Question

我想用Spring RestTemplate提供服务，在我的服务方面，代码是这样的：

@PostMapping(path="/savePersonList")
@ResponseBody
public List<Person> generatePersonList(@RequestBody List<Person> person){
    return iPersonRestService.generatePersonList(person);
}

如果我使用此代码调用服务，则在客户端：

List<Person> p = (List<Person>) restTemplate.postForObject(url, PersonList, List.class);

我无法将p对象用作List<Person>，它将成为LinkedHashList。经过一些研究后，我找到了一个解决方案，说我必须用交换方法调用服务：

ResponseEntity<List<Person>> rateResponse = restTemplate.exchange(url, HttpMethod.POST, personListResult, new ParameterizedTypeReference<List<Person>>() {});

使用此解决方案，服务器无法获取对象并引发异常，这是正确的方法吗？

Answer 1

检查您的代码是否如下所示。这应该有用。

//header
HttpHeaders headers = new HttpHeaders();
headers.setContentType(MediaType.APPLICATION_JSON);
//person list
List<Person> personList = new ArrayList<Person>();
Person person = new Person();
person.setName("UserOne");  
personList.add(person);
//httpEnitity       
HttpEntity<Object> requestEntity = new HttpEntity<Object>(personList,headers);
ResponseEntity<List<Person>> rateResponse = restTemplate.exchange(url, HttpMethod.POST, requestEntity,new ParameterizedTypeReference<List<Person>>() {});

Answer 2

这可能对您有帮助。

function duplicateProtectSheet(){
  var ss=SpreadsheetApp.getActive();
  var sh=ss.getSheetByName('Source').activate();
  var sh1=ss.duplicateActiveSheet();
  var r1=sh1.getRange('B6:F12');
  var r2=sh1.getRange('C3');
  var r3=sh1.getRange('K13');
  sh1.protect().setUnprotectedRanges([r1,r2,r3]);
}

Answer 3

我遇到了类似的挑战，这是我的 2 美分。

带有 POST 映射的控制器类

@PostMapping("/fetch/projects")
public ResponseEntity<List<Project>> getAllProjects(@RequestBody List<UserProject> userProject) {
    
    // Convert List of projectIds to List of Long using Lambda Functions
    List<Long> projectIds = userProject.stream().map(UserProject::getProjectId).collect(Collectors.toList());
    
    List<Project> projectList = projectService.getAllProjectsByProjectIds(projectIds);
    
    log.info("<< Fetching Projects based on Project Ids !");
    
    return new ResponseEntity<List<Project>>(projectList, HttpStatus.OK);
}

我上面控制器的静态 URL

final String projectBaseUri = "http://localhost:6013/api/v1/project/fetch/projects";

我的来电服务

protected List<Project> getProjectsByIds(List<UserProject> userProject) {

    String url = projectBaseUri;
    log.info("Project URL : " + url);

    HttpHeaders headers = new HttpHeaders();
    headers.setAccept(Arrays.asList(MediaType.APPLICATION_JSON));
    headers.setContentType(MediaType.APPLICATION_JSON);
    
    HttpEntity<Object> requestEntity = new HttpEntity<Object>(userProject,headers);

    ResponseEntity<List<Project>> response = restTemplate.exchange(url, HttpMethod.POST, requestEntity,new ParameterizedTypeReference<List<Project>>() {});

    return response.getBody();
}

Spring RestTemplate发送列表获取列表

3 个答案: