如果你更喜欢一个pojo列表，一种方法是这样做的:

class SomeObject {
    private int id;
    private String name;
}

public <T> List<T> getApi(final String path, final HttpMethod method) {     
    final RestTemplate restTemplate = new RestTemplate();
    final ResponseEntity<List<T>> response = restTemplate.exchange(
      path,
      method,
      null,
      new ParameterizedTypeReference<List<T>>(){});
    List<T> list = response.getBody();
    return list;
}

像这样使用它:

 List<SomeObject> list = someService.getApi("http://localhost:8080/some/api",HttpMethod.GET);

以上的解释可以在这里(https://www.baeldung.com/spring-rest-template-list)找到，并在下面进行解释。

“在上面的代码中发生了一些事情。首先，我们使用ResponseEntity作为返回类型，用它来包装我们真正需要的对象列表。其次，我们调用了RestTemplate.exchange()而不是getForObject()。

这是使用RestTemplate的最通用方式。它要求我们指定HTTP方法、可选请求体和响应类型。在本例中，我们为响应类型使用ParameterizedTypeReference的匿名子类。

最后一部分允许我们将JSON响应转换为适当类型的对象列表。当我们创建ParameterizedTypeReference的匿名子类时，它使用反射来捕获关于我们希望将响应转换为的类类型的信息。

它使用Java的Type对象保存这些信息，我们不再需要担心类型擦除。”

这里提到了3种检索对象列表的方法。所有这些都将完美地工作

@RequestMapping(value = "/emp2", produces = "application/json")
public List<Employee> getEmp2()
{
    HttpHeaders headers = new HttpHeaders();
    headers.setAccept(Arrays.asList(MediaType.APPLICATION_JSON));
    HttpEntity<String> entity = new HttpEntity<String>(headers);
    ResponseEntity<List<Employee>> response = restTemplate.exchange(
            "http://hello-server/rest/employees", HttpMethod.GET,entity, 
    new ParameterizedTypeReference<List<Employee>>() {});
    return response.getBody();
}

(或)

@RequestMapping(value = "/emp3", produces = "application/json")
public List<Employee> getEmp3()
{
    Employee[] empArray = restTemplate.getForObject("http://hello-server/rest/employees", Employee[].class);
    List<Employee> emp= Arrays.asList(empArray);
    return emp;
}

(或)

    @RequestMapping(value = "/emp4", produces = "application/json")
public Employee[] getEmp4()
{
    ResponseEntity<Employee[]> responseEntity = restTemplate.getForEntity("http://hello-server/rest/employees", Employee[].class);
    Employee[] empList = responseEntity.getBody();
    //MediaType contentType = responseEntity.getHeaders().getContentType();
    //HttpStatus statusCode = responseEntity.getStatusCode();
    return  empList;
}

Employee.class

public class Employee {

private Integer id;
private String name;
private String Designation;
private String company;

//getter setters and toString()

}

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

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>

你可以为每个条目创建POJO，

class BitPay{
private String code;
private String name;
private double rate;
}

然后使用BitPay列表的ParameterizedTypeReference，你可以使用:

RestTemplate restTemplate = new RestTemplate();
ResponseEntity<List<Employee>> response = restTemplate.exchange(
  "https://bitpay.com/api/rates",
  HttpMethod.GET,
  null,
  new ParameterizedTypeReference<List<BitPay>>(){});
List<Employee> employees = response.getBody();