取決於你正在嘗試做什麼。你可以設置一個POJO heirarchy,匹配你的json
,如here(首選方法)。或者,您可以提供自定義deserializer。我只處理了id
數據,因爲我認爲這是棘手的問題。只需使用gson
類型即可完成json
,然後構建您試圖表示的數據。 Data
和Id
類只是pojos
組成並反映了原始json
字符串中的屬性。
public class MyDeserializer implements JsonDeserializer<Data>
{
@Override
public Data deserialize(JsonElement je, Type type, JsonDeserializationContext jdc) throws JsonParseException
{
final Gson gson = new Gson();
final JsonObject obj = je.getAsJsonObject(); //our original full json string
final JsonElement dataElement = obj.get("data");
final JsonElement idElement = dataElement.getAsJsonObject().get("id");
final JsonArray idArray = idElement.getAsJsonArray();
final List<Id> parsedData = new ArrayList<>();
for (Object object : idArray)
{
final JsonObject jsonObject = (JsonObject) object;
//can pass this into constructor of Id or through a setter
final JsonObject stuff = jsonObject.get("stuff").getAsJsonObject();
final JsonArray valuesArray = jsonObject.getAsJsonArray("values");
final Id id = new Id();
for (Object value : valuesArray)
{
final JsonArray nestedArray = (JsonArray)value;
final Integer[] nest = gson.fromJson(nestedArray, Integer[].class);
id.addNestedValues(nest);
}
parsedData.add(id);
}
return new Data(parsedData);
}
}
測試:
@Test
public void testMethod1()
{
final String values = "[[123, 456], [987, 654]]";
final String id = "[ {stuff: { }, values: " + values + ", otherstuff: 'stuff2' }]";
final String jsonString = "{data: {id:" + id + "}}";
System.out.println(jsonString);
final Gson gson = new GsonBuilder().registerTypeAdapter(Data.class, new MyDeserializer()).create();
System.out.println(gson.fromJson(jsonString, Data.class));
}
結果:
Data{ids=[Id {nestedList=[[123, 456], [987, 654]]}]}
POJO:
public class Data
{
private List<Id> ids;
public Data(List<Id> ids)
{
this.ids = ids;
}
@Override
public String toString()
{
return "Data{" + "ids=" + ids + '}';
}
}
public class Id
{
private List<Integer[]> nestedList;
public Id()
{
nestedList = new ArrayList<>();
}
public void addNestedValues(final Integer[] values)
{
nestedList.add(values);
}
@Override
public String toString()
{
final List<String> formattedOutput = new ArrayList();
for (Integer[] integers : nestedList)
{
formattedOutput.add(Arrays.asList(integers).toString());
}
return "Id {" + "nestedList=" + formattedOutput + '}';
}
}
我走建議你http://json.parser.online.fr/的自由,它在PO提供的Json上也發現了一個語法錯誤。 – giampaolo
我認爲[鏈接] [http://www.jsonschema2pojo.org]在爲您提供JSON響應和創建Java對象時非常有用! –