我要将佣金百分比列添加到现有查询中。但是，佣金数据位于与查询中使用的主表（A）不相关的表中。但是，这两个表与表B具有相同的列/值。
我下面有三个表A，B和C
Table A
Reference Value_Name Renewal_Code
1 A N
2 A R
3 B N
4 A R
4 A N
Table B
Reference Value_Name Prod_Code
1 A 0016
2 A 0027
4 A 0032
4 A 0032
Table C
A_Prod_Code A_Tans_Code **Commission_Percent**
0016 Renewal 5
0027 Renewal 5
0032 New 10
0032 Renewal 5
我需要从Commission_Percent获取Table C，该Renewal_Code与来自Table A的相应A_Tans_Code有关。这与Table C中的Table C相同，除了Renewal拼写出New或Table A而R仅使用N或Commission_Percent。
通过使用Table B作为公用值，我已经能够将NULL列拉到输出中，但是所有值都显示为Renewal_code/A_Trans_Code。
我还尝试了使用解码语句来链接Tables A and C中的(
SELECT
distinct c.commision_percent
FROM
TableA a
JOIN TableB b ON a.reference = b.reference
AND b.value_name = 'A'
JOIN TableC c ON b.prod_code = c.a_prod_code
AND b.value_name = 'A'
JOIN TableC c ON a.renewal_code = decode(c.a_trans_code, 'Rewnal','R','New','N')
) Commission_Percent
列。
commission_percent对于每个参考，我都需要Renewal和New正确的NULLs业务。到目前为止，由于很难链接Tables A和C's Renewal_code和A_prod_code列，我只能得到int l1; //123456 for example
scanf("%d",&l1);
char s[sizeof(l1)];
sprintf(s,"%5d",l1);'
//This will give you separate digits of the number in char format inside s[0],s[1]
//and so on.
//If you want them in int format, declare a int array say int i[sizeof(l1)] and add
//the following code
for(int c=1;c<=sizeof(l1);c++){
i[c] = s[c] - '0';
}
//Now i[0], i[1] etc will have the digits in int format
。
非常感谢您的帮助！
答案 0 :(得分：2)
这通过清理临时表并重新插入数据来创建可重现的测试。 并可能成为其他解决方案的模型。
SQL'Select ...'连接到一个具有打开条件的两个部分的单个TableC-prod_code和renewal_code（而不是两个连接）。只需删除“ a。*”即可在您的sql中使用它。 （将解码功能更改为使用A_Trans_Code的第一个字符的下标）。
IF OBJECT_ID('tempdb..#TableA') IS NOT NULL DROP TABLE #TableA
GO
CREATE TABLE #TableA
( Reference INTEGER
, Value_Name VARCHAR(10)
, Renewal_Code VARCHAR(10) )
INSERT INTO #TableA VALUES( 1, 'A', 'N' );
INSERT INTO #TableA VALUES( 2, 'A', 'R' );
INSERT INTO #TableA VALUES( 3, 'B', 'N' );
INSERT INTO #TableA VALUES( 4, 'A', 'R' );
INSERT INTO #TableA VALUES( 4, 'A', 'N' );
IF OBJECT_ID('tempdb..#TableB') IS NOT NULL DROP TABLE #TableB
GO
CREATE TABLE #TableB
( Reference INTEGER
, Value_Name VARCHAR(10)
, Prod_Code VARCHAR(10) )
INSERT INTO #TableB VALUES( 1, 'A', '0016' );
INSERT INTO #TableB VALUES( 2, 'A', '0027' );
INSERT INTO #TableB VALUES( 4, 'A', '0032' );
INSERT INTO #TableB VALUES( 4, 'A', '0032' );
IF OBJECT_ID('tempdb..#TableC') IS NOT NULL DROP TABLE #TableC
GO
CREATE TABLE #TableC
( Prod_Code VARCHAR(10)
, A_Trans_Code VARCHAR(10)
, Commission_Percent INTEGER )
INSERT INTO #TableC VALUES( '0016', 'Renewal', 5 );
INSERT INTO #TableC VALUES( '0027', 'Renewal', 5 );
