如何从 JSON 而不是 LinkedHashMap 获取列表对象？

Question

我有这个方法，它应该 return 与 JSON 不同的对象，具体取决于 argument.I 中 class 的类型尝试了 return基于参数的对象列表，但我只将 LinkedHashMap 放入 ArrayList。

我搜索了很多，但在所有解决方案中 class 类型都是硬编码的。

有没有不用硬编码就能解决这个问题的方法？

 public static <T> List<T> getObjects(Class<T> c) {
    CloseableHttpClient rest =  HttpClientSessionSingleton.getInstance().getHttpClient();
    String urlRequest = (host + "/" +
            c.getSimpleName().toLowerCase() + "s");

    HttpGet httpGet = new HttpGet(urlRequest);
    try (CloseableHttpResponse response = rest.execute(httpGet)) {
        HttpEntity entity = response.getEntity();
        String jsonString = EntityUtils.toString(entity);

        List<T> listObjectFromJson = new ObjectMapper().readValue(jsonString, new TypeReference<List<T>>(){});

        return listObjectFromJson;
    } catch (IOException e) {
        e.printStackTrace();
    }
    return null;
}

我只想传递 class 类型并通过一种方法获取对象。

[
{
"id": "73cbc0b5-3dd5-49c4-97cb-6225a19122b5",
"name": "Management",
"fields": [
{
"id": "c2d740d5-4d47-42ae-b616-977b40327812",
"name": "newField1"
}
]
},
{
"id": "dd74384b-717d-4368-b0e4-3f441d5b1ffc",
"name": "IT",
"fields": []
},
{
"id": "03304335-d7d7-46ca-8075-8d5e9feb43c6",
"name": "hhh",
"fields": []
},
{
"id": "e11b4c3f-080e-490d-8ef4-ea301d551a5d",
"name": "NEWWWWW",
"fields": []
},
{
"id": "fec7eeb0-0845-49be-be14-6cdb5fcd3575",
"name": "NEWWWWW",
"fields": []
},
{
"id": "50dfea14-f30a-448c-99df-10bf01d088fa",
"name": "NEWWWWW",
"fields": []
},
{
"id": "a4a1224e-7c66-484c-ae87-dc2ecc058c36",
"name": "NEWWWWW",
"fields": []
}
]

当我的对象有关系时我得到这个异常

Unrecognized field "fields" (class model.orm.Department), not marked as ignorable (2 known properties: "id", "name"]) at [Source: (String)"[{"id":"73cbc0b5-3dd5-49c4-97cb-6225a19122b5","name":"Management","fields":[{"id":"c2d740d5-4d47-42ae-b616-977b40327812","name":"newField1"}]},{"id":"dd74384b-717d-4368-b0e4-3f441d5b1ffc","name":"IT","fields":[]},{"id":"03304335-d7d7-46ca-8075-8d5e9feb43c6","name":"hhh","fields":[]},{"id":"e11b4c3f-080e-490d-8ef4-ea301d551a5d","name":"NEWWWWW","fields":[]},{"id":"fec7eeb0-0845-49be-be14-6cdb5fcd3575","name":"NEWWWWW","fields":[]},{"id":"50dfea14-f30a-448c-99df-10bf01d088fa","name":"NEWWWWW","fie"[truncated 84 chars]; line: 1, column: 77] (through reference chain: java.util.ArrayList[0]->model.orm.Department["fields"]) at com.fasterxml.jackson.databind.exc.UnrecognizedPropertyException.from

Answer 1

您可以构造一个新的 JavaType parametric type passing as an argument the List.class to the ObjectMapper.html#getTypeFactory 方法，如下所示：

public static <T> List<T> getObjects(Class<T> c) throws IOException {
    //omitted the lines before creating the mapper including the jsonstring
    ObjectMapper mapper = new ObjectMapper();
    JavaType type = mapper.getTypeFactory().constructParametricType(List.class, c);
    
    return mapper.readValue(jsonString, type);
}

如何从 JSON 而不是 LinkedHashMap 获取列表对象？

How I can get List object from JSON instead of a LinkedHashMap?

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json

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