Question

我正在开发一种服务，该服务接收来自API的请求，并且标头向另一个API发出另一个请求。使用第二个API的数据，它完成了映射，然后我们用这些数据进行响应。在开发过程中，我发现此错误，无法解决。我无法使用针对mi first API的正确JSON进行响应。

我的代码很简单，但是很混乱。我有一个控制器和一个模型文件

控制器

@RestController
public class controller {


//Request of global API
@RequestMapping(value="orches", produces = MediaType.APPLICATION_JSON_VALUE, method=RequestMethod.GET)

public model globalrequest(
        @RequestHeader(value="Authorization") String Auth,
        @RequestHeader(value="X-Country") String Country,
        @RequestHeader(value="X-Global-Id") String LocalClid) throws JsonProcessingException, IOException {


    // Country testing
    String localApiUrl="";
    switch (Country) {
    case "SPA":
        localApiUrl = "https://myapilocal";
    default:
        //error
    };


    RestTemplate restTemplate = new RestTemplate();
    //Request of Local API   
    //header
    HttpHeaders headers = new HttpHeaders();
    headers.setContentType(MediaType.APPLICATION_JSON);
    headers.set("X-IBM-Client-Id", LocalClid);
    headers.set("Authorization", Auth);
    HttpEntity<String> entity = new HttpEntity<String>(headers);


    //Send the request as GET
    ResponseEntity<String> response = restTemplate.exchange(localApiUrl, HttpMethod.GET, entity, String.class);
    //System.out.println(response);
    String body = response.getBody();
    return new model(body);

}

另一边是模型

模型

public class model {

 private static String myJson = null;
ProfileGlo profileGlo = new ProfileGlo();
 public model(String body) throws JsonProcessingException  {
        //super();
        ProfileLoc profileLoc = new Gson().fromJson(body, ProfileLoc.class);
        mapping(profileGlo, profileLoc);
        ObjectMapper objectMapper = new ObjectMapper();
        objectMapper.setVisibility(PropertyAccessor.ALL, Visibility.NONE);
        objectMapper.setVisibility(PropertyAccessor.FIELD, Visibility.ANY);
        myJson = objectMapper.writeValueAsString(profileGlo);    
        System.out.println(myJson);
        System.out.println(profileGlo.customerBasicData.customerNameData.firstName);
     }  

public static void mapping(ProfileGlo profileGlo, ProfileLoc profileLoc) {

    if ( profileLoc.personType.equals("F"))  {
        // Mapping of the type of person
        profileGlo.customerType = "1";
        // Mapping of the Name
        profileGlo.customerBasicData.customerNameData.firstName = profileLoc.fullName.getName();
        //Mapping of the lastNames
        profileGlo.customerBasicData.customerNameData.middleName = profileLoc.fullName.getLastName();
        //Mapping companyName
        profileGlo.sMEBusinessCustomerBasicData.companyName = null;
        //Mapping birthDate
        profileGlo.customerBasicData.birthDate = profileLoc.birthDate;      
    }
    else if ( profileLoc.personType.equals("J")) {
        // Mapping of the type of person
        profileGlo.customerType = "2";
        // Mapping of the Name
        profileGlo.sMEBusinessCustomerBasicData.companyName = profileLoc.fullName.getName();
        //Mapping of the lastNames
        profileGlo.customerBasicData.customerNameData.middleName = null;
        //Mapping companyName
        profileGlo.sMEBusinessCustomerBasicData.companyName = profileLoc.fullName.getCompanyName();
        //Mapping birthDate
        profileGlo.customerBasicData.birthDate = null;

    }       
    else {
            //error500
    }

}

@ResponseBody
public String orches(HttpServletResponse response) {
  response.addHeader("content-type", "application/json");
  response.setStatus(200);
  return myJson;
}

}

还定义了ProfileGlo和ProfileLoc对象，但由于没有意义，因此在此不包括它们。我需要知道如何响应第一个API，因为我没有收到406错误。

Answer 1

您的第二个API将直接抛出HTTP错误406错误，并且第一个api将需要处理它，因为它是第一个api的客户端。
如果要设计json，可以创建一个Error对象，并使用代码和描述像这样在第一和第二个api之间进行通信

class Error { private String error; private String description; }

我该如何解决406错误“找不到可接受的表示形式”？

1 个答案: