Oracle Pivot/Unpivot
此特性只适用于Oracle 11g 以上版本
CREATE TABLE CUSTOMERS
(
CUST_ID NUMBER(10),
CUST_NAME VARCHAR2(20),
STATE_CODE VARCHAR2(2),
TIMES_PURCHASED NUMBER(3)
);
INSERT INTO CUSTOMERS VALUES(1,'ANKER','CT',1);
INSERT INTO CUSTOMERS VALUES(1,'ANKER','ET',2);
INSERT INTO CUSTOMERS VALUES(1,'ANKER','DT',3);
INSERT INTO CUSTOMERS VALUES(1,'ANKER','QD',4);
INSERT INTO CUSTOMERS VALUES(2,'JONH','ET',2);
INSERT INTO CUSTOMERS VALUES(2,'JONH','DT',3);
INSERT INTO CUSTOMERS VALUES(2,'JONH','QD',2);
INSERT INTO CUSTOMERS VALUES(3,'QIANG','DT',3);
INSERT INTO CUSTOMERS VALUES(3,'QIANG','QD',4);
COMMIT;
--该记录显示了客户所在的州以及该客户在商店购物的次数。当该客户从商店购买更多物品时，列 times_purchased 会进行更新。
SQL> SELECT * FROM CUSTOMERS;
CUST_ID CUST_NAME ST TIMES_PURCHASED
---------- -------------------- -- ---------------
1 Anker CT 1
1 Anker ET 2
1 Anker DT 3
1 Anker QD 4
2 Jonh ET 2
2 Jonh DT 3
2 Jonh QD 2
3 Qiang DT 3
3 Qiang QD 4
已选择9行。
PIVOT用法
--1.查看每个州，每个客户的购买频率
SELECT * FROM CUSTOMERS
PIVOT(
SUM(TIMES_PURCHASED) FOR STATE_CODE IN ('QD' AS "New York", 'CT' AS "Connecticut", 'ET' AS "New Jersey", 'DT' AS "Missouri")
);
--从结果可以看出,pivot将针对非pivot涉及的字段进行group by，pivot中的 in (...)子句中的值将作为列，聚合的值将作为列的值
CUST_ID CUST_NAME New York Connecticut New Jersey Missouri
---------- -------------------- ---------- ----------- ---------- ----------
1 Anker 4 1 2 3
2 Jonh 2 2 3
3 Qiang 4 3
--2.了解各个州的购买频率，即，各个州有多少客户只购物一次、两次、三次等等。如果使用常规 SQL
SQL> SELECT STATE_CODE, TIMES_PURCHASED, COUNT(*)
2 FROM CUSTOMERS
3 GROUP BY STATE_CODE, TIMES_PURCHASED;
ST TIMES_PURCHASED COUNT(*)
-- --------------- ----------
QD 4 2
QD 2 1
CT 1 1
ET 2 2
DT 3 3
--这样看起来不太直观
SELECT *
FROM
(
SELECT TIMES_PURCHASED, STATE_CODE FROM CUSTOMERS) T
PIVOT(
COUNT(STATE_CODE) FOR STATE_CODE IN ('QD', 'CT', 'ET', 'DT')
)
TIMES_PURCHASED 'QD' 'CT' 'ET' 'DT'
--------------- ---------- ---------- ---------- ----------
1 0 1 0 0
2 1 0 2 0
4 2 0 0 0
3 0 0 0 3
--但是假设您希望显示州名而非缩写
SELECT *
FROM
(
SELECT TIMES_PURCHASED as "Puchase Frequency", STATE_CODE FROM CUSTOMERS) T
PIVOT(COUNT(STATE_CODE) as CNT
FOR STATE_CODE IN ('QD' AS "New York", 'CT' AS "Connecticut", 'ET' AS "New Jersey", 'DT' AS "Missouri")
);
Puchase Frequency New York_CNT Connecticut_CNT New Jersey_CNT Missouri_CNT
----------------- ------------ --------------- -------------- ------------
1 0 1 0 0
2 1 0 2 0
4 2 0 0 0
3 0 0 0 3
-----------------------------------------------------------------------------------------------------------------------------------
Unpivot 讲解
--1.将上面的pivot后的结果进行反转回来
CREATE TABLE TAB_MATRIX1
AS
SELECT * FROM CUSTOMERS
PIVOT(
SUM(TIMES_PURCHASED) FOR STATE_CODE IN ('QD' AS "NEW YORK", 'CT' AS "CONNECTICUT", 'ET' AS "NEW JERSEY", 'DT' AS "MISSOURI")
);
--查询表
SQL> SELECT * FROM TAB_MATRIX1
2 ;
CUST_ID CUST_NAME New York Connecticut New Jersey Missouri
---------- -------------------- ---------- ----------- ---------- ----------
1 Anker 4 1 2 3
2 Jonh 2 2 3
3 Qiang 4 3
--从结果中可以看出,for .. in (..)中的列名将作为state_code列的值，字段state_counts中的列将由原来的列值来代替
SQL> SELECT * FROM TAB_MATRIX1
2 unpivot
3 (
4 state_counts
5 for state_code in ("New York","Connecticut","New Jersey","Missouri")
6 );
CUST_ID CUST_NAME STATE_CODE STATE_COUNTS
---------- -------------------- ----------- ------------
1 Anker New York 4
1 Anker Connecticut 1
1 Anker New Jersey 2
1 Anker Missouri 3
2 Jonh New York 2
2 Jonh New Jersey 2
2 Jonh Missouri 3
3 Qiang New York 4
3 Qiang Missouri 3
已选择9行。
--将上面第二个结果也进行反转
CREATE TABLE TAB_MATRIX2 AS
SELECT *
FROM
(
SELECT TIMES_PURCHASED as "Puchase Frequency", STATE_CODE FROM CUSTOMERS) T
PIVOT(COUNT(STATE_CODE) as CNT
FOR STATE_CODE IN ('QD' AS "New York", 'CT' AS "Connecticut", 'ET' AS "New Jersey", 'DT' AS "Missouri")
);
SQL> SELECT * FROM TAB_MATRIX2;
Puchase Frequency New York_CNT Connecticut_CNT New Jersey_CNT Missouri_CNT
----------------- ------------ --------------- -------------- ------------
1 0 1 0 0
2 1 0 2 0
4 2 0 0 0
3 0 0 0 3
SELECT * FROM TAB_MATRIX2
UNPIVOT
(
CNT_STATE_CODE
FOR STATE_CODE IN ("New York_CNT","Connecticut_CNT","New Jersey_CNT","Missouri_CNT")
);
Puchase Frequency STATE_CODE CNT_STATE_CODE
----------------- --------------- --------------
1 New York_CNT 0
1 Connecticut_CNT 1
1 New Jersey_CNT 0
1 Missouri_CNT 0
2 New York_CNT 1
2 Connecticut_CNT 0
2 New Jersey_CNT 2
2 Missouri_CNT 0
4 New York_CNT 2
4 Connecticut_CNT 0
4 New Jersey_CNT 0
4 Missouri_CNT 0
3 New York_CNT 0
3 Connecticut_CNT 0
3 New Jersey_CNT 0
3 Missouri_CNT 3
---------------------------------------------------------------------------------------------------------------------
Oracle 11g之前的版本如何实现行转列及列转行
CREATE TABLE T
AS
SELECT * FROM
(
SELECT OWNER,OBJECT_TYPE,COUNT(*) AS CNT FROM DBA_OBJECTS
GROUP BY OWNER,OBJECT_TYPE
) WHERE ROWNUM <= 3;
admin@ORCL> SELECT * FROM T;
OWNER OBJECT_TYPE CNT
------------------------------ ------------------- ----------
BI SYNONYM 8
HR VIEW 1
HR INDEX 19
--列转行
admin@ORCL> SELECT OWNER,
2 SUM(CASE
3 WHEN OBJECT_TYPE = 'SYNONYM' THEN
4 CNT
5 ELSE
6 0
7 END) AS TYPE_SYNONYM,
8 SUM(CASE
9 WHEN OBJECT_TYPE = 'VIEW' THEN
10 CNT
11 ELSE
12 0
13 END) AS TYPE_VIEW,
14 SUM(CASE
15 WHEN OBJECT_TYPE = 'INDEX' THEN
16 CNT
17 ELSE
18 0
19 END) AS TYPE_INDEX
20 FROM T
21 GROUP BY OWNER;
OWNER TYPE_SYNONYM TYPE_VIEW TYPE_INDEX
------------------------------ ------------ ---------- ----------
HR 0 1 19
BI 8 0 0
--列转行
DROP TABLE T PURGE;
CREATE TABLE T
(
OWNER VARCHAR2(10),
TYPE_SYNONYM INT,
TYPE_VIEW INT,
TYPE_INDEX INT
);
INSERT INTO T VALUES('HR','0','1','19');
INSERT INTO T VALUES('BI','8','0','0');
COMMIT;
--查看数据
admin@ORCL> SELECT * FROM T;
OWNER TYPE_SYNONYM TYPE_VIEW TYPE_INDEX
---------- ------------ ---------- ----------
HR 0 1 19
BI 8 0 0
--列转行
admin@ORCL> SELECT OWNER, 'SYNONYM' AS TYPE, TYPE_SYNONYM
2 FROM T
3 UNION ALL
4 SELECT OWNER, 'VIEW', TYPE_VIEW
5 FROM T
6 UNION ALL
7 SELECT OWNER, 'INDEX', TYPE_INDEX FROM T;
OWNER TYPE TYPE_SYNONYM
---------- ------- ------------
HR SYNONYM 0
BI SYNONYM 8
HR VIEW 1
BI VIEW 0
HR INDEX 19
BI INDEX 0
Oracle 行转列，列转行
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转载自yangeoo.iteye.com/blog/1917584
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