[英]PL/SQL wrapping hierarchical query in a stored procedure
感谢所有人的帮助,我对PL/SQL
还是陌生的,并且发现语法比T-SQL更具挑战性。 我有一个运行正常的PL/SQL cursor
,可以将我想要的确切插入表中。 下一步,我想将其包装在存储过程中,这样我就可以传递单个参数值,该值将代替脚本中显示的“ MD01”,用户可以输入任意4个字符串。
问题是当我这样做时(简单的CREATE OR REPLACE PROCEDURE test AS
),即使代码在两秒钟前起作用,我仍然遇到了很多错误。 我究竟做错了什么? 下面发布的代码部分功能正常,但是我不知道如何正确地将其包装到PL / SQL中的存储过程中。
ORA-00942:表或视图不存在PLS-00364:循环索引变量'EACH_REC'使用无效ORA-00984:此处不允许使用列
CREATE OR REPLACE PROCEDURE test IS
DECLARE
CURSOR c1 IS SELECT * FROM
(
SELECT
C.FEE_SCHEDULE
, C.PROC
, C.MODIFIER
, C.MODIFIER2
, C.PROVIDER
, C.YMDEFF
, C.YMDEND
, C.NEXT_SPAN_DATE
, C.SPAN
, C.SPAN_FLAG
, C.RATE
, TO_DATE(D.YMDTRANS,'YYYYMMDD') AS YMDTRANS
FROM
(
SELECT
B.FEE_SCHEDULE
, B.PROC
, B.MODIFIER
, B.MODIFIER2
, PROVIDER
, TO_DATE(B.YMDEFF,'YYYYMMDD') AS YMDEFF
, TO_DATE(B.YMDEND,'YYYYMMDD') AS YMDEND
, CASE WHEN RECURSION_LEVEL = 1 THEN NULL ELSE TO_DATE(B.T3,'YYYYMMDD')END AS NEXT_SPAN_DATE
, CASE WHEN B.YMDEND = '99991231' THEN NULL
WHEN B.RANK2 = '1' THEN NULL
ELSE TO_DATE(B.T3,'YYYYMMDD') - TO_DATE(B.YMDEND,'YYYYMMDD') END AS SPAN
, CASE WHEN TO_DATE(B.T3,'YYYYMMDD') - TO_DATE(B.YMDEND,'YYYYMMDD') = '1' THEN 'CORRECT_SPAN'
WHEN B.YMDEND = '99991231' THEN 'CORRECT_SPAN'
WHEN B.RANK2 = '1' THEN 'CORRECT_SPAN'
ELSE 'GAPPED_SPAN' END AS SPAN_FLAG
--, RANK1
--, RECURSION_LEVEL
, RATE
, YMDTRANS
FROM
(
SELECT
A.*
, CONNECT_BY_ISCYCLE AS T1
, sys_connect_by_path(YMDEFF,' ') AS T2
, SUBSTR(sys_connect_by_path(YMDEFF,' '),1,9) AS T3
, LEVEL AS RECURSION_LEVEL
FROM
(
SELECT
SUBSTR(FEE_KEY,3,4) AS FEE_SCHEDULE
, SUBSTR(FEE_KEY,7,5) AS PROC
, SUBSTR(FEE_KEY,19,2) AS MODIFIER
, SUBSTR(FEE_KEY,23,2) AS MODIFIER2
, SUBSTR(FEE_KEY,29,12) AS PROVIDER
, FEE_KEY
, YMDEFF
, YMDEND
, YMDTRANS
, DENSE_RANK () OVER (PARTITION BY SUBSTR(FEE_KEY,3,4)
, SUBSTR(FEE_KEY,7,5)
, SUBSTR(FEE_KEY,19,2)
, SUBSTR(FEE_KEY,23,2)
, SUBSTR(FEE_KEY,29,12)
ORDER BY YMDEFF) AS RANK1
, DENSE_RANK () OVER (PARTITION BY SUBSTR(FEE_KEY,3,4)
, SUBSTR(FEE_KEY,7,5)
, SUBSTR(FEE_KEY,19,2)
, SUBSTR(FEE_KEY,23,2)
, SUBSTR(FEE_KEY,29,12)
ORDER BY YMDEND DESC) AS RANK2
, RATE/100 AS RATE
FROM AMIOWN.FEE_SCHEDULE
WHERE 1 = 1
AND SUBSTR(FEE_KEY,3,4) = 'MD01'
) A
START WITH FEE_SCHEDULE IN('MD01')
CONNECT BY NOCYCLE
PRIOR RANK1 = RANK1 + 1
AND PRIOR FEE_SCHEDULE = SUBSTR(FEE_KEY,3,4)
AND PRIOR PROC = SUBSTR(FEE_KEY,7,5)
AND PRIOR MODIFIER = SUBSTR(FEE_KEY,19,2)
AND PRIOR MODIFIER2 = SUBSTR(FEE_KEY,23,2)
AND PRIOR PROVIDER = SUBSTR(FEE_KEY,29,12)
AND LEVEL = 2
ORDER BY PROC, YMDEFF, LEVEL
) B
WHERE 1 = 1
AND RECURSION_LEVEL = 2
OR (RECURSION_LEVEL = 1 AND CASE WHEN RECURSION_LEVEL = 1 THEN NULL ELSE TO_DATE(B.T3,'YYYYMMDD')END IS NOT NULL )
OR B.YMDEND = '99991231'
OR B.RANK2 = 1
) C
INNER JOIN
(
SELECT
SUBSTR(FEE_KEY,3,4) AS FEE_SCHEDULE
, MAX(YMDTRANS) AS YMDTRANS
FROM AMIOWN.FEE_SCHEDULE
WHERE SUBSTR(FEE_KEY,3,4) = 'MD01'
GROUP BY SUBSTR(FEE_KEY,3,4)
) D ON C.FEE_SCHEDULE = D.FEE_SCHEDULE
WHERE 1 = 1
);
i NUMBER:= 0;
BEGIN
FOR each_rec IN c1 LOOP
INSERT INTO SCHEMA.FEE_SCHEDULE_GAPS_DETAIL
(
FEE_SCHEDULE
, PROC
, MODIFIER
, MODIFIER2
, PROVIDER
, YMDEFF
, YMDEND
, NEXT_SPAN_DATE
, SPAN
, SPAN_FLAG
, RATE
, YMDTRANS
)
VALUES
( each_rec.FEE_SCHEDULE
, each_rec.PROC
, each_rec.MODIFIER
, each_rec.MODIFIER2
, each_rec.PROVIDER
, each_rec.YMDEFF
, each_rec.YMDEND
, each_rec.NEXT_SPAN_DATE
, each_rec.SPAN
, each_rec.SPAN_FLAG
, each_rec.RATE
, each_rec.YMDTRANS
);
i:= i+1;
END LOOP;
END;
/
Boneist是正确的:“具有authid定义程序(默认值)的过程不会查看角色来确定是否具有必要的特权-而是,授予必须是直接的-即,不是在schema1.table1上授予select some_role;向schema2授予some_role,必须在schema1.table1上向schema2授予select“
我接触了我们的Oracle DBA,他们迅速授予了我直接访问权限,并且存储过程完全正确地编译了。 非常感谢,很高兴我没有疯。
匿名块:
DECLARE
<declarations>
BEGIN
<actions>
END;
可以通过以下方式转换为存储过程:
CREATE OR REPLACE PROCEDURE test IS
<declarations>
BEGIN
<actions>
END;
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