原來在聚鯊的練習題（實際操作）(已經理解了)
主要需要理解的部分是：inner join 樂翻天join right join 的區別。二者之間的區別是以誰為主要的表，主要顯示誰的問題。
其次需要理解的是 select * from a inner join b on a.khdm = b.khdm inner join c on c.khdm = a.khdm 只要是可以繼續下去的就可以 不斷的寫下去
第三個需要注意的是：我的可以在我想加入的渠道中顯示任何我已經有的知識。如「北京」 tb
select
*from
(select
distinct a.khdm, a.zfl
from hqods.yg_res_ddyjb a
-- left join hqods.yg_ref_good_info b
-- on a.spbh = b.spbh
where a.zfl in
('紀念收藏品'
)and a.khdj in
('紫金會員'
,'**會員'
,'普通會員'
,'白金會員'
,'鑽石會員'
)--group by khdm, b.zfl
) m inner
join
(select cust_id khdm, cust_lvl_desc 會員等級, province_nm
from hqods.yg_dim_customer h
where cust_lvl_desc in
('紫金會員'
,'**會員'
,'普通會員'
,'白金會員'
,'鑽石會員'
)) x
on x.khdm = m.khdm
inner
join
(select e.customernumber,
u.name,
g.grpname,
e.bindstatus,
'北京' tb
from hqods.v_bf_crm_cp_customer e ---------------會員明細，繫結關係
inner
join hqods.crm_sys_user u
on e.assignedusr = u.usrid
inner
join hqods.crm_sys_uglink ug
on ug.usrid = u.usrid
inner
join hqods.crm_sys_grp g
on g.grpid = ug.grpid
inner
join
(select g.grpid,
g.grpname,
g.parentid,
rownum rn
from hqods.crm_sys_grp g
where g.
type
='1'
start
with g.grpid =
'mgrlv'
connect
by prior g.grpid = g.parentid) t1
on g.grpid = t1.grpid
inner
join
(select g.grpid,
g.grpname,
g.parentid,
rownum rn
from hqods.crm_sys_grp g
where g.
type
='1'
start
with g.grpid =
'agentgrp'
connect
by prior g.grpid = g.parentid) t
on g.grpid = t.grpid
where e.bindstatus =
1) h
--left join (select grpname 組, td from hqods.v_yg_dim_dsjzb) e
-- on h.grpname = e.組
-- group by customernumber, name, grpname, td
-- ) f
on m.khdm = h.customernumber
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