INSERT INTO #TableC VALUES( '0032', 'New', 10 );
INSERT INTO #TableC VALUES( '0032', 'Renewal', 5 );
SELECT distinct a.*, c.commission_percent
FROM #TableA a
JOIN #TableB b ON a.reference = b.reference
AND a.value_name = b.value_name
JOIN #TableC c ON b.prod_code = c.prod_code
AND a.renewal_code = SUBSTRING(c.a_trans_code,1,1)
结果是-
Reference Value_Name Renewal_Code commission_percent
2 A R 5
4 A N 10
4 A R 5
要放入您的sql中的代码
( SELECT distinct c.commission_percent
FROM TableA a
JOIN TableB b ON a.reference = b.reference
AND a.value_name = b.value_name
JOIN TableC c ON b.prod_code = c.prod_code
AND a.renewal_code = SUBSTRING(c.a_trans_code,1,1)
) Commission_Percent
答案 1 :(得分：0)
WITH table_a AS (
SELECT 1 AS reference, 'A' AS value_name, 'N' AS renewal_code FROM DUAL UNION ALL
SELECT 2 AS reference, 'A' AS value_name, 'R' AS renewal_code FROM DUAL UNION ALL
SELECT 3 AS reference, 'B' AS value_name, 'N' AS renewal_code FROM DUAL UNION ALL
SELECT 4 AS reference, 'A' AS value_name, 'R' AS renewal_code FROM DUAL UNION ALL
SELECT 4 AS reference, 'A' AS value_name, 'N' AS renewal_code FROM DUAL
),
table_b AS (
SELECT 1 AS reference, 'A' AS value_name, '0016' AS prod_code FROM DUAL UNION ALL
SELECT 2 AS reference, 'A' AS value_name, '0027' AS prod_code FROM DUAL UNION ALL
/* SELECT 4 AS reference, 'A' AS value_name, '0032' AS prod_code FROM DUAL UNION ALL duplicate row excluded */
SELECT 4 AS reference, 'A' AS value_name, '0032' AS prod_code FROM DUAL
),
table_c AS (
SELECT '0016' AS a_prod_code, 'Renewal' AS a_tans_code, 5 AS commission_percent FROM DUAL UNION ALL
SELECT '0027' AS a_prod_code, 'Renewal' AS a_tans_code, 5 AS commission_percent FROM DUAL UNION ALL
SELECT '0032' AS a_prod_code, 'New' AS a_tans_code, 10 AS commission_percent FROM DUAL UNION ALL
SELECT '0032' AS a_prod_code, 'Renewal' AS a_tans_code, 5 AS commission_percent FROM DUAL
)
SELECT a.reference AS ref_a,
a.value_name AS value_name_a,
b.reference AS ref_b,
b.value_name AS value_name_b,
a.renewal_code,
b.prod_code,
c.commission_percent
FROM table_a a
LEFT OUTER JOIN table_b b
ON a.value_name = b.value_name
AND a.reference = b.reference
LEFT OUTER JOIN table_c c
ON SUBSTR(c.a_tans_code, 1, 1) = a.renewal_code
AND c.a_prod_code = b.prod_code
;
公用表表达式（CTE）只是用于创建与发布相同的值（CTE是使用WITH构造的部分）。
结果：
REF_A VALUE_NAME_A REF_B VALUE_NAME_B RENEWAL_CODE PROD_CODE COMMISSION_PERCENT
---------- --------------- ---------- --------------- --------------- ---------- --------------------
2 A 2 A R 0027 5
4 A 4 A N 0032 10
4 A 4 A R 0032 5
1 A 1 A N 0016
3 B N
您可能必须检查发布的行中不导致联接的值。这还假定没有code：code_name关系违反代码名= code的首字母规则。