2

我有一个表,其中包含以逗号分隔的名称列表。我的目标是用房间分隔它们。

Room  | Name  
room1 | Anne,Amy  
room2 | Ben,Bryan

我的目标:

Room  | Name  
room1 | Anne   
room1 | Amy   
room2 | Ben  
room2 | Bryan  

我已经阅读了一些关于如何将字符串拆分为行的解决方案,但是是否有在 Oracle 8i 上运行的替代方案。我已经按照一些文章将它们拆分为这样的行:

create or replace function str2tbl( p_str IN varchar2 , p_delimiter in varchar2) return mytabletype
  as
    l_str      long default p_str || p_delimiter;
    l_n       number;
    l_data   mytabletype := mytabletype();
  begin
  loop
    l_n := instr( l_str, p_delimiter );
    exit when (nvl(l_n,0) = 0);
    l_data.extend;
    l_data( l_data.count ) := ltrim(rtrim(substr(l_str,1,l_n-1)));
    l_str := substr( l_str, l_n+1 );
  end loop;
   return l_data;
  end str2tbl;

然后我从我的表中进行 SELECT 如下所示:

select * from the ( select cast(str2tbl( Name, ',' ) as mytableType ) 
from   SPLITSTRING);

并得到以下结果,但不能带出 Room 列的值:

Name  
Anne  
Amy  
Ben  
Bryan  

有没有办法在 Oracle 8i 中拆分为行?

4

3 回答 3

1

您可以通过分层查询和字符串操作函数尝试使用稍微不同的方法,不使用函数。有点棘手,但这应该有效:

with test (Room, Name) as  
(
 select 'room2', 'Ben,Bryan' from dual
)
select room, 
       trim (',' from  substr( name,
                               decode ( level, 
                                        1, 1,
                                           instr(name, ',', 1, level -1) +1
                                      ),
                               decode ( instr( name, ',', 1, level), 
                                        0, length(name),
                                           instr( name, ',', 1, level) - 
                                                decode ( level,
                                                         1, 1,
                                                            instr(name, ',', 1, level -1)
                                                       )
                                      )
                              )        
            ) as name
from test
connect by level = 1 or instr(name, ',', 1, level-1) != 0
order by 1
于 2016-02-03T09:38:08.423 回答
0

我不确定 8i 是否支持此功能,我已在 12c 中对此进行了测试,并且工作正常:

create table test (room varchar2(20), names varchar2(40));

表“测试”已创建

insert into test values ('room1', 'anne');

INSERT INTO 测试成功 1 行受影响

insert into test values ('room2', 'amy,sheldon');

INSERT INTO 测试成功 1 行受影响

insert into test values ('room3', 'penny,leonard');

INSERT INTO 测试成功 1 行受影响

使用 XMLTABLE:

SELECT room,
trim(COLUMN_VALUE) names
FROM test,
xmltable(('"'
|| REPLACE(names, ',', '","')
|| '"'))
/

房间 | 名称


房间1 | 安妮

房间2 | 艾米

房间2 | 谢尔顿

房间3 | 一分钱

房间3 | 伦纳德

于 2016-02-03T09:30:34.000 回答
0

这是不使用 connect by 的替代方法

drop table pivot_t;
drop table rooms;

create table rooms (Room varchar2(30), Persons varchar2(30));

insert into rooms values ('room1',  'Anne,Amy');
insert into rooms values ('room2',  'Ben,Bryan,Paul');
insert into rooms values ('room3',  'John,Michael,Nik,Patrick');

create table pivot_t(num integer);

begin
for i in 1..10000 loop
 insert into pivot_t values(i);
end loop;
end;
/

commit;



select 
room
,substr(Persons, start_pos, case
  when 
    next_comma - start_pos < 0 then 999
  else 
    next_comma - start_pos 
end)
from
(
select r.room 
      ,r.persons
      ,nvl(instr(r.Persons,',',1,decode(pt.num-1,0,null,pt.num-1) ),0) +1 START_POS
      ,instr(r.Persons,',',1,pt.num) NEXT_COMMA
  from rooms   r
      ,pivot_t pt
where length(r.Persons) - length(replace(r.Persons,',')) +1 >= pt.num
order by r.room, pt.num
)
;
于 2016-02-03T11:03:21.393 回答