Question

EID     PID     Name

1       NULL    A
2       1       B
3       2       C
4       3       D
5       1       E
6       1       F
7       1       G
8       6       H

上面的信息显示了一个表的实际数据，我想在下表中显示该表中的数据。

即。使用EID和PID的数据层次结构显示。

这里PID表示ParentID，EID是EntityID，使用ParentID需要获取Hierarchy，如下所示。

Level1  Level2 Level3  Level4

 A      NULL    NULL    NULL
 A       B      NULL    NULL
 A       B       C      NULL
 A       B       C       D
 A       E      NULL    NULL
 A       F      NULL    NULL
 A       F       H      NULL
 A       G      NULL    NULL

Answer 1

因为您知道级别数，所以可以使用left join：

执行此操作

select l1.name as level1, l2.name as level2,
       l3.name as level3, l4.name as level4
from data l1 left join
     data l2
     on l2.pid = l1.id left join
     data l3
     on l3.pid = l2.id left join
     data l4
     on l4.pid = l3.id;

Answer 2

免责声明：这是（ MS-SQL ）中的一个示例

@Gordon Linoff回答几乎做到了，只是忘了基本情况（ Pid为空）

只需按照左连接+联合的这种模式覆盖固定数量级别的基本情况，对于不需要使用递归的动态数字操作系统级别

declare @Tree as
table (
    Eid int not null
   ,Pid int null
   ,Name char(1) not null
)

insert into @Tree
values
 (1, NULL, 'A')
,(2,    1, 'B')
,(3,    2, 'C')
,(4,    3, 'D')
,(5,    1, 'E')
,(6,    1, 'F')
,(7,    1, 'G')
,(8,    6, 'H')

(
select t1.Name as [Level 1], null as [Level 2], null as [Level 3], null as [Level 4]
from @Tree t1
where t1.Pid is null
union
select t1.Name as [Level 1], t2.Name as [Level 2], null as [Level 3], null as [Level 4]
from @Tree t1
left join @Tree t2 on t2.Pid = t1.Eid
where t1.Pid is null
union
select t1.Name as [Level 1], t2.Name as [Level 2], t3.Name as [Level 3], null as [Level 4]
from @Tree t1
left join @Tree t2 on t2.Pid = t1.Eid
left join @Tree t3 on t3.Pid = t2.Eid
where t1.Pid is null
union
select t1.Name as [Level 1], t2.Name as [Level 2], t3.Name as [Level 3], t4.Name as [Level 4]
from @Tree t1
left join @Tree t2 on t2.Pid = t1.Eid
left join @Tree t3 on t3.Pid = t2.Eid
left join @Tree t4 on t4.Pid = t3.Eid
where t1.Pid is null
) order by [Level 1], [Level 2], [Level 3], [Level 4]

编辑以及使用右连接而不是左连接的相同查询

declare @Tree as
table (
    Eid int not null
   ,Pid int null
   ,Name char(1) not null
)

insert into @Tree
values
 (1, NULL, 'A')
,(2,    1, 'B')
,(3,    2, 'C')
,(4,    3, 'D')
,(5,    1, 'E')
,(6,    1, 'F')
,(7,    1, 'G')
,(8,    6, 'H')

(
select t1.Name as [Level 1], null as [Level 2], null as [Level 3], null as [Level 4]
from @Tree t1
where t1.Pid is null
union
select t1.Name as [Level 1], t2.Name as [Level 2], null as [Level 3], null as [Level 4]
from @Tree t2
right join @Tree t1 on t2.Pid = t1.Eid
where t1.Pid is null
union
select t1.Name as [Level 1], t2.Name as [Level 2], t3.Name as [Level 3], null as [Level 4]
from @Tree t3
right join @Tree t2 on t3.Pid = t2.Eid
right join @Tree t1 on t2.Pid = t1.Eid
where t1.Pid is null
union
select t1.Name as [Level 1], t2.Name as [Level 2], t3.Name as [Level 3], t4.Name as [Level 4]
from @Tree t4
right join @Tree t3 on t4.Pid = t3.Eid
right join @Tree t2 on t3.Pid = t2.Eid
right join @Tree t1 on t2.Pid = t1.Eid
where t1.Pid is null
) order by [Level 1], [Level 2], [Level 3], [Level 4]

层次结构显示Sybase表数据

2 个答案: