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用SQL实现行列转换的一个例子

2007-11-20 13:53 423 查看
在CU上看到一个问题。 http://bbs.chinaunix.net/viewthread.php?tid=923817&extra=page%3D1
数据表中有下列记录：
xm sl
王一 2
王一 5
张二 4
张二 5
张二 8
求SQL输出结果为：
王一 2 王一 5
张二 4 张二 5 张二 8

类似的问题，其实经常可以在各BBS上看到有人提起，但从来没有人认为可以用一条SQL实现，一般回复者给出的解决方案都是用存储过程。不过既然自己搞DB2已近11个月，水平也应该逐步地提高了，不能始终停留在前人说不可能实现，自己就认为也不能实现的程度，有必要自己独立思索一下。

这个问题的难点在于：需要分组字段对应的所有值的最大个数（行数）不明确。如果明确知道这个最大值的话是可以用CASE WHEN语句来实现的；在不知道这个最大值的情况下，首先让人想到的就是必须用循环来实现，但FOR,LOOP,WHILE,REPEAT这四种循环，在DB2中只有存储过程支持，这就不难理解为什么一般的解决方案都是说要用存储过程了。

以前认真读过SQLLIB/samples下的很多例子，所以很自然地想到一个思路——递归。试了试，成功了，无论XM字段每种值对应多少行，都一样可用。实验过程如下：
DROP TABLE TEST;
CREATE TABLE TEST (XM VARCHAR(8), SL INTEGER);
INSERT INTO TEST
VALUES ('王一', 2),
('王一', 5),
('张二', 4),
('张二', 5),
('张二', 8),
('李三', 2),
('李三', 4),
('李三', 15),
('李三', 29);

WITH B (FATHER,SON,XM,CHAIN) AS
(SELECT A.XM || CHAR(ROW_NUMBER() OVER(PARTITION BY XM)) AS FATHER,A.XM || CHAR(ROW_NUMBER() OVER(PARTITION BY XM) + 1) AS SON, A.XM, CAST(CHAR(A.SL) AS VARCHAR(100))
FROM TEST A
UNION ALL
SELECT C.FATHER,C.SON,C.XM,B.CHAIN || C.XM || ' ' || C.SL
FROM (SELECT A.XM || CHAR(ROW_NUMBER() OVER(PARTITION BY XM)) AS FATHER,A.XM || CHAR(ROW_NUMBER() OVER(PARTITION BY XM) + 1) AS SON, A.XM, CHAR(A.SL) AS SL FROM TEST A) AS C, B
WHERE B.SON= C.FATHER)
SELECT
D.XM || ' ' || D.CHAIN
FROM
(SELECT ROW_NUMBER() OVER(PARTITION BY XM ORDER BY LENGTH(CHAIN) DESC) AS ROW_NUM, B.XM,B.CHAIN FROM B) AS D
WHERE
D.ROW_NUM = 1;

效果如下：
db2 => WITH B (FATHER,SON,XM,CHAIN) AS
db2 (cont.) => (SELECT A.XM || CHAR(ROW_NUMBER() OVER(PARTITION BY XM)) AS FATHER,A.XM || CHAR(ROW_NUMBER() OVER(PARTITION BY XM) + 1) AS SON, A.XM, CAST(CHAR(A.SL) AS VARCHAR(100))
db2 (cont.) => FROM TEST A
db2 (cont.) => UNION ALL
db2 (cont.) => SELECT C.FATHER,C.SON,C.XM,B.CHAIN || C.XM || ' ' || C.SL
db2 (cont.) => FROM (SELECT A.XM || CHAR(ROW_NUMBER() OVER(PARTITION BY XM)) AS FATHER,A.XM || CHAR(ROW_NUMBER() OVER(PARTITION BY XM) + 1) AS SON, A.XM, CHAR(A.SL) AS SL FROM TEST A) AS C, B
db2 (cont.) => WHERE B.SON= C.FATHER)
db2 (cont.) => SELECT
db2 (cont.) => D.XM || ' ' || D.CHAIN
db2 (cont.) => FROM
db2 (cont.) => (SELECT ROW_NUMBER() OVER(PARTITION BY XM ORDER BY LENGTH(CHAIN) DESC) AS ROW_NUM, B.XM,B.CHAIN FROM B) AS D
db2 (cont.) => WHERE
db2 (cont.) => D.ROW_NUM = 1;
WITH B (FATHER,SON,XM,CHAIN) AS (SELECT A.XM || CHAR(ROW_NUMBER() OVER(PARTITION BY XM)) AS FATHER,A.XM || CHAR(ROW_NUMBER() OVER(PARTITION BY XM) + 1) AS SON, A.XM, CAST(CHAR(A.SL) AS VARCHAR(100)) FROM TEST A UNION ALL SELECT C.FATHER,C.SON,C.XM,B.CHAIN || C.XM || ' ' || C.SL FROM (SELECT A.XM || CHAR(ROW_NUMBER() OVER(PARTITION BY XM)) AS FATHER,A.XM || CHAR(ROW_NUMBER() OVER(PARTITION BY XM) + 1) AS SON, A.XM, CHAR(A.SL) AS SL FROM TEST A) AS C, B WHERE B.SON= C.FATHER) SELECT D.XM || ' ' || D.CHAIN FROM (SELECT ROW_NUMBER() OVER(PARTITION BY XM ORDER BY LENGTH(CHAIN) DESC) AS ROW_NUM, B.XM,B.CHAIN FROM B) AS D WHERE D.ROW_NUM = 1
1
--------------------------------------------------------------------------------------------------------------
SQL0347W 递归公共表表达式 "DB2ADMIN.B" 可能包含无限循环。 SQLSTATE=01605
李三 2 李三 4 李三 15 李三 29
王一 2 王一 5
张二 4 张二 5 张二 8
已选择 3 条记录，打印 1 条警告消息。

db2 =>

注：
1、若想不看到SQL0347W这个警告，可以用db2 +w -t。
2、ORACLE的解决方法在这里：http://www.oracle-base.com/articles/10g/StringAggregationTechniques.php。它山之石，可以攻玉，有些思路也是可以借鉴的。

最后我还是认为：这样的问题，用SQL来实现实在是自找麻烦，文字处理本来就是SHELL SCRIPT的强项，所以用awk会感觉异常简单。
11:31:10 ewdbkjy:[/home/ewadmin]$cat 9
王一 2
王一 5
张二 4
张二 5
张二 8
李三 2
李三 4
李三 15
李三 29
11:31:25 ewdbkjy:[/home/ewadmin]$awk '{a[$1]=a[$1]" "$1" "$2} END{for(i in a) print a[i]}' 9
李三 2 李三 4 李三 15 李三 29
张二 4 张二 5 张二 8
王一 2 王一 5
11:31:45 ewdbkjy:[/home/ewadmin]$

========================================================================
任何形式的转载，请写明出处：
email: beginner@yeah.net
blog: http://blog.chinaunix.net/index.php?blogId=739 http://www.cublog.cn/u/739/
========================================================================

[align=left]我的实践例子：[/align]
[align=left] [/align]

