Question

我有一个树结构的表：

id parentId name
----------------
1  0        Category1
2  0        Category2
3  1        Category3
4  2        Category4
5  1        Category5
6  2        Category6
7  3        Category7

在sql查询结果中，我需要一个像：

这样的表

id parentId level name
----------------------
1  0        0     Category1
3  1        1     Category3
7  3        2     Category7
5  1        1     Category5
2  0        0     Category2
4  2        1     Category4
6  2        1     Category6

谁能帮我写mysql查询？谢谢！

Answer 1

WITH tree (id, parentid, level, name) as 
(
   SELECT id, parentid, 0 as level, name
   FROM your_table
   WHERE parentid = 0

   UNION ALL

   SELECT c2.id, c2.parentid, tree.level + 1, c2.name
   FROM your_table c2 
     INNER JOIN tree ON tree.id = c2.parentid
)
SELECT *
FROM tree;

我目前没有SQL Server来测试它，因此可能存在一些拼写错误（语法错误）

Answer 2

扩展a_horse_with_no_name的答案，这将展示如何结合row_number（）使用SQL Server的implementation of recursive CTE（递归单记录交叉应用）来生成问题中的确切输出。

declare @t table(id int,parentId int,name varchar(20))
insert @t select 1,  0        ,'Category1'
insert @t select 2,  0,        'Category2'
insert @t select 3,  1,        'Category3'
insert @t select 4 , 2,        'Category4'
insert @t select 5 , 1,        'Category5'
insert @t select 6 , 2,        'Category6'
insert @t select 7 , 3,        'Category7'
;

WITH tree (id, parentid, level, name, rn) as 
(
   SELECT id, parentid, 0 as level, name,
       convert(varchar(max),right(row_number() over (order by id),10)) rn
   FROM @t
   WHERE parentid = 0

   UNION ALL

   SELECT c2.id, c2.parentid, tree.level + 1, c2.name,
       rn + '/' + convert(varchar(max),right(row_number() over (order by tree.id),10))
   FROM @t c2 
     INNER JOIN tree ON tree.id = c2.parentid
)
SELECT *
FROM tree
order by RN

老实说，使用ID本身来生成树“路径”会起作用，因为我们直接按id排序，但我想我会在row_number（）函数中滑动。

树表的SQL查询

2 个答案: