我有一个像这样的SQL Server表
+----+-----------+------------+
| id | acoount | date |
+----+-----------+------------+
| | John | 2/6/2016 |
| | John | 2/6/2016 |
| | John | 4/6/2016 |
| | John | 4/6/2016 |
| | Andi | 5/6/2016 |
| | Steve | 4/6/2016 |
+----+-----------+------------+
我希望像这样插入id coloumn。
+-----------+-----------+------------+
| id | acoount | date |
+-----------+-----------+------------+
| 020616001 | John | 2/6/2016 |
| 020616002 | John | 2/6/2016 |
| 040616001 | John | 4/6/2016 |
| 040616002 | John | 4/6/2016 |
| 050616001 | Andi | 5/6/2016 |
| 040616003 | Steve | 4/6/2016 |
+-----------+-----------+------------+
我想生成像这样提供的日期的ID号。 02 + 06 + 16(自日期起)+001 = 020616001.如果日期相同,则为id + 1。 我试过但仍然失败了。 我希望在oracle sql开发中实现它。 谁来帮帮我。 感谢。
答案 0 :(得分:1)
根据给定的数据尝试以下SQL,它在SQL Server 2012中....
select REPLACE(CONVERT(VARCHAR(10),convert(date,t.[date]), 101), '/', '')
+'00'+convert(varchar(2),row_number()over(partition by account,[date] order by t.[date])) as ID,
t.account,
t.date
from (values ('John','2/6/2016'),
('John','2/6/2016'),
('John','4/6/2016'),
('John','4/6/2016'),
('Andi','5/6/2016'),
('Steve','4/6/2016'))T(account,[date])
答案 1 :(得分:0)
<强> MySQL的强>
我现在可以给你020616001
这部分的逻辑.......
对于相同的id +1我必须努力....我会在工作后告诉你
insert into table_name(id)
select concat
(
if(length (day(current_date))>1,day(current_date),Concat(0,day(current_date))),
if(length (month(current_date))>1,month(current_date),Concat(0,month(current_date))),
(right(year(current_date),2)),'001'
)as id
答案 2 :(得分:0)
使用statement更新表。
update table set id= replace(CONVERT(VARCHAR(10),CONVERT(datetime ,date,103),3) ,'/', '') + Right('00'+convert(varchar(2),row_number()over(partition by account,[date] order by t.[date])) ,3)
答案 3 :(得分:0)
您无法以正常方式将日期列转换为日期时间类型,因为它是dd / mm / yyyy。
试试这个,
declare @t table(acoount varchar(50),dates varchar(20))
insert into @t values
('John','2/6/2016')
,('John','2/6/2016')
,('John','4/6/2016')
,('John','4/6/2016')
,('Andi','5/6/2016')
,('Steve','4/6/2016')
;With CTE as
(select * , SUBSTRING(dates,0,charindex('/',dates)) dd
,SUBSTRING(stuff(dates,1,charindex('/',dates),''),0, charindex('/',stuff(dates,1,charindex('/',dates),''))) MM
,right(dates,2) yy
from @t
)
,CTE1 as
(
select *
,ROW_NUMBER()over(partition by yy,mm,dd order by yy,mm,dd)rn from cte c
)
select *, REPLICATE('0',2-len(dd))+cast(dd as varchar(2))
+REPLICATE('0',2-len(MM))+cast(MM as varchar(2))
+yy+REPLICATE('0',3-len(rn))+cast(rn as varchar(2))
from cte1