同一表中的行之间的差异(Oracle SQL)
我一直在比较Oracle SQL中同一个表中的行。我需要得到所有的前/后风格的变化 我有这样的桌子:同一表中的行之间的差异(Oracle SQL),sql,oracle,Sql,Oracle,我一直在比较Oracle SQL中同一个表中的行。我需要得到所有的前/后风格的变化 我有这样的桌子: id date name action 1 01-01-2011 Alex smth 1 05-01-2011 Alexx smth 1 07-01-2011 Alexa smth2 2 02-01-2012 Leo smth3 2 05-01-2012 Leon sm
id date name action
1 01-01-2011 Alex smth
1 05-01-2011 Alexx smth
1 07-01-2011 Alexa smth2
2 02-01-2012 Leo smth3
2 05-01-2012 Leon smth3
我需要得到这个:
id date field before after
1 05-01-2011 name Alex Alexx
1 07-01-2011 name Alexx Alexa
1 07-01-2011 action smth smth2
2 05-01-2012 name Leo Leon
我试着把桌子和它自己连在一起。我找到的方法请在这里找到,它应该可以帮助我将行与下一行进行内部连接,下一行返回无效数字错误
有没有更简单的方法来完成这项任务?
你能帮帮我吗
select t1.id from tablename t1
inner join tablename t3 on t1.id = t3.id + 1
我首先根据每个id的日期为每一行分配序号,然后将每一行与其下一行自连接。一旦有了这些,您就可以使用几个案例陈述来生成前后数据:
WITH cte AS (
SELECT id, date, name, action,
ROW_NUMBER() OVER (PARTITION BY id ORDER BY date ASC) AS rn
FROM mytable
)
SELECT a.id,
a.date
CASE WHEN b.name != a.name THEN 'name' ELSE 'action' END AS field,
CASE WHEN b.name != a.name THEN b.name ELSE b.action END AS before,
CASE WHEN b.name != a.name THEN a.name ELSE a.action END AS after
FROM cte b
JOIN cte a ON b.id = a.id AND b.rn = a.rn + 1
由于您有两个字段要比较,我使用UNION ALL进行了比较,如下所示:
WITH CTE AS (
SELECT
ID,
DATE1,
NAME1,
ACTION,
ROW_NUMBER() OVER(
PARTITION BY ID
ORDER BY
DATE1 ASC
) AS RN
FROM
MYTABLE
)
--
SELECT
A.ID,
B.DATE1,
'name' AS FIELD,
A.NAME1 AS BEFORE,
B.NAME1 AS AFTER
FROM
CTE A
JOIN CTE B ON B.ID = A.ID
AND B.RN = A.RN + 1
AND B.NAME1 != A.NAME1
UNION ALL
SELECT
A.ID,
B.DATE1,
'action' AS FIELD,
A.ACTION AS BEFORE,
B.ACTION AS AFTER
FROM
CTE A
JOIN CTE B ON B.ID = A.ID
AND B.RN = A.RN + 1
AND B.ACTION != A.ACTION
ORDER BY
ID,
DATE1
输出:
干杯 您可以使用分析函数:
with t( id, "date", name, action ) as
(
select 1, date'2011-01-01','Alex','smth' from dual union all
select 1, date'2011-01-05','Alexx','smth' from dual union all
select 1, date'2011-01-07','Alexa','smth2' from dual union all
select 2, date'2012-01-02','Leo','smth3' from dual union all
select 2, date'2012-01-05','Leon','smth3' from dual
), t2 as
(
select t.*,
lag(name,1,null) over (partition by id order by id, "date") as lg_name,
lag(action,1,null) over (partition by id order by id, "date") as lg_action
from t
), t3 as
(
select id, "date", 'name' as field, lg_name as before, name as after
from t2 where name != lg_name
union all
select id, "date", 'action', lg_action, action
from t2 where action != lg_action
)
select * from t3 order by id, "date";
ID date FIELD BEFORE AFTER
-- --------- ----- ------ ------
1 05-JAN-11 name Alex Alexx
1 07-JAN-11 action smth smth2
1 07-JAN-11 name Alexx Alexa
2 05-JAN-12 name Leo Leon
LAG显然是正确的方法。但是,我首先要取消Pivot:
select id, date, field, prev_value as before, value as after
from (select id, date, field, value,
lag(value) over (partition by id, field order by date) as prev_value
from ((select id, "date", 'name' as field, name as value
from t
) union all
(select id, "date", 'action' as field, action as value
from t
)
) t
) t
where prev_value <> value;
在较新版本的Oracle中,可以使用横向联接来简化这一过程:
select id, date, field, prev_value as before, value as after
from (select t.id, t.date, x.field, x.value,
lag(x.value) over (partition by t.id, x.field order by date) as prev_value
from t cross join lateral
(select 'name' as field, name as value from dual union all
select 'action', action from dual
) x
) t
where prev_value <> value;
您能解释一下为什么在结果中只有一行:field=action:action smth smth2,但缺少:smth和smth3吗?我无法获取逻辑。因为我只需要选择已更改的字段。如果对于同一条记录,两个字段都在更改给定数据,则此查询仅返回3条记录,则此操作无效。所有案例陈述中也缺少结尾关键字。非常感谢!这是我最需要的,因为我的表中有更多的字段。非常感谢!它也非常有用。抱歉,无法匹配作为第二个答案: