Question

我有一张桌子

create table abc (id integer, value varchar(20), date1 date);

数据是：

insert into abc values (1 , a, '2015-02-01');
insert into abc values (1 , b, '2015-02-02');
insert into abc values (1 , c, '2015-02-03');

我想要的结果如下：

old value || new value || updated date
a         ||  b        || 2015-02-02
b         || c         || 2015-02-03

我如何得到这个结果？

Answer 1

子选择和内连接的组合

SELECT 
old.value as "old value",
new.value as "new value",
new.date1 as "updated date"
FROM abc AS OLD
INNER JOIN abc as new
WHERE old.id = new.id
AND new.date1 = ( 
  SELECT MIN(date1) FROM abc
  WHERE old.id = id
  AND date1 > old.date1
)

see SQL Fiddle

Answer 2

DB2的OLAP功能可以在这里提供帮助：

select id,
       lag(value) over (partition by id order by date1) as "old value", 
       value as "new value",
       date1 as "updated date"
 from abc

我认为你可能还需要返回id列作为另一个id，你只需要它的旧值和新值。此查询还将返回另外的第一行，这也可能有用。如果您想避免它（如您所示的示例中），请使用以下SQL

with temp as (
select id,
       lag(value) over (partition by id order by date1) as "old value", 
       value as "new value",
       date1 as "updated date"
 from abc)
 select * from temp
  where "old value" is not null

如何将1列的2行数据提取到2列的行中

2 个答案: