我是SQL的新手,有问题希望你能帮助我:
ORACLE 10g
表ACCOUNT
+----------+----------+|ACCOUNTID|LBKEY|+----------+----------+|…|||254 |值254||…|||401 |值401||…|||405 |值405|+----------+----------+
交叉参考表
+----------+----------+----------+--------+|IDTABLE2|ACCOUNTID|OIDID|VALUE|+----------+----------+----------+--------+|…|…|……|||475 | 401 | 4 | 40000||476 | 405 | 4 | 35000||…|…|……|||3000 | 254 | 5 |巴黎||3001|401|5|伦敦||3002|405|5|悉尼||…|…|……||+----------+----------+----------+--------+
表OID
+----------+-------------+-------------+|OIDID|OID|DESCRIPTION|+----------+-------------+-------------+|1|x|x||2|x|x||3|x|x||4|1.3.6.1.4.1|邮编||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:在交叉引用表中具有一个邮政编码(OID 4)或城市编码(OIID 5)的所有ACCOUNT(LBKEY)
+----------+-------------+-------------+|LBKEY |邮编|城市|+----------+-------------+-------------+|value254 | null |巴黎||value401|40000|伦敦||value405 | 35000 |悉尼|+----------+-------------+-------------+
三种不同的方法:
SQL Fiddle
Oracle 11g R2架构设置:
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 |
我认为这对你有用:
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)