may*_*hal 0 java spring spring-mvc jersey-2.0
这是我的 get API 返回员工列表。
@GET
@Produces(MediaType.APPLICATION_JSON)
@Path("/get")
public List<Employee> getEmployee() {
return employeeFacade.getAllEmployee();
}
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我想Employee从ResponseEntity<String>.
RestTemplate restTemplate = new RestTemplate();
HttpHeaders headers = new HttpHeaders();
headers.setAccept(Arrays.asList(MediaType.APPLICATION_JSON));
//headers.setContentType(MediaType.APPLICATION_JSON);
Employee employee = new Employee();
HttpEntity<Employee> entity = new HttpEntity<Employee>(employee, headers);
ResponseEntity<String> result = restTemplate.exchange(uri, HttpMethod.GET, entity, String.class);
System.out.println("result:" + result);
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如何使这项工作?
如果 API 调用将对象作为 JSON返回List,Employee那么您可以直接将该 JSON 解析为List<Employee>对象ParameterizedTypeReference
ResponseEntity<List<Employee>> result = restTemplate.exchange(uri, HttpMethod.GET, entity, new ParameterizedTypeReference<List<Employee>>() {});
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getBody() 将返回响应正文
List<Employee> emps = result.getBody();
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