getValue() {
        return value;
    }

    public void setValue(List<String> value) {
        this.value = value;
    }

}

How to make Gson to parse it and use value as Array of String always?

Answer 1

My suggestion is as follows:-

Step 1: Please make following little changes to your Decode class.

Decode.java

import java.util.ArrayList;
import java.util.List;

public class Decode {

private String label;
private List<String> values = new ArrayList<>(); // in json it is "value"

public String getLabel() {
    return label;
}

public void setLabel(String label) {
    this.label = label;
}

public List<String> getValues() {
    return values;
 }

public void setValues(List<String> values) {
    this.values = values;
 }

@Override
public String toString() {
    return "Decode [label=" + label + ", values=" + values + "]";
 }

}

Step 2: Please create following JsonDeserializer .

MyDeserializer.java

import java.lang.reflect.Type;
import java.util.ArrayList;
import java.util.List;

import com.google.gson.Gson;
import com.google.gson.JsonDeserializationContext;
import com.google.gson.JsonDeserializer;
import com.google.gson.JsonElement;
import com.google.gson.JsonObject;
import com.google.gson.JsonParseException;
import com.google.gson.reflect.TypeToken;

public class MyDeserializer implements JsonDeserializer<Decode> {

@Override
public Decode deserialize(JsonElement arg0, Type arg1, JsonDeserializationContext arg2) throws JsonParseException {
    JsonObject decodeObj = arg0.getAsJsonObject();
    Gson gson = new Gson();
    Decode decode = gson.fromJson(arg0, Decode.class);
    List<String> values = null;
    if (decodeObj.get("value").isJsonArray()) {
        values = gson.fromJson(decodeObj.get("value"), new TypeToken<List<String>>() {
        }.getType());
    } else {
        String single = gson.fromJson(decodeObj.get("value"), String.class);
        values = new ArrayList<String>();
        values.add(single);
    }
    decode.setValues(values);
    return decode;
 }

}

Step 3: Now its time to deserialize your json as follows:

GsonMain.java

import java.io.File;
import java.io.IOException;
import java.nio.file.Files;
import java.util.Arrays;

import com.google.gson.GsonBuilder;

public class GsonMain{
public static void main(String[] args) throws IOException {
    String filename = "d:/test.json"; // contains the json
    String content = new String(Files.readAllBytes(new File(filename).toPath()));

    GsonBuilder b = new GsonBuilder();
    b.registerTypeAdapter(Decode.class, new MyDeserializer());
    Decode[] array = b.create().fromJson(content, Decode[].class);
    System.out.println(Arrays.toString(array));
  }

 }

Answer 2

The json is not in proper format, it should be :

[{
  "label": "Check Digit",
  "value": ["W"]
},
{
  "label": "Equipment",
  "value": [
    "With Standard Liftgate",
    "(-) Version Packages"
  ]
}]

Again your parser class should be :

public class Decode {

    private String label;
    private String[] value;

    public String getLabel() {
        return label;
    }

    public void setLabel(String label) {
        this.label = label;
    }

    public String[] getValue() {
        return value;
    }

    public void setValue(String[] value) {
        this.value = value;
    }

}

Hope it will help :)

Answer 3

I had the same problem, but the returned json results were more complex. I just defined those fields with uncertain types as Object type. Hope to help you.

Answer 4

the Decode solution is the cleanest, but if you don't plan on using the model for anything else than deserializing, you could just declare the field as

val myField: Any?

and then to read the value:

when (myField) {
    is List<*> -> {}
    is String -> {}
}