Question

我有一个jax-rs端点，该端点应该返回一个JSON对象，但是我想选择一些字段并隐藏其他一些字段，我的代码是这样的：

import javax.ws.rs.BadRequestException;
import javax.ws.rs.GET;
import javax.ws.rs.Path;
import javax.ws.rs.PathParam;
import javax.ws.rs.Produces;
import javax.ws.rs.QueryParam;

import org.apache.commons.lang3.StringUtils;
import org.springframework.http.MediaType;
import org.springframework.stereotype.Component;
import io.swagger.annotations.Api;
import io.swagger.annotations.ApiOperation;
import io.swagger.annotations.ApiParam;
import io.swagger.annotations.ApiResponse;
import io.swagger.annotations.ApiResponses;

@Component
@Path("/customers")
@Api(value = "Customers resource", produces = MediaType.APPLICATION_JSON_VALUE)
public class CustomersEndpoint{
    private final CustomersService customersService;

    public CustomersEndpoint(CustomersService customersService) {
        this.customersService = customersService;
    }
@GET
    @Path("{customerResourceId}")
    @Produces(MediaType.APPLICATION_JSON_VALUE)
    @ApiOperation(value = "Get customer details")
    @ApiResponses(value = { @ApiResponse(code = 200, message = "Listing the customer details", response = **Customer**.class)") })
    public **Customer** getCustomerDetails(@ApiParam(value = "ID of customer to fetch") @PathParam("customerResourceId") String customerId,
                                      @QueryParam(value = "Retrieve only selected fields [by comma]") String fields )
            throws ApiException {       

        return this.customersService.getCustomerDetails(customerId,fields);
    }

在这里，我只想为选定的字段返回自定义“ 客户”。

我正在使用jax-rs，jackson解组/将对象编组为JSON。

请提供任何解决方案。

客户类的示例：

import java.util.ArrayList;
import java.util.List;

import com.fasterxml.jackson.annotation.JsonInclude;
import com.fasterxml.jackson.annotation.JsonProperty;
import com.fasterxml.jackson.annotation.JsonPropertyDescription;
@JsonInclude(JsonInclude.Include.NON_NULL)
public class Customer {

        public Customer() {

            }

    public Customer(String customerId,String phoneNumber) {

        this.customerId=customerId;
        this.phoneNumber=phoneNumber;
    }

    /**
     * customer identifier
     */
    @JsonPropertyDescription("customer identifier")
    @JsonProperty("customerId")
    private String customerId;

    /**
     * customer phone number
     */
    @JsonPropertyDescription("customer phone number")
    @JsonProperty("phoneNumber")
    private String phoneNumber;
    /**
     * customer first number
     */
    @JsonPropertyDescription("customer first number")
    @JsonProperty("firstName")
    private String firstName;
    /**
     * customer last number
     */
    @JsonPropertyDescription("customer last number")
    @JsonProperty("lastName")
    private String lastName;
public String getCustomerId() {
        return customerId;
    }
    public Customer setCustomerId(String customerId) {
        this.customerId = customerId;
        return this;
    }

    public String getPhoneNumber() {
        return phoneNumber;
    }
    public Customer setPhoneNumber(String phoneNumber) {
        this.phoneNumber = phoneNumber;
        return this;
    }
    public String getFirstName() {
        return firstName;
    }
    public Customer setFirstName(String firstName) {
        this.firstName = firstName;
        return this;
    }
    public String getLastName() {
        return lastName;
    }
    public Customer setLastName(String lastName) {
        this.lastName = lastName;
        return this;
    }
}

输出：

{
  "customerId": "string",
  "phoneNumber": "string",
  "firstName": "string",
  "lastName": "string",
}

=>选择后的结果：字段= phoneNumber，customerId

{
  "customerId": "string",
  "phoneNumber": "string"
}

我知道在实例化对象时不设置'hide'属性并且不包含此批注 @JsonInclude（JsonInclude.Include.NON_NULL）是一种解决方案，但它需要太多代码和维护

Answer 1

有几种不同的方法来指示杰克逊不序列化属性。其中之一是在正在序列化的java类中注释您的ignore属性。在下面的示例中，intValue不会在json中序列化。

private String stringValue;

@JsonIgnore
private int intValue;

private boolean booleanValue;

这是一篇不错的文章，内容涉及忽略json序列化字段的其他策略。

http://www.baeldung.com/jackson-ignore-properties-on-serialization

Answer 2

我认为您应该在该过滤器中添加一个组件：

@Component
public class CustomerFilterConfig {


    public static  Set<String> fieldNames = new HashSet<String>();

      @Bean
      public ObjectMapper objectMapper() {
        ObjectMapper objectMapper = new ObjectMapper();
        SimpleFilterProvider simpleFilterProvider = new SimpleFilterProvider().setFailOnUnknownId(false);
        FilterProvider filters =simpleFilterProvider.setDefaultFilter(SimpleBeanPropertyFilter.filterOutAllExcept(fieldNames)).addFilter("customerFilter", SimpleBeanPropertyFilter.filterOutAllExcept(fieldNames));    
        objectMapper.setFilterProvider(filters);
        return objectMapper;
      } 


}

然后在模型中添加该过滤器：

@JsonFilter("customerFilter")
public class Customer {...}

最后使用该过滤器：

String fields = "A,B,C,D";
CustomerFilterConfig.fieldNames.clear();
CustomerFilterConfig.fieldNames.addAll(Arrays.asList(fields.split(",")));

编组之前动态隐藏一些字段（java到json）

2 个答案: