Oracle SQL创建函数 - 混淆编译错误

Question

基于如何在Oracle函数中返回动态结果集

我正在尝试创建一个函数,它将从几个表中返回一行包含多个计数.这是我到目前为止所拥有的:

CREATE OR REPLACE TYPE RESULT_ROW is OBJECT
(LOC_TABLE_ENTRY_KY VARCHAR2(50),
LOCATION_NAME VARCHAR2(50),
A_ASSIGN_CNT VARCHAR2(50),
B_ASSIGN_CNT VARCHAR2(50),
C_ASSIGN_CNT VARCHAR2(50),
D_ASSIGN_CNT VARCHAR2(50),
E_ASSIGN_CNT VARCHAR2(50),
F_ASSIGN_CNT VARCHAR2(50),
G_ASSIGN_CNT VARCHAR2(50),
H_ASSIGN_CNT VARCHAR2(50));
/
CREATE OR REPLACE TYPE RESULT_TABLE AS TABLE OF RESULT_ROW;
/
CREATE OR REPLACE FUNCTION LOCATION_RULE_LOOKUP(P_LOCATION_VAR IN NUMBER)
RETURN RESULT_TABLE
IS
OUT_REC RESULT_TABLE;
BEGIN
WITH LOC AS
(SELECT LOC_TABLE_ENTRY_KY,
LOCATION_NAME
FROM LOCATION_CODE
WHERE LOC_TABLE_ENTRY_KY = P_LOCATION_VAR
),
ONE AS
(SELECT COUNT(*) AS A_ASSIGN_CNT
FROM COLLECTOR_ASSIGNMENT
WHERE LOCATION_CODE     = P_LOCATION_VAR
AND FUNCTION_STATE_CODE = '   '
OR FUNCTION_STATE_CODE  = '***'
),
TWO AS
(SELECT COUNT(*) AS B_ASSIGN_CNT
FROM COMM_PLAN_ASSGN_RULE
WHERE LOCATION_CODE = P_LOCATION_VAR
),
THREE AS
(SELECT COUNT(*) AS C_ASSIGN_CNT
FROM INPUT_TRANS_ASGN
WHERE LOCATION_CODE = P_LOCATION_VAR
),
FOUR AS
(SELECT COUNT(*) AS D_ASSIGN_CNT
FROM RECALL_DAYS_ASSGN_RULE
WHERE LOCATION_CODE = P_LOCATION_VAR
),
FIVE AS
(SELECT COUNT(*) AS E_ASSIGN_CNT
FROM SCRIPT_VIEW_ASSIGNMENT
WHERE LOCATION_CODE     = P_LOCATION_VAR
AND FUNCTION_STATE_CODE = '   '
OR FUNCTION_STATE_CODE  = '***'
AND SCRIPT_TYPE         = 'V'
),
SIX AS
(SELECT COUNT(*) AS F_ASSIGN_CNT
FROM SCRIPT_VIEW_ASSIGNMENT
WHERE LOCATION_CODE     = P_LOCATION_VAR
AND FUNCTION_STATE_CODE = '   '
OR FUNCTION_STATE_CODE  = '***'
AND SCRIPT_TYPE         = 'D'
),
SEVEN AS
(SELECT COUNT(*) AS G_ASSIGN_CNT
FROM TSR_ASSGN_RULE
WHERE LOCATION_CODE = P_LOCATION_VAR
),
EIGHT AS
(SELECT COUNT(*) AS H_ASSIGN_CNT
FROM TRAN_STATE_ASSGN_RULE
WHERE LOCATION_CODE = P_LOCATION_VAR
)
SELECT * INTO OUT_REC
FROM ONE, TWO, THREE, FOUR, FIVE, SIX, SEVEN, EIGHT, LOC;
RETURN OUT_REC;
END LOCATION_RULE_LOOKUP;

但无论我做什么使它看起来像现有的函数示例,无论是在这里还是在其他站点上,它都不接受BEGIN块中的任何内容.BEGIN块内的SQL有效; 我可以运行所有那些WITH AS SELECT等,它会给我一行作为结果与我正在寻找的计数.但是这个函数不会编译.根据我放置空格或分号的位置,错误会发生变化.目前的错误:

Error(7,3): PL/SQL: SQL Statement ignored
Error(62,3): PL/SQL: ORA-00947: not enough values

救命？d:

Answer 1

你非常接近,但你并没有非常忠实地遵循这个相关的问题.not enough values消息的原因是你的INTO子句中只有一个值,但Oracle期望你有八个,因为那是你的(有效)多少个表达式SELECT *.您需要将这八个表达式转换为单个值.要做到这一点,您需要根据需要将其更改WITH ... SELECT * INTO out_rec ...为SELECT CAST(MULTISET(WITH ... SELECT * ...) AS result_table) INTO out_rec FROM dual或WITH ... SELECT CAST(MULTISET(SELECT * ...) AS result_table) INTO out_rec FROM dual.

此外,您需要以FROM正确的顺序放置您的条款,并在LOC前面ONE通过EIGHT; 基于表达式的相对位置执行强制转换,而不是基于字段别名.(或者,您可以更改您SELECT *以正确的顺序明确标识字段,而不是依赖于FROM-clause顺序.)