如何在 ResponseEntity 中不返回空值？

Question

我使用ResponseEntity為 GET “api/v1/name” 和 POST “api/v1/name” 請求返回響應。

我的目標不是返回帶有空值的響應，例如在 POST“api/v1/name”請求中，當前響應正文是：

{
    "id": null,
    "name": "who",
    "newid": "A8C90A53-78F6-4BD6-9628-CBA8FC349C08"
}

我希望它看起來像：

{
    "name": "who",
    "newid": "A8C90A53-78F6-4BD6-9628-CBA8FC349C08"
}

在我看來，使用下面的代碼重新創建對象只會使代碼的可讀性降低，並且可能會使用更多的內存（我不確定，如果我錯了，請告訴我）：

...
Map<String, String> responseBody = new HashMap<>();
responseBody.put("name", nameModel.getName());
responseBody.put("newid", nameModel.getNewId());

return new ResponseEntity<>(responseBody, HttpStatus.OK);

==== 下面是完整的存儲庫，如果您想查看更新的存儲庫： https : //github.com/kidfrom/g2_java/tree/main/etc/mssqlserver

控制器/名稱控制器.java

package com.example.mssqlserver.controller;

import com.example.mssqlserver.mapper.NameMapper;
import com.example.mssqlserver.model.NameModel;
import org.springframework.beans.factory.annotation.Autowired;
import org.springframework.http.HttpStatus;
import org.springframework.http.ResponseEntity;
import org.springframework.web.bind.annotation.*;

@RestController
public class NameController {

  @Autowired
  private NameMapper nameMapper;

  @GetMapping("api/v1/name")
  public ResponseEntity<?> selectAll() {
    return new ResponseEntity<>(nameMapper.selectAll(), HttpStatus.OK);
  }

  @PostMapping("api/v1/name")
  public ResponseEntity<?> insert(@RequestBody NameModel nameModel) {

    // validator
    if (!nameModel.requestIsValid()) {
      return ResponseEntity.badRequest().build();
    }

    if (nameMapper.insert(nameModel) == 1) {
      return new ResponseEntity<>(nameModel, HttpStatus.OK);
    } else {
      return ResponseEntity.badRequest().build();
    }
  }
}

映射器/NameMapper.java

package com.example.mssqlserver.mapper;

import com.example.mssqlserver.model.NameModel;
import org.apache.ibatis.annotations.*;

import java.util.List;

@Mapper
public interface NameMapper {

  @Select("SELECT * FROM name")
  public List<NameModel> selectAll();

  @SelectKey(statement = "SELECT NEWID()", keyProperty = "newid", resultType = String.class, before = true)
  @Insert("INSERT INTO name (name, newid) VALUES (#{name}, #{newid})")
//  @Options(useGeneratedKeys = true, keyProperty = "id")
  int insert(NameModel nameModel);
}

模型/名稱Model.java

package com.example.mssqlserver.model;

public class NameModel {
  private Integer id;
  private String name;
  private String newid;

  public NameModel(Integer id, String name, String newid) {
    this.id = id;
    this.name = name;
    this.newid = newid;
  }

  public Integer getId() {
    return id;
  }

  public void setId(Integer id) {
    this.id = id;
  }

  public String getName() {
    return name;
  }

  public String getNewid() {
    return newid;
  }

  public void setNewid(String newid) {
    this.newid = newid;
  }

  public boolean requestIsValid() {
    if (this.name.isEmpty()) return false;

    return true;
  }
}

Answer 1

我認為這個簡單的解決方案可以幫助你

https://stackoverflow.com/a/36515285/5108695

由於正在使用 Jackson，您必須將其配置為 Jackson 屬性。 對於 Spring Boot REST 服務，您必須在 application.properties 或 application.yml 中進行配置：

spring.jackson.default-property-inclusion = NON_NULL

Answer 2

在 spring 引導視圖https://docs.spring.io/spring-boot/docs/2.4.0/reference/htmlsingle並搜索spring.jackson.default-property-inclusion