我正在呼叫另一個 api 並獲得以下 json 回應
{ "metadata": {}, "data": { "productid": 102001, "productname": "p101", "branddetail": { "brandid": 3840, "brandname": "abc", "brandcode": "x01" } } }
如何解開品牌詳細資訊並將其作為類實體讀取,如下所示?
HttpGet httpGet = buildHttpGet("/externalApiURL"); HttpResponse response = getHttpClient().execute(httpGet); HttpEntity entity = response.getEntity(); if (entity != null && response.getStatusLine().getStatusCode() == HttpStatus.OK.value()) { ObjectMapper objectMapper = new ObjectMapper(); BrandDetail brandDetail = objectMapper.readValue(entity.getContent(), BrandDetail.class); }
提前致謝
使用convertvalue()
,這裡有測驗。
@data public class branddetail { private int brandid; private string brandname; private string brandcode; }
@Test public void demo() throws Exception { ObjectMapper mapper = new ObjectMapper(); var data = """ { "metadata": {}, "data": { "productId": 102001, "productName": "P101", "brandDetail": { "brandId": 3840, "brandName": "ABC", "brandCode": "X01" } } } """; JsonNode node = mapper.readTree(data); JsonNode brandNode = node.get("data").get("brandDetail"); BrandDetail brandDetail = mapper.convertValue(brandNode, BrandDetail.class); // BrandDetail(brandId=3840, brandName=ABC, brandCode=X01) System.out.println(brandDetail); }
以上是如何將嵌套 json 物件解包為實體的詳細內容。更多資訊請關注PHP中文網其他相關文章!