1

我是 SQL 新手，遇到问题希望您能帮助我：

甲骨文 10g

表帐户

+----------+----------+
| 帐号| LBKEY |
+----------+----------+
| ... | ... |
| 254 | 价值254 |
| ... | ... |
| 401 | 价值401 |
| ... | ... |
| 405 | 价值405 |
+----------+----------+

交叉引用表

+----------+----------+----------+--------+
| IDTABLE2 | 帐号| OIDID | 价值 |
+----------+----------+----------+--------+
| ... | ... | ... | ... |
| 第475章 401 | 4 | 40000 |
| 第476章 405 | 4 | 35000 |
| ...| ... | ... | ... |
| 3000 | 254 | 5 | 巴黎 |
| 3001 | 401 | 5 | 伦敦 |
| 3002 | 405 | 5 | 悉尼 |
| ...| ... | ... | ... |
+----------+----------+----------+--------+

表 OID

+----------+-------------+-------------+
| OIDID | 标识 | 说明 |
+----------+-------------+-------------+
| 1 | x | x |
| 2 | x | x |
| 3 | x | x |
| 4 | 1.3.6.1.4.1 | 邮政编码 |
| 5 | 1.3.6.1.4.2 | 城市 |
| 6 | x | x |
| 7 | x | x |
| 8 | x | x |
| 9 | x | x |
| 10 | x | x |
+----------+-------------+-------------+

预期结果

约束：交叉引用表中有一个邮政编码（OID 4）或城市编码（OID 5）的所有ACCOUNT（LBKEY）

+----------+-------------+-------------+
| LBKEY | 邮政编码 | 城市 |
+----------+-------------+-------------+
| 价值254 | 空 | 巴黎 |
| 价值401 | 40000 | 伦敦 |
| 价值405 | 35000 | 悉尼 |
+----------+-------------+-------------+
4

2 回答 2

0

我认为这可能对你有用：

select
        lbkey,
        cross_post.value as postcode,
        cross_city.value as city
from
        ACCOUNT a,
        cross cross_city,
        cross cross_post,
where
        a.accountid=cross_city.accountid(+) and
        a.accountid=cross_post.accountid(+) and
        nvl(cross_city.oidid,5)=5 and
        nvl(cross_post.oidid,4)=4 and
        (cross_city.oidid is not null or cross_post.oidid is not null)
于 2013-12-23T17:22:17.870 回答
0

三种不同的方法：

SQL小提琴

Oracle 11g R2 模式设置：

CREATE TABLE ACCOUNT ( ACCOUNTID, LBKEY ) AS
          SELECT 254, 'value254' FROM DUAL
UNION ALL SELECT 401, 'value401' FROM DUAL
UNION ALL SELECT 405, 'value405' FROM DUAL
UNION ALL SELECT 406, 'value406' FROM DUAL;

CREATE TABLE CrossReference ( IDTABLE2, ACCOUNTID, OIDID, VALUE ) AS
          SELECT  475, 401, 4, '40000' FROM DUAL
UNION ALL SELECT  476, 405, 4, '35000' FROM DUAL
UNION ALL SELECT 3000, 254, 5, 'PARIS' FROM DUAL
UNION ALL SELECT 3001, 401, 5, 'LONDON' FROM DUAL
UNION ALL SELECT 3002, 405, 5, 'SYDNEY' FROM DUAL
UNION ALL SELECT 4000, 406, 6, 'x' FROM DUAL;


CREATE TABLE OID (OIDID, OID, DESCRIPTION ) AS
          SELECT  1, 'x', 'x' FROM DUAL
UNION ALL SELECT  2, 'x', 'x' FROM DUAL
UNION ALL SELECT  3, 'x', 'x' FROM DUAL
UNION ALL SELECT  4, '1.3.6.1.4.1', 'Post Code' FROM DUAL
UNION ALL SELECT  5, '1.3.6.1.4.2', 'City' FROM DUAL
UNION ALL SELECT  6, 'x', 'x' FROM DUAL
UNION ALL SELECT  7, 'x', 'x' FROM DUAL
UNION ALL SELECT  8, 'x', 'x' FROM DUAL
UNION ALL SELECT  9, 'x', 'x' FROM DUAL
UNION ALL SELECT 10, 'x', 'x' FROM DUAL;

查询 1：

SELECT LBKEY,
       MAX( CASE OIDID WHEN 4 THEN VALUE END ) AS "Post Code",
       MAX( CASE OIDID WHEN 5 THEN VALUE END ) AS "City"
FROM   ACCOUNT a
       INNER JOIN
       CrossReference c
       ON ( a.ACCOUNTID = c.ACCOUNTID )
WHERE  c.OIDID IN ( 4, 5 )
GROUP BY LBKEY

结果：

|    LBKEY | POST CODE |   CITY |
|----------|-----------|--------|
| value254 |    (null) |  PARIS |
| value405 |     35000 | SYDNEY |
| value401 |     40000 | LONDON |

查询 2：

WITH data AS (
  SELECT LBKEY,
         ( SELECT VALUE
           FROM   CrossReference c
           WHERE  c.ACCOUNTID = a.ACCOUNTID
           AND    c.OIDID = 4 ) AS "Post Code",
         ( SELECT VALUE
           FROM   CrossReference c
           WHERE  c.ACCOUNTID = a.ACCOUNTID
           AND    c.OIDID = 5 ) AS "City"
  FROM   ACCOUNT a
)
SELECT *
FROM   data
WHERE  "Post Code" IS NOT NULL
OR     "City"      IS NOT NULL

结果：

|    LBKEY | POST CODE |   CITY |
|----------|-----------|--------|
| value254 |    (null) |  PARIS |
| value401 |     40000 | LONDON |
| value405 |     35000 | SYDNEY |

查询 3：

SELECT LBKEY,
       c1.VALUE AS "Post Code",
       c2.VALUE AS City
FROM   ACCOUNT a
       LEFT OUTER JOIN
       ( SELECT ACCOUNTID, VALUE FROM CrossReference WHERE OIDID = 4 ) c1
       ON ( c1.ACCOUNTID = a.ACCOUNTID )
       LEFT OUTER JOIN
       ( SELECT ACCOUNTID, VALUE FROM CrossReference WHERE OIDID = 5 ) c2
       ON ( c2.ACCOUNTID = a.ACCOUNTID )
WHERE  c1.VALUE IS NOT NULL
OR     c2.VALUE IS NOT NULL

结果：

|    LBKEY | POST CODE |   CITY |
|----------|-----------|--------|
| value254 |    (null) |  PARIS |
| value401 |     40000 | LONDON |
| value405 |     35000 | SYDNEY |
于 2013-12-23T22:15:00.710 回答