list转tree结构时如何加上排序？

Z时代
2024-02-28
分类：IT

Entity

@Data
public class MenuNode {
    private Integer id;
    private String name;
    private String menuCode;
    private String parentCode;
    private String path;
    private Integer sort;
    private List<MenuNode> children;}

Controller

@RequestMapping("/trees")
public List<MenuNode> trees() {
    // return  service.trees();
    return listToTree(service.trees());
}
// list转tree结构
private List<MenuNode> listToTree(List<MenuNode> list) {
    List<MenuNode> tree = new ArrayList<>();
    for (MenuNode menu : list) {
        //找到根节点
        if (menu.getParentCode().length() == 0) {
            tree.add(menu);
        }
        //再次遍历list，找到子节点
        List<MenuNode> children = new ArrayList<>();
        for (MenuNode node : list) {
            if (node.getParentCode().equals(menu.getMenuCode())) {
                children.add(node);
            }
        }
        menu.setChildren(children);
    }
    return tree;}

id	menu_code	parent_code	name	path	sort
1	ME00001		菜单1	/a	3
2	ME00002		菜单2	/b	2
3	ME00003		菜单3	/c	1
4	ME00004	ME00001	菜单11	/a/i	4
5	ME00005	ME00001	菜单12	/a/j	0
6	ME00006	ME00004	菜单111	/a/i/x	0
7	ME00007	ME00004	菜单112	/a/i/y	0

结果没排序

sort数据库默认值0，排序不同层级的要设置成唯一吗？
再写一个方法，递归遍历数组排序？

[ { "id": 1, "name": "菜单1", "menuCode": "ME00001", "parentCode": "", "path": "/a", "sort": 1, "children": [ { "id": 4, "name": "菜单11", "menuCode": "ME00004", "parentCode": "ME00001", "path": "/a/i", "sort": 4, "children": [ { "id": 6, "name": "菜单111", "menuCode": "ME00006", "parentCode": "ME00004", "path": "/a/i/x", "sort": 0, "children": [] }, { "id": 7, "name": "菜单112", "menuCode": "ME00007", "parentCode": "ME00004", "path": "/a/i/y", "sort": 0, "children": [] } ] }, { "id": 5, "name": "菜单12", "menuCode": "ME00005", "parentCode": "ME00001", "path": "/a/j", "sort": 0, "children": [] } ] }, { "id": 2, "name": "菜单2", "menuCode": "ME00002", "parentCode": "", "path": "/b", "sort": 2, "children": [] }, { "id": 3, "name": "菜单3", "menuCode": "ME00003", "parentCode": "", "path": "/c", "sort": 3, "children": [] }

]

回答：

楼主可以用list的stream流来进行转换
例如：

//查询数据库所有的数据
List<AllSpaceVo> spaceInfoAllVos = mapperFacade.mapAsList(list, AllSpaceVo.class);
        //查询出一级空间
        List<AllSpaceVo> collect = spaceInfoAllVos.stream()
                .filter(info -> {
                    if (id == null) {
                        return info.getSpaceParentId().equals(0L);
                    } else {
                        return info.getSpaceParentId().equals(id);
                    }
                })
                .map(menu -> {
                    menu.setChildSpaces(getChild(menu, spaceInfoAllVos));
                    return menu;
                })
                .sorted(Comparator.comparing(AllSpaceVo::getDefOrder)
                        .thenComparing(AllSpaceVo::getCreateTime, Comparator.reverseOrder()))
                .collect(Collectors.toList());

 private List<AllSpaceVo> getChild(AllSpaceVo root, List<AllSpaceVo> all) {
        return all.stream()
                .filter((allSpaceVo -> root.getId().equals(allSpaceVo.getSpaceParentId())))
                .map(menu -> {
                    menu.setChildSpaces(getChild(menu, all));
                    return menu;
                })
                .sorted(Comparator.comparing(AllSpaceVo::getDefOrder)
                        .thenComparing(AllSpaceVo::getCreateTime, Comparator.reverseOrder()))
                .collect(Collectors.toList());    }

用sorted进行排序