如何从JSON而不是LinkedHashMap获取List对象？

Question

我有一个方法，它应该从JSON返回不同的对象，这取决于参数中的类类型。我试着根据参数返回一个对象列表，但是我只把LinkedHashMap放入ArrayList。

我搜索了很多，但是在解决方案中，类类型的所有地方都是硬编码的。

有没有一种不用硬代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;
}

我只想传递类类型并通过一个方法获取对象。

[
{
"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，我得到这个例外

无法识别的字段"字段"（类model.orm.部门），在[Source：（String）"[{"id"："73cbc0b5-3dd5-49c4-97cb-6225a19122b5"，"name"："Management"，"字段"：[{"id"："c2d740d5-4d47-42ae-b616-977b40327812"，"name"："newField1"}]},{"id"："dd74384b-717d-4368-b0e4-3f441d5b1ffc"，"name"："IT"，"字段":[]},{"id"："03304335-d7d7-46ca-8075-8d5e9feb43c6"，"name"："hhhh"，"字段":[]},{"id"："e11b4c3f-080e-490d-8ef4-ea30c"："50dfe14-f30a-448c-99df-10bf01d088fa"，"name"："NEWWWW"，"fie"[截断84个字符]；行： 1，列： 77]（通过引用链：java.util.ArrayList[0]-

Answer 1

您可以构造一个新的JavaType参数化类型，将列表作为参数传递。类到对象映射器。html#getTypeFactory方法如下：

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);
}