java - Spring/RestTemplate-将实体放置到服务器

Question

请看下面的简单代码:

final String url = String.format("%s/api/shop", Global.webserviceUrl);

RestTemplate restTemplate = new RestTemplate();
restTemplate.getMessageConverters().add(new MappingJackson2HttpMessageConverter());

HttpHeaders headers = new HttpHeaders();
headers.set("X-TP-DeviceID", Global.deviceID);
HttpEntity entity = new HttpEntity(headers);

HttpEntity<Shop[]> response = restTemplate.exchange(url, HttpMethod.GET, entity, Shop[].class);
shops = response.getBody();

如您所见，以上代码旨在从服务器(以json格式)获取商店列表，并将响应映射到Shop对象数组。
现在，我需要新建一个商店，例如/ api / shop / 1。请求实体应具有与返回实体完全相同的格式。

我应该在网址中添加/ 1，创建新的Shop类对象，并用我要放置的值填充所有字段，然后与HttpMethod.PUT进行交换吗？

请为我澄清一下，我是Spring的初学者。代码示例将不胜感激。

[编辑]
我很困惑，因为我也注意到了RestTemplate.put()方法。那么，我应该使用哪一个呢？交换还是put()？

Answer 1

您可以尝试类似:

    final String url = String.format("%s/api/shop/{id}", Global.webserviceUrl);

    RestTemplate restTemplate = new RestTemplate();
    restTemplate.getMessageConverters().add(new MappingJackson2HttpMessageConverter());

    HttpHeaders headers = new HttpHeaders();
    headers.set("X-TP-DeviceID", Global.deviceID);
    Shop shop= new Shop();
    Map<String, String> param = new HashMap<String, String>();
    param.put("id","10")
    HttpEntity<Shop> requestEntity = new HttpEntity<Shop>(shop, headers);
    HttpEntity<Shop[]> response = restTemplate.exchange(url, HttpMethod.PUT, requestEntity, Shop[].class, param);

    shops = response.getBody();

认沽权无效，而交易所会给您回应，最好的检查地点是文件https://docs.spring.io/spring/docs/current/javadoc-api/org/springframework/web/client/RestTemplate.html