利用Oracle分析函数row_number和sys_connect_by_path实现多行数据合并为一行

emo场景，以oracle自带库中的表emp为例：

select ename,deptno from emp order by deptno;

ENAME	DEPTNO
CLARK	10
KING	10
MILLER	10
SMITH	20
ADAMS	20
FORD	20
SCOTT	20
JONES	20
ALLEN	30
BLAKE	30
MARTIN	30
JAMES	30
TURNER	30
WARD	30

现在想要将同一部门的人给合并成一行记录，如何做呢？如下：

ENAME	DEPTNO
CLARK,KING,MILLER	10
ADAMS,FORD,JONES,SCOTT,SMITH	20
ALLEN,BLAKE,JAMES,MARTIN,TURNER,WARD	30

通常我们都是自己写函数或在程序中处理，这里我们利用oracle自带的分析函数row_number()和sys_connect_by_path来进行sql语句层面的多行到单行的合并，并且效率会非常高。

基本思路：

1、对deptno进行row_number()按ename排位并打上排位号

select deptno,ename,row_number() over(partition by deptno order by deptno,ename) rank
from emp order by deptno,ename;

DEPTNO	ENAME	RANK
10	CLARK	1
10	KING	2
10	MILLER	3
20	ADAMS	1
20	FORD	2
20	JONES	3
20	SCOTT	4
20	SMITH	5
30	ALLEN	1
30	BLAKE	2
30	JAMES	3
30	MARTIN	4
30	TURNER	5
30	WARD	6

可看出，经过row_number()后，部门人已经按部门和人名进行了排序，并打上了一个位置字段rank

2、利用oracle的递归查询connect by进行表内递归，并通过sys_connect_by_path进行父子数据追溯串的构造，这里要针对ename字段进行构造，使之合并在一个字段内（数据很多，只截取部分）

select deptno,ename,rank,level as curr_level,
ltrim(sys_connect_by_path(ename,','),',') ename_path from (
select deptno,ename,row_number() over(partition by deptno order by deptno,ename) rank
from emp order by deptno,ename) connect by deptno = prior deptno and rank-1 = prior rank;

各部门递归后的数据量都是：(1+n)/2 * n 即：deptno=10 数据量：(1+3)/2 * 3 = 6;

deptno=20 数据量：(1+5)/2 * 5 = 15;      deptno=30 数据量：(1+6)/2 * 6 = 21;

DEPTNO	ENAME	RANK	CURR_LEVEL	ENAME_PATH
10	CLARK	1	1	CLARK
10	KING	2	2	CLARK,KING
10	MILLER	3	3	CLARK,KING,MILLER
10	KING	2	1	KING
10	MILLER	3	2	KING,MILLER
10	MILLER	3	1	MILLER

DEPTNO	ENAME	RANK	CURR_LEVEL	ENAME_PATH
20	ADAMS	1	1	ADAMS
20	FORD	2	2	ADAMS,FORD
20	JONES	3	3	ADAMS,FORD,JONES
20	SCOTT	4	4	ADAMS,FORD,JONES,SCOTT
20	SMITH	5	5	ADAMS,FORD,JONES,SCOTT,SMITH
20	FORD	2	1	FORD
20	JONES	3	2	FORD,JONES
20	SCOTT	4	3	FORD,JONES,SCOTT
20	SMITH	5	4	FORD,JONES,SCOTT,SMITH
20	JONES	3	1	JONES
20	SCOTT	4	2	JONES,SCOTT
20	SMITH	5	3	JONES,SCOTT,SMITH
20	SCOTT	4	1	SCOTT
20	SMITH	5	2	SCOTT,SMITH
20	SMITH	5	1	SMITH

这里我们仅列出deptno=10、20的，至此我们应该能否发现一些线索了，即每个部门中，curr_level最高的那行，有我们所需要的数据。那后面该怎么办，取出那个数据？ 对了，继续用row_number()进行排位标记，然后再按排位标记取出即可。

3、 对deptno继续进行row_number()按curr_level排位

select deptno,ename_path,row_number() over(partition by deptno order by deptno,curr_level desc) ename_path_rank from (select deptno,ename,rank,level as curr_level,
ltrim(sys_connect_by_path(ename,','),',') ename_path from (
select deptno,ename,row_number() over(partition by deptno order by deptno,ename) rank
from emp order by deptno,ename) connect by deptno = prior deptno and rank-1 = prior rank);

DEPTNO	ENAME_PATH	ENAME_PATH_RANK
10	CLARK,KING,MILLER	1
10	CLARK,KING	2
10	KING,MILLER	3
10	CLARK	4
10	KING	5
10	MILLER	6

DEPTNO	ENAME_PATH	ENAME_PATH_RANK
20	ADAMS,FORD,JONES,SCOTT,SMITH	1
20	ADAMS,FORD,JONES,SCOTT	2
20	FORD,JONES,SCOTT,SMITH	3
20	ADAMS,FORD,JONES	4
20	FORD,JONES,SCOTT	5
20	JONES,SCOTT,SMITH	6
20	ADAMS,FORD	7
20	FORD,JONES	8
20	SCOTT,SMITH	9
20	JONES,SCOTT	10
20	ADAMS	11
20	JONES	12
20	SMITH	13
20	SCOTT	14
20	FORD	15

这里还是仅列出deptno为10、20的，至此应该很明了了，在进行一次查询，取ename_path_rank为1的即可获得我们想要的结果。

4、获取想要排位的数据，即得部门下所有人多行到单行的合并

select deptno,ename_path from (select deptno,ename_path,
row_number() over(partition by deptno order by deptno,curr_level desc) ename_path_rank
from (select deptno,ename,rank,level as curr_level,
ltrim(sys_connect_by_path(ename,','),',') ename_path from (
select deptno,ename,row_number() over(partition by deptno order by deptno,ename) rank
from emp order by deptno,ename) connect by deptno = prior deptno and rank-1 = prior rank))
where ename_path_rank=1;