开发中经常会遇到树形结构的场景,比如:导航菜单、组织机构等等,但凡是有这种父子层级结构的都是如此,一级类目、二级类目、三级类目。。。
对于这种树形结构的表要如何设计呢?接下来一起探讨一下
首先,想一个问题,用非关系型数据库存储可不可以?
答案是肯定可以的,比如用mongoDB,直接将整棵树存成json。但是,这样不利于按条件查询,当然也取决于具体的需求,抛开需求谈设计都是耍流氓。
在菜单这个场景下,一般还是用关系型数据库存储,可以将最终的查询结构缓存起来。
常用的方法有四种:
- 每一条记录存parent_id
- 每一条记录存整个tree path经过的node枚举
- 每一条记录存 nleft 和 nright
- 维护一个表,所有的tree path作为记录进行保存
第一种:每条记录存储parent_id
这种方式简单明了,但是想要查询某个节点的所有父级和子级的时候比较困难,势必需要用到递归,在mysql里面就得写存储过程,太麻烦了。
当然,如果只有两级的话就比较简单了,自连接就搞定了,例如:
第四种:单独用一种表保存节点之间的关系
CREATE TABLE `city` ( `id` int(11) NOT NULL AUTO_INCREMENT, `name` varchar(16), PRIMARY KEY (`id`) USING BTREE) ENGINE = InnoDB AUTO_INCREMENT = 1 CHARACTER SET = utf8mb4;CREATE TABLE `city_tree_path_info` ( `id` int(11) NOT NULL AUTO_INCREMENT, `city_id` int(11) NOT NULL, `ancestor_id` int(11) NOT NULL COMMENT '祖先ID', `level` tinyint(4) NOT NULL COMMENT '层级', PRIMARY KEY (`id`) USING BTREE) ENGINE = InnoDB AUTO_INCREMENT = 1 CHARACTER SET = utf8mb4;
上面这个例子中,city表代表城市,city_tree_path_info代表城市之间的层级关系,ancestor_id表示父级和祖父级ID,level是当前记录相对于ancestor_id而言的层级。这样就把整个层级关系保存到这张表中了,以后想查询某个节点的所有父级和子级就很容易了。
最后,我发现构造这种层级树最简单的还是用java代码
java递归生成菜单树
Menu.java
1 package com.example.demo.model; 2 3 import lombok.AllArgsConstructor; 4 import lombok.Data; 5 import lombok.NoArgsConstructor; 6 7 import java.util.List; 8 9 @AllArgsConstructor10 @NoArgsConstructor11 @Data12 public class Menu {13 14 /**15 * 菜单ID16 */17 private Integer id;18 19 /**20 * 父级菜单ID21 */22 private Integer pid;23 24 /**25 * 菜单名称26 */27 private String name;28 29 /**30 * 菜单编码31 */32 private String code;33 34 /**35 * 菜单URL36 */37 private String url;38 39 /**40 * 菜单图标41 */42 private String icon;43 44 /**45 * 排序号46 */47 private int sort;48 49 /**50 * 子级菜单51 */52 private List<Menu> children;53 54 public Menu(Integer id, Integer pid, String name, String code, String url, String icon, int sort) {55 this.id = id;56 this.pid = pid;57 this.name = name;58 this.code = code;59 this.url = url;60 this.icon = icon;61 this.sort = sort;62 }63 64 }
Test.java
1 package com.example.demo.model; 2 3 import com.faster; 4 import com.faster; 5 6 import java.util.ArrayList; 7 import java.util.Comparator; 8 import java.util.List; 9 import java.util.stream.Collectors;10 11 public class Hello {12 public static void main(String[] args) throws JsonProcessingException {13 List<Menu> allMenuList = new ArrayList<>();14 allMenuList.add(new Menu(1, 0, "湖北", "HuBei", "/a", "a", 3));15 allMenuList.add(new Menu(2, 0, "河南", "HeNan", "/b", "b", 2));16 allMenuList.add(new Menu(3, 1, "宜昌", "YiChang", "/c", "c", 2));17 allMenuList.add(new Menu(4, 2, "信阳", "XinYang", "/d", "d", 1));18 