VALUES OF语句实例¶
如下实例使用VALUES OF语法进行批量执行。参数集合INDICE的全部有效元素值为1、3、5,因此相应FORALL的循环变量分别取值1、3、5,对表中相应数据进行更新操作。
-- 清理环境
DROP PROCEDURE proc;
DROP TABLE tab;
CREATE TABLE tab
(
ID INTEGER,
NUM INTEGER,
NAME VARCHAR2(15)
);
INSERT INTO tab VALUES(1, 11,'1A');
INSERT INTO tab VALUES(2, 22,'2B');
INSERT INTO tab VALUES(3, 33,'3C');
INSERT INTO tab VALUES(4, 44,'4D');
INSERT INTO tab VALUES(5, 55,'5E');
SELECT * FROM tab ORDER BY NAME;
ID(int) |NUM(int) |NAME(varchar) |
----------------------------------------------
1 |11 |1A |
----------------------------------------------
2 |22 |2B |
----------------------------------------------
3 |33 |3C |
----------------------------------------------
4 |44 |4D |
----------------------------------------------
5 |55 |5E |
总数目:5
CREATE OR REPLACE PROCEDURE proc AS
TYPE IDTAB IS VARRAY(20) OF tab.ID%TYPE;
TYPE NUMTAB IS TABLE OF tab.NUM%TYPE INDEX BY PLS_INTEGER;
TYPE NAMETAB IS TABLE OF tab.NAME%TYPE INDEX BY PLS_INTEGER;
TYPE INDICESTAB IS TABLE OF PLS_INTEGER INDEX BY PLS_INTEGER;
IDS IDTAB;
NUMS NUMTAB;
NAMES NAMETAB;
INDICE INDICESTAB;
ITERATIONS CONSTANT PLS_INTEGER := 5;
K INT;
BEGIN
INDICE(100) := 1;
INDICE(300) := 3;
INDICE(500) := 5;
FORALL II IN VALUES OF INDICE
UPDATE tab SET NAME = 'NEW' WHERE ID = II RETURNING ID,NUM,NAME
BULK COLLECT INTO IDS,NUMS,NAMES;
--输出集合数据
K := IDS.FIRST;
DBMS_OUTPUT.PUT_LINE('== IDS ==');
DBMS_OUTPUT.PUT_LINE(IDS.COUNT);
WHILE K IS NOT NULL LOOP
DBMS_OUTPUT.PUT_LINE('DATA OF '|| K || ' IS '||IDS(K));
K := IDS.NEXT(K);
END LOOP;
K := NUMS.FIRST;
DBMS_OUTPUT.PUT_LINE('== NUMS ==');
DBMS_OUTPUT.PUT_LINE(NUMS.COUNT);
WHILE K IS NOT NULL LOOP
DBMS_OUTPUT.PUT_LINE('DATA OF '|| K || ' IS '||NUMS(K));
K := NUMS.NEXT(K);
END LOOP;
K := NAMES.FIRST;
DBMS_OUTPUT.PUT_LINE('== NAMES ==');
DBMS_OUTPUT.PUT_LINE(NAMES.COUNT);
WHILE K IS NOT NULL LOOP
DBMS_OUTPUT.PUT_LINE('DATA OF '|| K || ' IS '||NAMES(K));
K := NAMES.NEXT(K);
END LOOP;
END;
/
EXEC proc;
== IDS ==
3
DATA OF 1 IS 1
DATA OF 2 IS 3
DATA OF 3 IS 5
== NUMS ==
3
DATA OF 1 IS 11
DATA OF 2 IS 33
DATA OF 3 IS 55
== NAMES ==
3
DATA OF 1 IS NEW
DATA OF 2 IS NEW
DATA OF 3 IS NEW
SELECT * FROM tab ORDER BY NAME;
ID(int) |NUM(int) |NAME(varchar) |
----------------------------------------------
2 |22 |2B |
----------------------------------------------
4 |44 |4D |
----------------------------------------------
1 |11 |NEW |
----------------------------------------------
3 |33 |NEW |
----------------------------------------------
5 |55 |NEW |
总数目:5
--删除
DROP PROCEDURE proc;
DROP TABLE tab;