机构树的数据库设计与查询构想-使用递归算法

需求

• 设计机构树，查询实验室所有机构
• 以某一节点为根节点，查询局部机构树

实现

• 思路：

1. 自身嵌套查询
2. 代码递归查询

• 数据库设计

CREATE TABLE `sys_dept` (
  `id` bigint NOT NULL AUTO_INCREMENT COMMENT '编号',
  `name` varchar(20) CHARACTER SET utf8mb4 COLLATE utf8mb4_0900_ai_ci NOT NULL COMMENT '机构名',
  `dept_logo` varchar(200) CHARACTER SET utf8mb4 COLLATE utf8mb4_0900_ai_ci NOT NULL COMMENT '机构ID',
  `parent_id` bigint NOT NULL COMMENT '上级机构',
  `create_by` varchar(20) CHARACTER SET utf8mb4 COLLATE utf8mb4_0900_ai_ci NOT NULL COMMENT '创建人',
  `create_time` datetime NOT NULL COMMENT '创建时间',
  `last_update_time` datetime NOT NULL COMMENT '更新人',
  `last_update_by` varchar(20) CHARACTER SET utf8mb4 COLLATE utf8mb4_0900_ai_ci NOT NULL COMMENT '更新时间',
  `del_flag` tinyint NOT NULL DEFAULT '0' COMMENT '删除标志，-1删除，0正常',
  PRIMARY KEY (`id`),
  UNIQUE KEY `UK_DEPT_NAME` (`name`) USING BTREE COMMENT '机构名唯一性注解'
) ENGINE=InnoDB AUTO_INCREMENT=6 DEFAULT CHARSET=utf8mb4 COLLATE=utf8mb4_0900_ai_ci;

2.设计对象

@JsonIgnoreProperties(value = {"handler"})
public class SysDept implements Serializable {
    private Long id;

    private String name;

    private String deptLogo;
    //父机构节点，第一级节点设置为0
    private Long parentId;

    private String createBy;
    @JsonFormat(pattern = "yyyy-MM-dd",timezone = "GMT+8")
    @DateTimeFormat(pattern = "yyyy-MM-dd")
    private Date createTime;
    @JsonFormat(pattern = "yyyy-MM-dd",timezone = "GMT+8")
    @DateTimeFormat(pattern = "yyyy-MM-dd")
    private Date lastUpdateTime;

    private String lastUpdateBy;

    private Byte delFlag;
    //三级权限嵌套查询使用
    private List<SysUser> userList;
    //构建机构树使用，子机构节点
    private List<SysDept> deptList;

3.遍历获取子节点

  private List<SysDept> getChildNodes(Long id, List<SysDept> rootList){
        //新建字节点列表
        List<SysDept> chileList = new ArrayList<>();
        //根据父节点ID填充对应的字节点
        for(SysDept sysDept : rootList){
            if(sysDept.getParentId() != null){
                if(sysDept.getParentId() == id){
                    chileList.add(sysDept);
                }
            }
        }
        if(chileList.size() == 0){
            return null;
        }
        //便利子节点并且进行对应的填充
        for(SysDept sysDept: chileList){
            sysDept.setDeptList(getChildNodes(sysDept.getId(), rootList));
        }
        return chileList;
    }

4.具体实现代码

   @ApiOperation(value = "查询机构树",notes = "查询机构树")
    @ApiImplicitParams({
            @ApiImplicitParam(type = "query", name = "id", value = "机构树根目录ID,规定根目录的ID为1",required = true)

    })
    @PostMapping("/find/tree")
    public HttpResult findDeptNodes(Long id){
        try{
            SysDept sysDept = sysDeptService.findById(id);
            List<SysDept> rootList = sysDeptService.findRootTree();
            List<SysDept> depts = getChildNodes(id, rootList);
            sysDept.setDeptList(depts);
            return HttpResult.ok(sysDept);
        }catch (Exception e){
            e.printStackTrace();
            return HttpResult.error("查询失败");
        }
    }

-------改进一下–

• 根据父ID进行机构树查询
  因为我们不知道顶级机构树的ID到底是哪一个，但是我们可以知道顶级机构的父机构ID，人为设置为-1就可以

  @ApiOperation(value = "查询机构树",notes = "查询机构树")
    @ApiImplicitParams({
            @ApiImplicitParam(type = "query", name = "parentId", value = "规定根目录的父ID为-1，",required = true)

    })
    @PostMapping("/find/tree")
    public HttpResult findDeptNodes(Long parentId){
        try{
            System.out.println("parentId:"+parentId);
            List<SysDept> sysDepts = sysDeptService.findByParentId(parentId);
            //SysDept sysDept = sysDeptService.findById(id);
            List<SysDept> rootList = sysDeptService.findRootTree();
            for(SysDept sysDept: sysDepts){
                List<SysDept> depts = getChildNodes(sysDept.getParentId(), rootList);
                sysDept.setDeptList(depts);
            }
            return HttpResult.ok(sysDepts);
        }catch (Exception e){
            e.printStackTrace();
            return HttpResult.error("查询失败");
        }
    }