JDK8中Stream流的使用

[韩_师兄]

问题如下:我有一个部门,部门里面分子部门,子部门里面有孙部门,....可能还有更多层

实体类:
//lombok插件
@Data
public class Dept {

//部门id
private Integer id;
//部门名称
private String deptName;

//子部门集合
private List<Dept> deptList;
}

问题1:通过一次查询.根据父部门id查询 ,把父部门的所有子部门,孙部门,...(可能有孙子部门等),全部查询出来?

问题2:查询出来的父部门,我想获取所有的部门id(父部门id,子部门id,孙部门id,....)放到一个List<Integer>集合中?

[韩_师兄]

@Test public void testtree(){ //模拟从数据库查询出来 List<Menu> menus = Arrays.asList( new Menu(1,"根节点",0), new Menu(2,"子节点1",1), new Menu(3,"子节点1.1",2), new Menu(4,"子节点1.2",2), new Menu(5,"根节点1.3",2), new Menu(6,"根节点2",1), new Menu(7,"根节点2.1",6), new Menu(8,"根节点2.2",6), new Menu(9,"根节点2.2.1",7), new Menu(10,"根节点2.2.2",7), new Menu(11,"根节点3",1), new Menu(12,"根节点3.1",11) ); System.out.println(JSON.toJSON(menus)); System.out.println("===================================="); //获取父节点 List<Menu> collect = menus.stream().filter(m -> m.getParentId() == 0).map( (m) -> { m.setChildList(getChildrens(m, menus)); return m; } ).collect(Collectors.toList()); System.out.println("-------转json输出结果-------"); System.out.println(JSON.toJSON(collect)); } /** * 递归查询子节点 * @param root 根节点 * @param all 所有节点 * @return 根节点信息 */ private List<Menu> getChildrens(Menu root, List<Menu> all) { List<Menu> children = all.stream().filter(m -> { return Objects.equals(m.getParentId(), root.getId()); }).map( (m) -> { m.setChildList(getChildrens(m, all)); return m; } ).collect(Collectors.toList()); return children; }

[qybao]

写个递归方法不就可以了吗？为啥要把问题复杂化
for example

public class Sample {
    public static class Dept {
        private Integer id;
        private String deptName;
        private List<Dept> deptList;
        public Integer getId() {return id;}
        public String getDeptName( ) {return deptName;}
        public List<Dept> getDeptList() {return deptList;}
        public void setId(Integer id) {this.id = id;}
        public void setDeptName(String deptName) {this.deptName = deptName;}
        public void setDeptList(List<Dept> deptList) {this.deptList = deptList;}
        public String toString() {return String.format("id=%s, deptName=%s, deptList=[%s]", id, deptName, deptList);}
    }

    // 测试数据（一共3层，每层2个子部门）
    public static Dept generateSampleDept(int id, int deep) {
        Dept dept = new Dept();
        dept.setId(id);
        dept.setDeptName(String.format("部门%d", id));
        dept.setDeptList(new ArrayList<>());
        if (deep < 3) {
            for (int i=1; i<3; i++) {
                dept.getDeptList().add(generateSampleDept(id*10 + i, deep+1));
            }
        }
        return dept;
    }

    // 查询部门
    public static Dept searchDept(Dept dept, int id) {
        Dept result = null;;
        if (dept.getId().equals(id)) {
            return dept;
        }
        for (Dept sub : dept.getDeptList()) {
            result = searchDept(sub, id);
        }
        return result;
    }

    // 递归获取所有部门
    public static List<Dept> getAllDept(Dept dept) {
        List<Dept> list = new ArrayList<>();
        list.add(dept);
        for (Dept sub : dept.getDeptList()) {
            list.addAll(getAllDept(sub));
        }
        return list;
    }

    public static void main(String[] args) {
        try {
            Dept dept = generateSampleDept(0, 0); //测试数据
            System.out.println(dept); //打印所有部门

            Dept father = searchDept(dept, 22); //查找父部门
            if (father != null) {
                List<Dept> allDept = getAllDept(father); //递归获得所有部门
                System.out.printf("一共有%d个部门\n", allDept.size()); //确认一共有几个部门
                for (Dept sub : allDept) {
                    System.out.println(sub);
                }
                List<Integer> allId = allDept.stream().map(e->e.getId()).collect(Collectors.toList()); //获得所有部门的id
                System.out.println(allId);
            }
        } catch (Throwable e) {
            e.printStackTrace();
        }
    }
}

[韩_师兄]

问题1: 查询出来时,不进行封装,使用stream流+递归完成封装. 问题2:直接使用stream流, List<Integer> collect1 = menus.stream().distinct().map(str -> str.getId()).collect(Collectors.toList());