Question

Oracle 11g - 选择查询帮助。

表＆＃34; data_table＆＃34;有如下三个字段，并保持子父关系，如下所示 -

user_code   -> description  -> parent_code

A          -> Root ID   ->    B

B        ->   Level1    ->    C

C      ->     Level2    ->    D

D    ->       Level3    ->    E

E       ->    Level4    ->     F

要求是将此层次结构发布在一行中，如下所示。

**
Root ID -> Level1 ->    Level2  -> Level3->     Level4

A   -> B    -> C    -> D    -> E **

我有什么选择？请帮忙..

现在我使用下面的查询，但看起来并不好。

 SELECT e.user_code   user_code,
    d.description user_description,
    e.parent_code level1,
    a.parent_code level2 ,
    b.parent_code level3 ,
    c.parent_code level4
FROM    data_table e,
    data_table a,
    data_table b,
    data_table c,
    data_table d
WHERE   e.user_code = d.user_code
AND     b.user_code = a.parent_code
AND     c.user_code = b.parent_code
AND     e.parent_code = a.user_code

Answer 1

您必须使用通过user_code和parent_code连接父级和子级的层次结构查询。像这样：

$sth = $dbh->prepare($sql, array(PDO::ATTR_CURSOR => PDO::CURSOR_FWDONLY));

SYS_CONNECT_BY_PATH显示路径，但如果您想要将整个路径放在一行中，则需要获取MAX。

您可以阅读有关等级查询here

的更多信息

Answer 2

如果您知道有固定数量的关卡，则可以执行以下操作：

with sample_data as (select 'A' user_code, 'Root ID' description, 'B' parent_code from dual union all
                     select 'B' user_code, 'Level 1' description, 'C' parent_code from dual union all
                     select 'C' user_code, 'Level 2' description, 'D' parent_code from dual union all
                     select 'D' user_code, 'Level 3' description, 'E' parent_code from dual union all
                     select 'E' user_code, 'Level 4' description, 'F' parent_code from dual union all
                     select 'G' user_code, 'Root ID' description, 'H' parent_code from dual union all
                     select 'H' user_code, 'Level 1' description, 'I' parent_code from dual union all
                     select 'I' user_code, 'Level 2' description, 'J' parent_code from dual union all
                     select 'J' user_code, 'Level 3' description, 'K' parent_code from dual),
             res as (select sd.user_code,
                            sd.description,
                            connect_by_root(sd.user_code) top_node
                     from   sample_data sd
                     connect by sd.user_code = prior sd.parent_code
                     start with sd.description = 'Root ID')
select root_id,
       level1,
       level2,
       level3,
       level4
from   res
pivot  (max(user_code) for description in ('Root ID' as root_id,
                                           'Level 1' as level1,
                                           'Level 2' as level2,
                                           'Level 3' as level3,
                                           'Level 4' as level4))
order by top_node;

ROOT_ID LEVEL1 LEVEL2 LEVEL3 LEVEL4
------- ------ ------ ------ ------
A       B      C      D      E     
G       H      I      J

如果级别数量未知，那么您将不得不调查动态数据透视 - 如果您搜索Google，有很多关于如何进行动态数据透视的示例。

在单行中选择子父层次结构pl {s}

2 个答案: