Question

尝试将几个查询组合在一起但遇到问题，无论我尝试将它们连接在一起。这是两个的模型：

查询1（获取所有G和相关R）：

select distinct 
    a.g_id, 
    a.g_name,
    b.r_name, 
    b.r_id
from 
    g_tab a, 
    r_tab b,
    f_tab c
where
    a.g_id = b.g_id
    and b.r_id = c.r_id
    and c.f_id = 7
order by a.g_id

查询2（遍历层次结构并过滤到特定组）：

with parentGs as 
(
    select 
        f.g_id,
        f.g_name
    from 
        g_tab f
    where  
        f.delete_indicator is null
    connect by
       prior f.parent_g_id = f.g_id
    start with
        f.g_id = 22 --DO NOT WANT THIS HARDCODED
)

select 
    d.g_name
from
    parentGs d,
    g_group_members_tab e
where 
    d.g_id = e.member_g_id
    and e.g_group_description = 'test'

我不想更改查询1的返回计数，但是想要从查询2加入g_name并将其添加为新的别名列。

以下是我尝试获得的一个示例，但由于g_id被调用而无法正常工作＆＃34;无效的标识符＆＃34;在＆＃34;以＆＃34;开头。我知道父值只有一个子查询有效，但如果仅使用一个子查询进行重组，仍会得到同样的错误（在＆＃34;来自＆＃34;在这种情况下）：

select distinct 
    a.g_id, 
    a.g_name,
    b.r_name, 
    b.r_id,
    (select 
        d.g_name
    from
        (select 
            f.g_id,
            f.g_name
        from 
            g_tab f
        where  
            f.delete_indicator is null
        connect by
           prior f.parent_g_id = f.g_id
        start with
            f.g_id = a.g_id) d, --THIS GIVES INVALID IDENTIFIER
        g_group_members_tab e
    where 
        d.g_id = e.member_g_id
        and e.g_group_description = 'test') as parent_g_name
from 
    g_tab a, 
    r_tab b,
    f_tab c
where
    a.g_id = b.g_id
    and b.r_id = c.r_id
    and c.f_id = 7
order by a.g_id

非常感谢任何帮助加入这两个问题的帮助。我的目标平台是Oracle 11g。非常感谢！

*更新*

以下是一些示例SQL Fiddle数据：

create table g_tab
( 
  g_id number(9),
  g_name varchar2(50),
  parent_g_id number(9),
  delete_indicator char(1)
)
/
create table g_group_members_tab
( 
  member_g_id number(9),
  g_group_description varchar2(50)
)
/
create table r_tab
( 
  r_id number(9),
  r_name varchar2(50),
  g_id number(9)
)
/
create table f_tab
( 
  f_id number(9),
  f_name varchar2(50),
  r_id number(9)
)
/
insert into g_tab (g_id, g_name, parent_g_id) values (0, 'a', null)
/
insert into g_tab (g_id, g_name, parent_g_id) values (1, 'b', 0)
/
insert into g_tab (g_id, g_name, parent_g_id) values (2, 'c', 1)
/
insert into g_tab (g_id, g_name, parent_g_id) values (3, 'd', null)
/
insert into g_tab (g_id, g_name, parent_g_id) values (4, 'e', 3)
/
insert into g_tab (g_id, g_name, parent_g_id) values (5, 'f', 4)
/
insert into g_tab (g_id, g_name, parent_g_id) values (6, 'g', null)
/
insert into g_group_members_tab (member_g_id, g_group_description) values (1, 'test')
/
insert into g_group_members_tab (member_g_id, g_group_description) values (3, 'test')
/
insert into r_tab (r_id, r_name, g_id) values (0, 'r', 0)
/
insert into r_tab (r_id, r_name, g_id) values (0, 'r', 1)
/
insert into r_tab (r_id, r_name, g_id) values (0, 'r', 2)
/
insert into r_tab (r_id, r_name, g_id) values (0, 'r', 3)
/
insert into r_tab (r_id, r_name, g_id) values (0, 'r', 4)
/
insert into r_tab (r_id, r_name, g_id) values (1, 'rr', 5)
/
insert into r_tab (r_id, r_name, g_id) values (0, 'r', 6)
/
insert into f_tab (f_id, f_name, r_id) values (7, 'f', 0)
/
insert into f_tab (f_id, f_name, r_id) values (7, 'f', 1)

期望的结果：

g_id    g_name    r_name    r_id    parent_g_name
0       a         r         0       null
1       b         r         0       b   
2       c         r         0       b
3       d         r         0       d
4       e         r         0       d
5       f         rr        1       d
6       g         r         0       null

Answer 1

只需将子查询转换为内嵌视图，即可完成大部分工作：

select distinct 
    a.g_id, 
    a.g_name,
    b.r_name, 
    b.r_id,
    d.g_name as parent_g_name
from 
    g_tab a, 
    r_tab b,
    f_tab c,
    (
        select 
            f.g_id,
            f.g_name,
            connect_by_root f.g_id as root_g_id
        from 
            g_tab f
        where  
            f.delete_indicator is null
        connect by
            prior f.parent_g_id = f.g_id
    ) d,
    g_group_members_tab e
where
    a.g_id = b.g_id
    and b.r_id = c.r_id
    and c.f_id = 7
    and a.g_id = d.root_g_id
    and d.g_id = e.member_g_id
    and e.g_group_description = 'test'
order by a.g_id

给出了：

      G_ID G_NAME R_NAME       R_ID PARENT_G_NAME
---------- ------ ------ ---------- -------------
         1 b      r               0 b             
         2 c      r               0 b             
         3 d      r               0 d             
         4 e      r               0 d             
         5 f      rr              1 d

您错过了空值，因此您需要将其转换为外部联接;但是你需要加入d到e。使用ANSI连接时，所有这些都更令人愉快：

select distinct 
    a.g_id, 
    a.g_name,
    b.r_name, 
    b.r_id,
    d.g_name as parent_g_name
from g_tab a
join r_tab b
on   b.g_id = a.g_id
join f_tab c
on   c.r_id = b.r_id
left join (
    select
        e.g_id,
        e.g_name,
        e.root_g_id
    from (
        select 
            f.g_id,
            f.g_name,
            connect_by_root f.g_id as root_g_id
        from 
            g_tab f
        where  
            f.delete_indicator is null
        connect by
            prior f.parent_g_id = f.g_id) e
    join g_group_members_tab f
    on   f.member_g_id = e.g_id
    where f.g_group_description = 'test'
    ) d
on  d.root_g_id = a.g_id
where c.f_id = 7
order by a.g_id

给出了：

      G_ID G_NAME R_NAME       R_ID PARENT_G_NAME
---------- ------ ------ ---------- -------------
         0 a      r               0               
         1 b      r               0 b             
         2 c      r               0 b             
         3 d      r               0 d             
         4 e      r               0 d             
         5 f      rr              1 d             
         6 g      r               0

SQL Fiddle demo

SQL子查询使用父值在＆＃34;开始于＆＃34;

1 个答案: