一、hive中實現方法
基表：
組表：
| gt |
|
| gid |
gname |
| 1001 |
g1 |
| 1002 |
g2 |
| 1003 |
g3 |
create table g(
gid int,
gname string
)row format delimited fields terminated by '\t'
stored as textfile;
用戶表：
| ut |
|
| uid |
uname |
| 10001 |
u1 |
| 10002 |
u2 |
| 10003 |
u3 |
| 10004 |
u4 |
| 10005 |
u5 |
| 10006 |
u6 |
| 10007 |
u7 |
| 10008 |
u8 |
| 10009 |
u9 |
| 10010 |
u10 |
create table u(
uid int,
uname string
)row format delimited fields terminated by '\t'
stored as textfile;
權限表：
| gu |
|
| gid |
uid |
| 1001 |
10002,10001,10003,10009 |
| 1002 |
10004,10005,10006 |
| 1003 |
10007,10008,10010 |
create table gu(
gid int,
uid string
)row format delimited fields terminated by '\t'
stored as textfile;
組表gt中記錄了組的信息組id和組名稱，用戶表記錄了用戶基本信息用戶id和用戶名稱，gu是組表和用戶表的關系，記錄了每一個組內與用戶對應關系，其中僅記錄id信息。題目是根據gt和ut表將gu表中的所有id轉換為名稱？
我寫的SQL是：
select t.gname,concat_ws(',',collect_set(t.uname)) from (
select g.gname,u.uname
from (
select gid,s_uid from gu
lateral view explode(split(uid,',')) uid as s_uid) temp,g,u
where temp.gid=g.gid and temp.s_uid=u.uid) t
group by t.gname;
運行結果如下：
hive> select t.gname,concat_ws(',',collect_set(t.uname)) from (
> select g.gname,u.uname
> from (
> select gid,s_uid from gu
> lateral view explode(split(uid,',')) uid as s_uid) temp,g,u
> where temp.gid=g.gid and temp.s_uid=u.uid) t
> group by t.gname;
OK
g1 u2,u1,u3,u9
g2 u4,u5,u6
g3 u7,u8,u10
二、oracle中實現方法
1、建立基表
create table g(
gid number(10),
gname varchar2(20)
)
create table u(
usrid number(10),
uname varchar2(20)
)
create table gu(
gid number(10),
usrid varchar2(200)
)
我所實現的sql方法如下：
select g.gname,
(select wm_concat(uname) from u where instr(gu.usrid, u.usrid) > 0)
from gu, g
where gu.gid = g.gid;
執行結果：
SQL> select g.gname,
2 (select wm_concat(uname) from u where instr(gu.usrid, u.usrid) > 0)
3 from gu, g
4 where gu.gid = g.gid;
GNAME (SELECTWM_CONCAT(UNAME)FROMUWH
-------------------- --------------------------------------------------------------------------------
g1 u1,u2,u3,u9
g2 u4,u5,u6
g3 u7,u8,u10