SQL语句行变列，带；号的

[LGame]

CASE_NO ANKEN_NO
12 AN01
2 AN01
3 AN01
3 AN02
3 AN03
4 AN03
5 AN02
5 AN03
6 AN01

怎样使SQL语句得到这样的结果
12 AN01
2 AN01
3 AN01;AN02;AN03
......
......
5 AN02;AN03
......

[HelloWorld_001]

csdn出现过N回,简单的就用存储过程解决.
给你个我在csdn看到的,自己修改下就可以了

表结构：
create table TBL_TEST (TID VARCHAR2(2), NM VARCHAR2(2))
表数据：
insert into TBL_TEST2 (TID, NM) values ('1', 'a');
insert into TBL_TEST2 (TID, NM) values ('2', 'b');
insert into TBL_TEST2 (TID, NM) values ('3', 'c');
insert into TBL_TEST2 (TID, NM) values ('1', 'd');
即为：
TIDNM
1a
2b
3c
1d
----------------------
现在想要得出下面的集合该怎么写：
TIDNM
1ad
2b
3c
----------------------
即：按TID分类，如果TID一样则num列字符串并在一起（不是sum)
注：TID、NM中的值是不固定的
我想了半天想不出来，只好找大家帮忙了

存储过程:
CREATE OR REPLACE function link(v_id varchar2)
return varchar2
is
union_nm varchar2(200);
begin
for cur in (select nm from TBL_TEST where tid=v_id) loop
union_nm := union_nm||cur.nm;
end loop;
union_nm := rtrim(union_nm,1);
return union_nm;
end;
///
select tid,link(tid) from tbl_test group by tid

sql直接实现:
select tid, replace(max(sys_connect_by_path(nm,',')),',','') from
(SELECT tid, nm, MIN(nm) over(PARTITION BY tid) minnm
,(row_number() over(ORDER BY tid, nm)) + (dense_rank() over(ORDER BY tid)) no
FROM ttt)
start with nm=minnm
connect by no-1 = prior no
group by tid

[LGame]

谢谢了

[wdswcy]

with x AS (
 SELECT '张三' NAME, 'KB101' userid FROM dual UNION ALL
 SELECT '张三', 'KB002' FROM dual UNION ALL
 SELECT '张三', 'KB113' FROM dual UNION ALL
 SELECT '张三', 'KB003' FROM dual UNION ALL
 SELECT '张三', 'KB004' FROM dual UNION ALL
 SELECT '张三', 'KB005' FROM dual UNION ALL
 SELECT '张三', 'KB013' FROM dual UNION ALL
 SELECT '张三', 'KB103' FROM dual UNION ALL
 SELECT '李四', 'KB001' FROM dual UNION ALL
 SELECT '李四', 'KB002' FROM dual UNION ALL
 SELECT '王五', 'KB002' FROM dual UNION ALL
 SELECT '王五', 'KB006' FROM dual UNION ALL
 SELECT '李四', 'KB003' FROM dual)
--//--直接提取法
SELECT NAME, ltrim(MAX(sys_connect_by_path(userid, ',')), ',') userid
  FROM (SELECT NAME, userid, row_number() over(PARTITION BY NAME ORDER BY userid) rn FROM x)
 START WITH rn = 1 CONNECT BY rn - 1 = PRIOR rn AND NAME = PRIOR NAME GROUP BY NAME ORDER BY NAME;

[LGame]

select M_CASE,replace(max(sys_connect_by_path(ANKEN_NO , ', ')), ', ', ' ') 
  from   
(SELECT   
     M_CASE
    ,ANKEN_NO
    ,MIN(ANKEN_NO) over(PARTITION   BY   M_CASE)   minnm 
    ,(row_number()   over(ORDER   BY   M_CASE,   ANKEN_NO))+(dense_rank()   over(ORDER   BY   M_CASE))   no 
FROM   
    TBL_MITUMORI_CASE2
) 
start   with M_CASE = minnm 
connect by no-1 = prior no   
group   by M_CASE

没成功

[LGame]

1楼的哥哥！！！

用你数据也没有成功

[wdswcy]

SELECT CASE_NO, ltrim(MAX(sys_connect_by_path(trim(ANKEN_NO), ',')), ',') userid
FROM 
(SELECT CASE_NO, ANKEN_NO, row_number() over(PARTITION BY CASE_NO ORDER BY CASE_NO) rn FROM a)
START WITH rn = 1 CONNECT BY rn - 1 = PRIOR rn AND CASE_NO = PRIOR CASE_NO
GROUP BY CASE_NO ORDER BY CASE_NO;

[Well]

用动态语句实现，

[Well]

写一个自定义函数，进行字符串连接，然后再进行group by求得

[HelloWorld_001]

楼主没成功?
我现在没oracle环境,你可以用存储过程实现,那个很好理解的,很容易懂的.

那个数据我以前测试过的,是在正确的啊!

[fxianxian]

呵呵
這種問題已經有很多了！
建議你看哈這個博客，里面關于行列轉換的各種方式都有啊！
http://jack198409.itpub.net/