今天遇到了行转为列的问题,在网上搜到一个经典的例子
又一个示例：
create table dd_app.bb AS(
select '01' no,'ss' name, 10 jiner,'麦子' you from dual union all
select '02' no,'dd' name, 11 jiner,'萝卜' you from dual union all
select '03' no,'cc' name, 9 jiner,'大豆' you from dual union all
select '01' no,'ss' name, 10 jiner,'白菜' you from dual)
;
WITH tab AS(
select '01' no,'ss' name, 10 jiner,'麦子' you from dual union all
select '02' no,'dd' name, 11 jiner,'萝卜' you from dual union all
select '03' no,'cc' name, 9 jiner,'大豆' you from dual union all
select '01' no,'ss' name, 10 jiner,'白菜' you from dual)
SELECT a.no, a.name, a.jiner, ltrim(MAX(sys_connect_by_path(you, ',')), ',') you
FROM (SELECT row_number() over(PARTITION BY a.no, a.name, a.jiner
ORDER BY a.no, a.name, a.jiner) rn,a.*
FROM tab a) a
START WITH rn = 1
CONNECT BY PRIOR rn = rn - 1 AND
a.no = PRIOR a.no AND
a.name = PRIOR a.name AND
a.jiner = PRIOR a.jiner
GROUP BY a.no, a.name, a.jiner
经过在oracle中测试,把分解说明分享一下:
--1,下面的这个sql求得的rn 是按no,name,jiner排序,第n次出现的次数统计
select row_number() over(partition by no,name,jiner order by no,name,jiner) rn,dd_app.bb.* from dd_app.bb;
--结果
RN NO NAME JINER YOU
1 01 ss 10 麦子
2 01 ss 10 白菜
1 02 dd 11 萝卜
1 03 cc 9 大豆
;
--2,行连接为列
select no,name,jiner,sys_connect_by_path(name||'.'||you,',')
from (
select row_number() over(partition by no,name,jiner order by no,name,jiner) rn,dd_app.bb.* from dd_app.bb
)t
connect by prior rn=rn-1 and t.no=prior t.no
;
--结果
NO NAME JINER SYS_CONNECT_BY_PATH(NAME||'.'|
01 ss 10 ,ss.麦子
01 ss 10 ,ss.麦子,ss.白菜
03 cc 9 ,cc.大豆
01 ss 10 ,cc.大豆,ss.白菜
02 dd 11 ,dd.萝卜
01 ss 10 ,dd.萝卜,ss.白菜
01 ss 10 ,ss.白菜
;
--3,去重,其实这里只是选取了长度最大的记录
select no,name,jiner,max(sys_connect_by_path(name||'.'||you,','))
from (
select row_number() over(partition by no,name,jiner order by no,name,jiner) rn,dd_app.bb.* from dd_app.bb
)
connect by prior rn=rn-1
group by no,name,jiner
另一种简的方法:运用wm_concat直接把行变为一列,并用","分隔
select wm_concat(name) from test group by type;
填写不合要求，退回重新编写计划,需求分析
另附列转为行的方法:
decode(clumn_name,条件1,结果1,条件2,结果2,条件3,结果3,...)