allMenuList.add(new Menu(5, 1, "随州", "SuiZhou", "/e", "e", 1));19 allMenuList.add(new Menu(6, 5, "随县", "SuiXian", "/f", "f", 2));20 allMenuList.add(new Menu(7, 3, "枝江", "ZhiJiang", "/g", "g", 2));21 22 // 一级菜单23 List<Menu> parentList = allMenuList.stream().filter(e->e.getPid()==0).sorted(Comparator.comparing(Menu::getSort)).collect(Collectors.toList());24 // 递归调用,为所有一级菜单设置子菜单25 for (Menu menu : parentList) {26 menu.setChildren(getChild(menu.getId(), allMenuList));27 }28 29 ObjectMapper objectMapper = new ObjectMapper();30 System.out.println(objectMapper.writeValueAsString(parentList));31 }32 33 /**34 * 递归查找子菜单35 * @param id 当前菜单ID36 * @param allList 查找菜单列表37 * @return38 */39 public static List<Menu> getChild(Integer id, List<Menu> allList) {40 // 子菜单41 List<Menu> childList = new ArrayList<>();42 for (Menu menu : allList) {43 if (menu.getPid().equals(id)) {44 childList.add(menu);45 }46 }47 48 // 为子菜单设置子菜单49 for (Menu nav : childList) {50 nav.setChildren(getChild(nav.getId(), allList));51 }52 53 // 排序54 childList = childList.stream().sorted(Comparator.comparing(Menu::getSort)).collect(Collectors.toList());55 56 if (childList.size() == 0) {57 // return null;58 return new ArrayList<>();59 }60 return childList;61 }62 }
结果:
1 [ 2 { 3 "id":2, 4 "pid":0, 5 "name":"河南", 6 "code":"HeNan", 7 "url":"/b", 8 "icon":"b", 9 "sort":2,10 "children":[11 {12 "id":4,13 "pid":2,14 "name":"信阳",15 "code":"XinYang",16 "url":"/d",17 "icon":"d",18 "sort":1,19 "children":[]20 }21 ]22 },23 {24 "id":1,25 "pid":0,26 "name":"湖北",27 "code":"HuBei",28 "url":"/a",29 "icon":"a",30 "sort":3,31 "children":[32 {33 "id":5,34 "pid":1,35 "name":"随州",36 "code":"SuiZhou",37 "url":"/e",38 "icon":"e",39 "sort":1,40 "children":[41 {42 "id":6,43 "pid":5,44 "name":"随县",45 "code":"SuiXian",46 "url":"/f",47 "icon":"f",48 "sort":2,49 "children":[]50 }51 ]52 },53 {54 "id":3,55 "pid":1,56 "name":"宜昌",57 "code":"YiChang",58 "url":"/c",59 "icon":"c",60 "sort":2,61 "children":[62 {63 "id":7,64 "pid":3,65 "name":"枝江",66 "code":"ZhiJiang",67 "url":"/g",68 "icon":"g",69 "sort":2,70 "children":[]71 }72 ]73 }74 ]75 }76 ]
参考:
https://www.cnblogs.com/w2206/p/10490208.html
https://www.cnblogs.com/mokingone/p/9109021.html
https://www.cnblogs.com/makai/p/12301707.html
https://www.cnblogs.com/zhifengge/p/6910881.html
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开发中经常会遇到树形结构的场景,比如:导航菜单、组织机构等等,但凡是有这种父子层级结构的都是如此,一级类目、二级类目、三级类目。。。对于这种树形结构的表要如何设计呢?接下来一起探讨一下首先,想一个问题,用非关系型数据库存储可不可以?答案是肯定可以的,比如用mongoDB,直接将整棵树存成json。但是,这样不利于按条件查询,当然也取决于具体的需求,抛开需求谈设计都是耍流氓。在菜单这个场景下,一般还
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