[align=left]table:
CREATE TABLE TEST (XM VARCHAR(8), SL INTEGER);
INSERT INTO TEST
VALUES ('王一', 2),
('王一', 5),
('张二', 4),
('张二', 5),
('张二', 8),
('李三', 2),
('李三', 4),
('李三', 15),
('李三', 29);[/align]
[align=left] [/align]
[align=left]
case 1:[/align]
[align=left] [/align]
[align=left]WITH B (FATHER,SON,XM,CHAIN) AS
(SELECT A.XM || CHAR(ROW_NUMBER() OVER(PARTITION BY XM)) AS FATHER,A.XM || CHAR(ROW_NUMBER() OVER(PARTITION BY XM) + 1) AS SON,
A.XM, CAST(CHAR(A.SL) AS VARCHAR(100))
FROM TEST A
UNION ALL
SELECT C.FATHER,C.SON,C.XM,B.CHAIN || C.XM || ' ' || C.SL
FROM (SELECT A.XM || CHAR(ROW_NUMBER() OVER(PARTITION BY XM)) AS FATHER,A.XM || CHAR(ROW_NUMBER() OVER(PARTITION BY XM) + 1) AS SON, A.XM, CHAR(A.SL) AS SL FROM TEST A) AS C, B
WHERE B.SON= C.FATHER)[/align]
[align=left] [/align]
[align=left] SELECT D.XM || ' ' || D.CHAIN
FROM (SELECT ROW_NUMBER() OVER(PARTITION BY XM ORDER BY LENGTH(CHAIN) DESC) AS ROW_NUM, B.XM,B.CHAIN FROM B) AS D
WHERE D.ROW_NUM = 1;[/align]
[align=left] [/align]
[align=left]result:
李三 2 李三 4 李三 15 李三 29
王一 2 王一 5
张二 4 张二 5 张二 8[/align]
[align=left] [/align]
[align=left]case 2:[/align]
[align=left] [/align]
[align=left]WITH B (FATHER,SON,XM,CHAIN) AS
(SELECT A.XM || CHAR(ROW_NUMBER() OVER(PARTITION BY XM)) AS FATHER,A.XM || CHAR(ROW_NUMBER() OVER(PARTITION BY XM) + 1) AS SON,
A.XM, rtrim(ltrim(CHAR(A.SL)))
FROM TEST A
UNION ALL
SELECT C.FATHER,C.SON,C.XM,rtrim(ltrim(B.CHAIN))||','||rtrim(ltrim(C.SL))
FROM (SELECT A.XM || CHAR(ROW_NUMBER() OVER(PARTITION BY XM)) AS FATHER,
A.XM || CHAR(ROW_NUMBER() OVER(PARTITION BY XM) + 1) AS SON,
A.XM, rtrim(ltrim(CHAR(A.SL))) AS SL FROM TEST A) AS C, B
WHERE B.SON= C.FATHER) [/align]
[align=left] [/align]
[align=left] SELECT D.XM,D.CHAIN
FROM (SELECT ROW_NUMBER() OVER(PARTITION BY XM ORDER BY LENGTH(CHAIN) DESC) AS ROW_NUM, B.XM,B.CHAIN FROM B) AS D
WHERE D.ROW_NUM = 1;[/align]
[align=left] [/align]
[align=left]result:
XM CHAIN
-------- -----------
李三 2,4,15,29
王一 2,5
张二 4,5,8[/align]
[align=left] [/align]
[align=left]
case 3:[/align]
[align=left] [/align]
[align=left]WITH B (FATHER,SON,XM,CHAIN) AS
(SELECT A.XM || CHAR(ROW_NUMBER() OVER(PARTITION BY XM)) AS FATHER,A.XM || CHAR(ROW_NUMBER() OVER(PARTITION BY XM) + 1) AS SON,
A.XM, rtrim(ltrim(CHAR(A.SL)))
FROM TEST A
UNION ALL
SELECT C.FATHER,C.SON,C.XM,rtrim(ltrim(B.CHAIN))||rtrim(ltrim(C.SL))
FROM (SELECT A.XM || CHAR(ROW_NUMBER() OVER(PARTITION BY XM)) AS FATHER,
A.XM || CHAR(ROW_NUMBER() OVER(PARTITION BY XM) + 1) AS SON,
A.XM, rtrim(ltrim(CHAR(A.SL))) AS SL FROM TEST A) AS C, B
WHERE B.SON= C.FATHER)
-- end with[/align]
[align=left] [/align]
[align=left] SELECT D.XM,D.CHAIN
FROM (SELECT ROW_NUMBER() OVER(PARTITION BY XM ORDER BY LENGTH(CHAIN) DESC) AS ROW_NUM, B.XM,B.CHAIN FROM B) AS D
WHERE D.ROW_NUM = 1;[/align]
[align=left] [/align]
[align=left]result:
XM CHAIN
-------- -----------
李三 241529
王一 25
张二 458 [/align]
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