在查询中选择特定列时如何将数组结果转换为 JSON 格式

Question

用于将查询结果（数组）转换为JSON格式，

我尝试使用 @ResponseBody 注释和 produces="application/json" 但输出格式没有改变。我也尝试过 ObjectMapper() 但字符串格式的输出不在键 value.I 中删除了我的试用而不是 working.Now 我在此处添加了我当前的代码。

这是我的代码片段。

InvestigatorModel

package com.demo.model;

import java.io.Serializable;
import javax.persistence.Entity;
import javax.persistence.Id;
import javax.persistence.Table;

@Entity
@Table(name="investigator")
public class InvestigatorModel implements Serializable{

        private static final long serialVersionUID = 1L;

        @Id
        public Integer ninvestigatorid;
        public String sinvestigatorname;
        public Integer ninstid ;
        public String stitle;
        public Integer ntempinvestigatorid;
        public Integer nempid;
        //getter and setter

调查员控制器

package com.demo.controller;

import java.util.List;
import org.springframework.beans.factory.annotation.Autowired;
import org.springframework.web.bind.annotation.CrossOrigin;
import org.springframework.web.bind.annotation.GetMapping;
import org.springframework.web.bind.annotation.RequestMapping;
import org.springframework.web.bind.annotation.RestController;
import com.demo.model.InvestigatorModel;
import com.demo.services.InvestigatorService;

@RestController
@RequestMapping("/SpaceStudy/SpaceAdmin")

public class InvestigatorController {

    @Autowired
    InvestigatorService investService;

  @CrossOrigin(origins="*") 
  @GetMapping("/AccountMaintenance/LoadInvestigator")   

  public List<InvestigatorModel> findInvestigatorName()
  {
    return investService.findInvestigatorName();

  }
}

调查员服务

package com.demo.services;
import java.util.List;
import org.springframework.beans.factory.annotation.Autowired;
import org.springframework.stereotype.Service;
import com.demo.model.InvestigatorModel;
import com.demo.repository.InvestigatorRepository;

@Service
public class InvestigatorService 
{
    @Autowired
    InvestigatorRepository investRepo;

    public List<InvestigatorModel> findInvestigatorName()
    {
        return investRepo.findBySinvestigatorname();

    }

}

InvestigatorRepository

package com.demo.repository;
import java.util.List;
import org.springframework.data.jpa.repository.JpaRepository;
import org.springframework.data.jpa.repository.Query;
import org.springframework.stereotype.Repository;

import com.demo.model.InvestigatorModel;
@Repository
public interface InvestigatorRepository extends JpaRepository<InvestigatorModel, Integer>
{
    @Query("select invest.sinvestigatorname from InvestigatorModel as invest where invest.ninstid=60")
    List<InvestigatorModel> findBySinvestigatorname();
}

我的输出是这样的

示例输出

[
  {
    "sinvestigatorname": "Bargman",

   }
]

如何将此输出转换为 JSON 格式（键，值）任何人都可以帮助我如何做到这一点

Answer 1

当您使用 JPA 时，最简单的解决方案是：

@Query("select new map(invest.sinvestigatorname as sinvestigatorname) from InvestigatorModel invest where invest.ninstid=60")
List<InvestigatorModel> findBySinvestigatorname();

这会给你想要的结果。

查看文档 here

在查询中选择特定列时如何将数组结果转换为 JSON 格式

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