更简单的方法: 我会给你们看Authorization heard和unauthorization header:

未经授权: a.进行依赖注入(构造函数注入): 你也可以选择字段注入。我考虑了构造函数注入。

public class RestTemplateService {
private final RestTemplate template;
public RestTemplateService(RestTemplate template) {
    this.template = template;
  }
}

b.调用getList()方法:

public ResponseEntity<List> getResponseList(String url, HttpMethod type) {
    return template.exchange(url, type, new HttpEntity<>(new HttpHeaders()), List.class);
  }

授权:我喜欢小方法。所以我把这些功能分开:

 public ResponseEntity<List> getResponse(String url, HttpMethod type) {
    return template.exchange(url, type, getRequest(getHeaders(USERNAME, PASS)), List.class);
  }

 private HttpEntity<String> getRequest(HttpHeaders headers) {
    return new HttpEntity<>(headers);
  }

 private HttpHeaders getHeaders(String username, String password) {
    HttpHeaders headers = new HttpHeaders();
    headers.add("Authorization", "Basic " + new String(Base64.encodeBase64((username + ":" + password).getBytes())));
    return headers;
  }

希望问题能得到解决!

考虑一下这个答案，特别是如果你想在列表中使用泛型 Spring RestTemplate和泛型类型ParameterizedTypeReference集合，如List<T>

我从这个帖子https://jira.spring.io/browse/SPR-8263找到了工作。

基于这篇文章，你可以返回一个类似这样的列表:

ResponseEntity<? extends ArrayList<User>> responseEntity = restTemplate.getForEntity(restEndPointUrl, (Class<? extends ArrayList<User>>)ArrayList.class, userId);

实际上，我之前为我的一个项目开发了一些功能，以下是代码:

/**
 * @param url             is the URI address of the WebService
 * @param parameterObject the object where all parameters are passed.
 * @param returnType      the return type you are expecting. Exemple : someClass.class
 */

public static <T> T getObject(String url, Object parameterObject, Class<T> returnType) {
    try {
        ResponseEntity<T> res;
        ObjectMapper mapper = new ObjectMapper();
        RestTemplate restTemplate = new RestTemplate();
        restTemplate.getMessageConverters().add(new MappingJackson2HttpMessageConverter());
        restTemplate.getMessageConverters().add(0, new StringHttpMessageConverter(Charset.forName("UTF-8")));
        ((SimpleClientHttpRequestFactory) restTemplate.getRequestFactory()).setConnectTimeout(2000);
        HttpHeaders headers = new HttpHeaders();
        headers.setContentType(MediaType.APPLICATION_JSON);
        HttpEntity<T> entity = new HttpEntity<T>((T) parameterObject, headers);
        String json = mapper.writeValueAsString(restTemplate.exchange(url, org.springframework.http.HttpMethod.POST, entity, returnType).getBody());
        return new Gson().fromJson(json, returnType);
    } catch (Exception e) {
        e.printStackTrace();
        return null;
    }
}

/**
 * @param url             is the URI address of the WebService
 * @param parameterObject the object where all parameters are passed.
 * @param returnType      the type of the returned object. Must be an array. Exemple : someClass[].class
 */
public static <T> List<T> getListOfObjects(String url, Object parameterObject, Class<T[]> returnType) {
    try {
        ObjectMapper mapper = new ObjectMapper();
        RestTemplate restTemplate = new RestTemplate();
        restTemplate.getMessageConverters().add(new MappingJackson2HttpMessageConverter());
        restTemplate.getMessageConverters().add(0, new StringHttpMessageConverter(Charset.forName("UTF-8")));
        ((SimpleClientHttpRequestFactory) restTemplate.getRequestFactory()).setConnectTimeout(2000);
        HttpHeaders headers = new HttpHeaders();
        headers.setContentType(MediaType.APPLICATION_JSON);
        HttpEntity<T> entity = new HttpEntity<T>((T) parameterObject, headers);
        ResponseEntity<Object[]> results = restTemplate.exchange(url, org.springframework.http.HttpMethod.POST, entity, Object[].class);
        String json = mapper.writeValueAsString(results.getBody());
        T[] arr = new Gson().fromJson(json, returnType);
        return Arrays.asList(arr);
    } catch (Exception e) {
        e.printStackTrace();
        return null;
    }
}

我希望这能帮助到别人!

对我来说这很有效

Object[] forNow = template.getForObject("URL", Object[].class);
    searchList= Arrays.asList(forNow);

Object是你想要的类的位置

我在这里遇到的最大问题是构建将RestTemplate匹配到兼容类所需的Object结构。幸运的是，我找到了http://www.jsonschema2pojo.org/(在浏览器中获得JSON响应并将其作为输入)，我非常推荐这个功能!