是否有可能构造SQL来连接列值 多行吗?
举例如下:
表一个
PID A B C
表B
PID SEQ Desc A 1 Have A 2 a nice A 3 day. B 1 Nice Work. C 1 Yes C 2 we can C 3 do C 4 this work!
SQL的输出应该是-
PID Desc A Have a nice day. B Nice Work. C Yes we can do this work!
所以基本上输出表的Desc列是来自表B的SEQ值的连接?
SQL有什么帮助吗?
当前回答
SQL模型子句:
SQL> select pid
2 , ltrim(sentence) sentence
3 from ( select pid
4 , seq
5 , sentence
6 from b
7 model
8 partition by (pid)
9 dimension by (seq)
10 measures (descr,cast(null as varchar2(100)) as sentence)
11 ( sentence[any] order by seq desc
12 = descr[cv()] || ' ' || sentence[cv()+1]
13 )
14 )
15 where seq = 1
16 /
P SENTENCE
- ---------------------------------------------------------------------------
A Have a nice day
B Nice Work.
C Yes we can do this work!
3 rows selected.
我在这里写过。如果链接到otn线程，您将找到更多内容，包括性能比较。
其他回答
有几种方法取决于你有什么版本-请参阅oracle文档关于字符串聚合技术。一个非常常见的方法是使用LISTAGG:
SELECT pid, LISTAGG(Desc, ' ') WITHIN GROUP (ORDER BY seq) AS description
FROM B GROUP BY pid;
然后加入A，挑出你想要的皮皮。
注意:开箱即用，LISTAGG只对VARCHAR2列正确工作。
在运行select查询之前，运行以下命令:
将服务器设置为6000大小
SELECT XMLAGG(XMLELEMENT(E,SUPLR_SUPLR_ID||',')).EXTRACT('//text()') "SUPPLIER"
FROM SUPPLIERS;
SQL模型子句:
SQL> select pid
2 , ltrim(sentence) sentence
3 from ( select pid
4 , seq
5 , sentence
6 from b
7 model
8 partition by (pid)
9 dimension by (seq)
10 measures (descr,cast(null as varchar2(100)) as sentence)
11 ( sentence[any] order by seq desc
12 = descr[cv()] || ' ' || sentence[cv()+1]
13 )
14 )
15 where seq = 1
16 /
P SENTENCE
- ---------------------------------------------------------------------------
A Have a nice day
B Nice Work.
C Yes we can do this work!
3 rows selected.
我在这里写过。如果链接到otn线程，您将找到更多内容，包括性能比较。
试试下面的代码:
SELECT XMLAGG(XMLELEMENT(E,fieldname||',')).EXTRACT('//text()') "FieldNames"
FROM FIELD_MASTER
WHERE FIELD_ID > 10 AND FIELD_AREA != 'NEBRASKA';
对于那些必须使用Oracle 9i(或更早版本)解决这个问题的人，您可能需要使用SYS_CONNECT_BY_PATH，因为LISTAGG不可用。
为了回答OP，下面的查询将显示表A中的PID，并连接表B中的所有DESC列:
SELECT pid, SUBSTR (MAX (SYS_CONNECT_BY_PATH (description, ', ')), 3) all_descriptions
FROM (
SELECT ROW_NUMBER () OVER (PARTITION BY pid ORDER BY pid, seq) rnum, pid, description
FROM (
SELECT a.pid, seq, description
FROM table_a a, table_b b
WHERE a.pid = b.pid(+)
)
)
START WITH rnum = 1
CONNECT BY PRIOR rnum = rnum - 1 AND PRIOR pid = pid
GROUP BY pid
ORDER BY pid;
也可能存在键和值都包含在一个表中的情况。下面的查询可以在没有表A，只有表B的情况下使用:
SELECT pid, SUBSTR (MAX (SYS_CONNECT_BY_PATH (description, ', ')), 3) all_descriptions
FROM (
SELECT ROW_NUMBER () OVER (PARTITION BY pid ORDER BY pid, seq) rnum, pid, description
FROM (
SELECT pid, seq, description
FROM table_b
)
)
START WITH rnum = 1
CONNECT BY PRIOR rnum = rnum - 1 AND PRIOR pid = pid
GROUP BY pid
ORDER BY pid;
所有值都可以按需要重新排序。各个连接的描述可以在PARTITION BY子句中重新排序，而pid列表可以在最后的ORDER BY子句中重新排序。
另外，有时您可能希望将整个表中的所有值连接到一行中。
这里的关键思想是为要连接的描述组使用一个人工值。
在下面的查询中，使用常量字符串'1'，但任何值都可以:
SELECT SUBSTR (MAX (SYS_CONNECT_BY_PATH (description, ', ')), 3) all_descriptions
FROM (
SELECT ROW_NUMBER () OVER (PARTITION BY unique_id ORDER BY pid, seq) rnum, description
FROM (
SELECT '1' unique_id, b.pid, b.seq, b.description
FROM table_b b
)
)
START WITH rnum = 1
CONNECT BY PRIOR rnum = rnum - 1;
各个连接的描述可以在PARTITION BY子句中重新排序。
本页上的其他几个回答也提到了这个非常有用的参考: https://oracle-base.com/articles/misc/string-aggregation-techniques
