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!

首先定义一个对象来保存返回数组的实体。如。

@JsonIgnoreProperties(ignoreUnknown = true)
public class Rate {
    private String name;
    private String code;
    private Double rate;
    // add getters and setters
}

然后你可以使用该服务并通过以下方式获得一个强类型列表:

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

上面的其他解决方案也可以工作，但我喜欢得到一个强类型列表，而不是Object[]。

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()

}

对于那些使用Spring + Kotlin的人来说，下面是翻译:

val rates = restTemplate.exchange("https://bitpay.com/api/rates", HttpMethod.GET, null, object : ParameterizedTypeReference<List<Rate>>() {}).body!!

作为一个通用模块，Page<?>对象可以被模块反序列化，就像JodaModule, Log4jJsonModule等。参考我的回答。使用Pageable字段测试端点时出现JsonMappingException

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>