Oracle同行合并分组
Oracle同行合并分组使用函数sys_connect_by_path(column,'')的例子^^。 表结构为：create table test(bookid char(3) not null,author varchar2(10) not null); insert into test values('001','jack');insert into test values('001','tom');insert into test values('002','wang');insert into test values('002','zhang');insert into test values('002','li'); commit; select * from test;显示结果为：BOO AUTHOR-----------------001 jack001 tom002 wang002 zhang002 li 我们想得到的结果为：BOO AUTHOR-----------------------------001 jack&&tom002 wang&&zhang&&li SQL文为： select bookid,substr(max(sys_connect_by_path(author,'&&')),3) author
from
(select bookid,author,id,lag(id) over(partition by bookid order by>--(最后一列或者为)lead(id) over(partition by bookid order by>
from (select bookid,author,rownum>start with pid is nullconnect by prior>group by bookid; 详细解释：sys_connect_by_path(column,'')//column为列名，''中间加要添加的字符这个函数本身不是用来给我们做结果集连接的（合并行），而是用来构造树路径的，所以需要和connect by一起使用。 test只是张普通表，怎样才能变成树结构呢？我们需要加一个pid和id。 id我们只需加一个rownum就好。select bookid,author,rownum>BOO AUTHOR >----------------------------001 jack 1001 tom 2002 wang 3002 zhang 4002 li 5 而pid上一条记录不就是下一条记录的父节点了。这里我们需要函数lag()取前记录，和lead()相对。//把lag(id) over(order by>
select bookid,author,id,lag(id) over(order by>
from (select bookid,author,rownum>BOO AUTHOR >-------------------------------------------001 jack 1001 tom 2 1002 wang 3 2002 zhang 4 3002 li 5 4 由于要按bookid分我们的pid，在分析函数over中我们需要加上partition by，一看下面结果我们就知道有什么不同了。
select bookid,author,id,lag(id) over(partition by bookid order by>
from (select bookid,author,rownum>BOO AUTHOR >-------------------------------------------001 jack 1001 tom 2 1002 wang 3002 zhang 4 3002 li 5 4 继续，把上述看成一张虚拟表，用到我们的sys_connect_by_path函数取出想要的值。格式：sys_connect_by_path(column,'')start with 条件1connect by 条件2(prior 子节点=父节点) select bookid,sys_connect_by_path(author,'&&') author
from
(select bookid,author,id,lag(id) over(partition by bookid order by>
from (select bookid,author,rownum>start with pid is nullconnect by prior>BOO AUTHOR-----------------------------------001 &&jack001 &&jack&&tom002 &&wang002 &&wang&&zhang 002 &&wang&&zhang&&li
OK，离我们的结果越来越近了，现在就是一般函数的应用了。1，以bookid分组，取author的最大值。2，用substr(string,start,length)截掉前面多余的字符。//没第三参数默认取到结束 select bookid,substr(max(sys_connect_by_path(author,'&&')),3) author
......group by bookid;//详细sql文，一开始已给出!BOO AUTHOR------------------------------001 jack&&tom 002 wang&&zhang&&li 大功告成，^_^!drop table test;
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