如何在不创建存储过程的情况下在oracle中实现以下功能?
数据集:
question_id element_id
1 7
1 8
2 9
3 10
3 11
3 12
期望的结果:
question_id element_id
1 7,8
2 9
3 10,11,12
答案 0 :(得分:97)
从Oracle 11gR2开始,LISTAGG子句可以解决这个问题:
SELECT question_id,
LISTAGG(element_id, ',') WITHIN GROUP (ORDER BY element_id)
FROM YOUR_TABLE
GROUP BY question_id;
请注意结果字符串是否太大(例如,对于VARCHAR2超过4000个字符):从版本12cR2开始,我们可以使用ON OVERFLOW TRUNCATE/ERROR来处理此问题。
答案 1 :(得分:36)
易:
SELECT question_id, wm_concat(element_id) as elements
FROM questions
GROUP BY question_id;
在10g上进行Pesonally测试; - )
来自http://www.oracle-base.com/articles/10g/StringAggregationTechniques.php
答案 2 :(得分:28)
有许多方法可以进行字符串聚合,但最简单的方法是用户定义的函数。 Try this for a way that does not require a function.作为一个注释,没有这个功能就没有简单的方法。
这是没有自定义函数的最短路径:(它使用ROW_NUMBER()和SYS_CONNECT_BY_PATH函数)
SELECT questionid,
LTRIM(MAX(SYS_CONNECT_BY_PATH(elementid,','))
KEEP (DENSE_RANK LAST ORDER BY curr),',') AS elements
FROM (SELECT questionid,
elementid,
ROW_NUMBER() OVER (PARTITION BY questionid ORDER BY elementid) AS curr,
ROW_NUMBER() OVER (PARTITION BY questionid ORDER BY elementid) -1 AS prev
FROM emp)
GROUP BY questionid
CONNECT BY prev = PRIOR curr AND questionid = PRIOR questionid
START WITH curr = 1;
答案 3 :(得分:2)
此OTN线程包含几种进行字符串聚合的方法,包括性能比较:http://forums.oracle.com/forums/message.jspa?messageID=1819487#1819487