Question

有一种表类型：

CREATE TABLE categories
    (`id` int, `pid` int, `name` varchar(35),
     PRIMARY KEY (id)
)
;

INSERT INTO categories
    (`id`, `pid`, `name`)
VALUES
    (1, 0, 'a1'),
    (2, 0, 'a2'),
    (3, 0, 'a3'),
    (4, 0, 'a4'),
    (5, 0, 'a5'),
    (6, 3, 'a6'),
    (7, 3, 'a7'),
    (8, 5, 'a8'),
    (9, 5, 'a9')
;

其中： id - 类别的标识符， pid 是父类别的标识符，名称 - 类别的名称。

如何编写查询，以便像这样：

|          id |                           path |
|-------------|--------------------------------|
|           1 |                             a1 |
|           2 |                             a2 |
|           3 |                             a3 |
|           4 |                             a4 |
|           5 |                             a5 |
|           6 |                        a3 > a6 |
|           7 |                        a3 > a7 |
|           8 |                        a5 > a8 |
|           9 |                        a5 > a9 |

不可能有多项投资。

Answer 1

请参阅JOINS上的文档

您可以使用joins来获得结果。

<强>示例：

mysql> SELECT c.id AS child_id
    ->      , c.name AS child_name
    ->      , c.pid AS parent_id
    ->      , p.name AS parent_name
    ->      , IF( c.pid, CONCAT( p.name, ' > ', c.NAME ), c.NAME ) AS path
    ->   FROM categories AS c
    ->   LEFT JOIN categories AS p ON c.pid = p.id
    ->  ORDER BY child_id;
+----------+------------+-----------+-------------+---------+
| child_id | child_name | parent_id | parent_name | path    |
+----------+------------+-----------+-------------+---------+
|        1 | a1         |         0 | NULL        | a1      |
|        2 | a2         |         0 | NULL        | a2      |
|        3 | a3         |         0 | NULL        | a3      |
|        4 | a4         |         0 | NULL        | a4      |
|        5 | a5         |         0 | NULL        | a5      |
|        6 | a6         |         3 | a3          | a3 > a6 |
|        7 | a7         |         3 | a3          | a3 > a7 |
|        8 | a8         |         5 | a5          | a5 > a8 |
|        9 | a9         |         5 | a5          | a5 > a9 |
+----------+------------+-----------+-------------+---------+
9 rows in set (0.00 sec)

注意：
父 - 子树可以更长。在这种情况下，您最好依赖存储过程或仅将树限制在某个级别。

Answer 2

Sql server：

select a.id,
case when b.id is null then concat('a',a.name)
else concat('a',b.name,' > ','a',a.name) end as path
from categories a
left outer join 
(
select id,pid,name from categories
where id in (select pid from categories)) as b
on a.pid=b.id

此致

Saravanan

Concat类别和子类别

2 个答案: