【SpringBoot Web开发】如何构建树形结构数据

树形结构数据

应用场景

比如我们需要构建菜单、机构树、其他业务类型树形结构

工具类

1. 我们可以把菜单列表返回，获取parent_id字段等于0的节点，称为根节点，这样的节点代表一级菜单
2. 再通过根节点的主键去寻找子菜单，因为要有多及菜单，所以要用递归构建子树，直到没有子菜单为止
3. 最后通过构建完整的菜单树

代码示例

public class MenuTree {

    private final List<Menu> menuList;

    public MenuTree(List<Menu> menuList) {
        this.menuList = menuList;
    }

    /**
     * 获取根结点
     *
     * @return
     */
    private List<Menu> getRootNode() {
        List<Menu> rootNode = new ArrayList<>();
        menuList.forEach(item -> {
            if (item.getParentId() == 0) {
                rootNode.add(item);
            }
        });
        return rootNode;
    }

    /**
     * 构建子树
     *
     * @param rootNode
     * @return
     */
    private Menu builderChildrenNode(Menu rootNode) {
        List<Menu> childrenList = new ArrayList<>();
        menuList.forEach(item -> {
            if (Objects.equals(item.getParentId(), rootNode.getId())) {
                // 还需要遍历三级菜单以后的
                Menu menu = builderChildrenNode(item);
                childrenList.add(menu);
            }
        });
        rootNode.setChildrenList(childrenList);
        return rootNode;
    }

    /**
     * 构建树
     *
     * @return
     */
    public List<Menu> buildTree() {
        List<Menu> menus = getRootNode();
        menus.forEach(this::builderChildrenNode);
        return menus;
    }
}

返回结果示例

只展示部分数据

{
  "code": 0,
  "data": [
    {
      "id": 1,
      "parentId": 0,
      "menuName": "系统管理",
      "menuIcon": "el-icon-setting",
      "childrenList": [
        {
          "id": 2,
          "parentId": 1,
          "menuName": "用户管理",
          "menuIcon": "el-icon-service",
          "childrenList": [
            {
              "id": 3,
              "parentId": 2,
              "menuName": "查看",
              "menuIcon": null,
              "childrenList": []
            },
            {
              "id": 4,
              "parentId": 2,
              "menuName": "新增",
              "menuIcon": null,
              "childrenList": []
            },
            {
              "id": 5,
              "parentId": 2,
              "menuName": "修改",
              "menuIcon": null,
              "childrenList": []
            },
            {
              "id": 6,
              "parentId": 2,
              "menuName": "删除",
              "menuIcon": null,
              "childrenList": []
            }
          ]
        }
      ]
    }
  ],
  "msg": "执行成功"
}

SQL

通过SQL的方式的话，只能查询出两层，如下所示：

代码示例

DOCTYPE mapper PUBLIC "-//mybatis.org//DTD Mapper 3.0//EN" "http://mybatis.org/dtd/mybatis-3-mapper.dtd">
<mapper namespace="xx.tree.dao.MenuDao">
    
    <resultMap type="xx.tree.entity.Menu" id="MenuMap">
        <result property="id" column="id" jdbcType="INTEGER"/>
        <result property="parentId" column="parent_id" jdbcType="INTEGER"/>
        <result property="menuName" column="menu_name" jdbcType="VARCHAR"/>
        <result property="menuIcon" column="menu_icon" jdbcType="VARCHAR"/>
        <collection property="childrenList" ofType="xx.tree.entity.Menu" select="queryChildrenMenuInfo" column="id"/>
    resultMap>

    
    <select id="queryMenuTreeBySQL" parameterType="int" resultMap="MenuMap">
        select *
        from menu
        where parent_id  = 0
    select>
    
    
    <select id="queryChildrenMenuInfo" parameterType="int" resultType="xx.tree.entity.Menu">
        select *
        from menu
        where parent_id = #{id}
    select>
mapper>

返回结果示例

只展示部分数据

{
    "code": 0,
    "data": [
        {
            "id": 1,
            "parentId": 0,
            "menuName": "系统管理",
            "menuIcon": "el-icon-setting",
            "childrenList": [
                {
                    "id": 2,
                    "parentId": 1,
                    "menuName": "用户管理",
                    "menuIcon": "el-icon-service",
                    "childrenList": null
                }
            ]
        }
    ],
    "msg": "执行成功"
}