[英]Union null in Oracle
我正在尝试在oracle中合并2个查询,这些行具有相同的值,但需要一个字段。
例如:
SELECT NAME, AGE, EMAIL, DATE FROM table_a WHERE NAME = 'JOAO' AND FLAG = '0'
UNION
SELECT NAME, AGE, EMAIL, DATE FROM table_a WHERE NAME = 'JOAO' AND FLAG = '1'
结果:
NAME AGE EMAIL DATE
JOAO 23 a@a.com 20150414
JOAO 23 a@a.com null
我如何将这些行分组? 我正在寻找可以给我类似结果的东西:
NAME AGE EMAIL DATE
JOAO 23 a@a.com 20150414
谢谢你(抱歉我的英语..)
如果您只是想忽略NULL值:
SELECT NAME, AGE, EMAIL, DATE
FROM table_a
WHERE NAME = 'JOAO'
AND FLAG in ( '0', '1' )
and date is not null
/
或者如果您想保留空值,但要遵循可用的非空值:
with w_data as (
SELECT NAME, AGE, EMAIL, DATE ,
row_number() over ( partition by name
order by date desc nulls last ) rnum
FROM table_a
WHERE NAME = 'JOAO'
AND FLAG in ( '0', '1' )
)
select name, age, email, date
from w_data
where rnum = 1
/
[edit]回应评论:
如果您想保留联合,那很好,联合和OR大多可以互换(在这种情况下):
SELECT NAME, AGE, EMAIL, DATE
FROM table_a
WHERE NAME = 'JOAO'
AND FLAG = '0'
and date is not null
union
SELECT NAME, AGE, EMAIL, DATE
FROM table_a
WHERE NAME = 'JOAO'
AND FLAG = '1'
and date is not null
/
甚至:
select * from (
SELECT NAME, AGE, EMAIL, DATE
FROM table_a
WHERE NAME = 'JOAO'
AND FLAG = '0'
union
SELECT NAME, AGE, EMAIL, DATE
FROM table_a
WHERE NAME = 'JOAO'
AND FLAG = '1'
)
where date is not null
/
如果您只想将NULL条件放在1个位置;)相同的逻辑可以应用于我上面使用row_number()分析编写的第二个查询...
with w_data as (
SELECT NAME, AGE, EMAIL, DATE
FROM table_a
WHERE NAME = 'JOAO'
AND FLAG = '0'
union
SELECT NAME, AGE, EMAIL, DATE
FROM table_a
WHERE NAME = 'JOAO'
AND FLAG = '1'
),
w_sub as (
select name, age, email, date,
row_number() over ( partition by name
order by date desc nulls last ) rnum
from w_data
)
select name, age, email, date
from w_sub
where rnum = 1
/
在这种情况下,您可以在“ WHERE”子句中更改您的条件
WHERE NAME = 'JOAO' AND FLAG = '0'
至
WHERE NAME = 'JOAO' AND FLAG IN('0','1')
因此,您的选择将在“ FLAG”列中显示所有值为“ 0”或“ 1”的寄存器。
您可以使用COALESCE()。 http://docs.oracle.com/cd/B28359_01/server.111/b28286/functions023.htm#SQLRF00617/ms190349.aspx
该查询应适用于每个名称,并应合并其他行。
SELECT
NAME1 AS NAME,
COALESCE(AGE1, AGE2) AS AGE,
COALESCE(EMAIL1, EMAIL2) AS EMAIL,
COALESCE(DATE1, DATE2) AS DATE
FROM(
SELECT
t1.NAME AS NAME1,
t1.AGE AS AGE1,
t1.EMAIL AS EMAIL1,
t1.DATE AS DATE1,
t2.NAME AS NAME2,
t2.AGE AS AGE2,
t2.EMAIL AS EMAIL2,
t2.DATE AS DATE2
FROM table_a AS t1
INNER JOIN table_a AS t2
ON t2.FLAG = 1 AND t1.FLAG = 0 AND t1.NAME = t2.NAME
) AS t3;
声明:本站的技术帖子网页,遵循CC BY-SA 4.0协议,如果您需要转载,请注明本站网址或者原文地址。任何问题请咨询:yoyou2525@163.com.