android - 如何反序列化 JSON 对象但将特定字段保留为字符串而不是嵌套对象？

Question

我在下面粘贴了一个 json 结构。我想使用 Gson 将 json 反序列化为 java POJO，这非常简单，只是我想保留其中一个字段，data，作为字符串类型而不是嵌套对象。

JSON 结构

{
  "created_on": "2015-06-04T16:12:04-0700",
  "modified_on": "2015-06-04T16:12:09-0700",
  "identifier": "sample",
  "name": "some name",
  "summary": "some summary",
  "data": {
    "$type": "a_type",
    "some_other_stuff": {
        "more_stuff": "lorem ipsum"
    },
    "type2": {
        "$type": "another_type",
        "other_stuff": {
            "event_more_stuff": "lorem ipsum"
        }
    }
  }
}

我的 POJO 看起来像这样:

public class Sample {
  private String identifier; // "sample"
  private String created_on; // "2015-06-04T16:12:04-0700"
  private String modified_on; // "2015-06-04T16:12:09-0700"
  private String name; // "some name"
  private String summary; // "some summary"
  private String data; // "{ \"$type\": ... }"

  // getters and setters
}

data 字段应保留为 JSON 格式的字符串。

我已经尝试实现自定义 TypeAdapter 并将该字段作为字符串读取，但它失败了 Expected a string but was BEGIN_OBJECT。

另请注意，我也希望在序列化时保持结构 - 这样我就可以将 POJO 序列化回原始 JSON 结构。

编辑自定义TypeAdapter:

public class SampleTypeAdapter extends TypeAdapter<Sample> {
@Override
public void write(JsonWriter out, Sample sample) throws IOException {
    out.beginObject();

    out.name("identifier").value(sample.getIdentifier());
    out.name("name").value(sample.getName());
    out.name("data").value(sample.getData());
    out.name("summary").value(sample.getSummary());
    out.name("modified_on").value(sample.getModifiedOn());
    out.name("created_on").value(sample.getCreatedOn());

    out.endObject();
}

@Override
public Sample read(JsonReader in) throws IOException {
    final Sample sample = new Sample();

    in.beginObject();
    while (in.hasNext()) {
        String nextName = in.nextName();
        switch (nextName) {
            case "identifier":
                sample.setIdentifier(in.nextString());
                break;
            case "name":
                sample.setName(in.nextString());
                break;
            case "data":
                sample.setData(in.nextString()); // <-- fails here
                break;
            case "summary":
                sample.setSummary(in.nextString());
                break;
            case "modified_on":
                sample.setModifiedOn(in.nextString());
                break;
            case "created_on":
                sample.setCreatedOn(in.nextString());
                break;
            default:
                in.skipValue();
                break;
        }
    }
    in.endObject();

    return sample;
}
}

Answer 1

您可以创建自定义 JsonDeserializer像这样:

public class SampleGSONParserAdapter implements
        JsonDeserializer<Sample> {

    @Override
    public Sample deserialize(JsonElement json, Type typeOfT, JsonDeserializationContext context) throws JsonParseException {

        Sample sample = new Sample();
        JsonObject sampleJsonObject = json.getAsJsonObject();

        sample.setName(sampleJsonObject.get("name").getAsString());

        // do the other parsing stuff here..

        // getting the data object as a string
        sample.setJsonString(sampleJsonObject.get("data").toString());

        return sample;
    }

}

然后你像这样使用它:

GsonBuilder gsonBuilder = new GsonBuilder();
        gsonBuilder.registerTypeAdapter(Sample.class, new SampleGSONParserAdapter());
Gson gson = gsonBuilder.create();

不好的部分是它没有你写的那么快，但至少你可以像这样进行自定义解析。