Question

我目前正在开发一个项目，我需要对外部API进行休息调用，并将JSON响应解析为POJO，并将POJO作为JSON返回给另一个休息请求。我能够解析JSON响应，但我的要求是只从中解析一个特定节点。我怎样才能实现这一目标？我使用Spring Boot和Spring Rest Template进行外部休息呼叫。请帮忙!!!

@RestController
public class ProductsController {

    private static final Logger LOGGER = LoggerFactory.getLogger(ProductsController.class);

     @RequestMapping(value = "/myRetail/product/{id}", method = RequestMethod.GET, produces = {
                MediaType.APPLICATION_JSON_UTF8_VALUE, MediaType.APPLICATION_XML_VALUE })
        @ResponseBody
        public Item getSchedule(@Valid Payload payload) {

         String URL = "<External API>";

         LOGGER.info("payload:{}", payload);
         Item response = new Item();
         RestTemplate restTemplate = new RestTemplate();

         Item item = restTemplate.getForObject(URL, Item.class);

         LOGGER.info("Response:{}", item.toString());
            return response;
        }

}

JSONResponse (This is a part of whole i receive)
{
    "ParentNode": {
        "childNode": {
            "att": "13860428",
            "subchildNode 1": {
                "att1": false,
                "att2": false,
                "att3": true,
                "att4": false
            },
            "att4": "058-34-0436",
            "att5": "025192110306",
            "subchildenode2": {
                "att6": "hello",
                "att7": ["how are you", "fine", "notbad"],
                "is_required": "yes"
            },
            ............
}

Required JSONpart from the above whole response:
"subchildenode2": {
                    "att6": "hello",
                    "att7": ["how are you", "fine", "notbad"],
                    "is_required": "yes"
                }

Answer 1

使用JSON-String库。使用此库，您可以将有效负载解析为JSONObject并导航到文档所需的子部分。

因此，您必须将有效负载作为JSONObject并将其解析为库中的spyOn。之后，您可以导航到文档所需的子部分并提取值，然后将其解析为所需的Java POJO。

请查看：How to parse JSON

Answer 2

只需将路径映射到所需对象：

{
    "ParentNode": {
        "childNode": {
            "subchildenode2": {
                "att6": "hello",
                "att7": ["how are you", "fine", "notbad"],
                "is_required": "yes"
            }
        }
}

然后简单地说：

Response responseObject= new Gson().fromJson(json, Response.class);
SubChildNode2 att6 = responseObject.getParentNode().getChildNode().getSubChildNode2();

如何将一部分JSON HTTP响应检索为POJO

2 个答案: