RestTemplate GET请求请求参数

2 回答

• 31

  对于任何类型的url，占位符的工作方式都相同

  String result = restTemplate.getForObject("http://example.com/hotels?state={state}&country={country}", String.class,"NY","USA");
  

  或者更好的是，使用哈希映射进行实名匹配 -

  回复于 2024-04-19T05:27:58+08:00
• 0

  在向 ReSTful 服务器发出请求时，它需要在许多情况下发送查询参数，请求正文（在 POST 和 PUT 请求方法的情况下），以及请求到服务器的标头 .

  在这种情况下，可以使用UriComponentsBuilder.build()构建 URI 字符串，如果需要，使用UriComponents.encode()编码，并使用RestTemplate.exchange()发送，如下所示：

  public ResponseEntity<String> requestRestServerWithGetMethod()
  {
      HttpEntity<?> entity = new HttpEntity<>(requestHeaders); // requestHeaders is of HttpHeaders type
      UriComponentsBuilder builder = UriComponentsBuilder.fromUriString(rawValidUrl) // rawValidURl = http://example.com/hotels
              .queryParams(
                      (LinkedMultiValueMap<String, String>) allRequestParams); // The allRequestParams must have been built for all the query params
      UriComponents uriComponents = builder.build().encode(); // encode() is to ensure that characters like {, }, are preserved and not encoded. Skip if not needed.
      ResponseEntity<Object> responseEntity = restTemplate.exchange(uriComponents.toUri(), HttpMethod.GET,
              entity, String.class);
      return responseEntity;
  }
  
  public ResponseEntity<String> requestRestServerWithPostMethod()
  {
      HttpEntity<?> entity = new HttpEntity<>(requestBody, requestHeaders); // requestBody is of string type and requestHeaders is of type HttpHeaders
      UriComponentsBuilder builder = UriComponentsBuilder.fromUriString(rawValidUrl) // rawValidURl = http://example.com/hotels
              .queryParams(
                      (LinkedMultiValueMap<String, String>) allRequestParams); // The allRequestParams must have been built for all the query params
      UriComponents uriComponents = builder.build().encode(); // encode() is to ensure that characters like {, }, are preserved and not encoded. Skip if not needed.
      ResponseEntity<Object> responseEntity = restTemplate.exchange(uriComponents.toUri(), HttpMethod.POST,
              entity, String.class);
      return responseEntity;
  }
  

  回复于 2024-04-19T05:27:58+08:00