Question

尝试设置一个查询，该查询返回多年内一年中最早的日期。

例如：

06-apr-1990
07-may-1991
03-apr-1992
07-jun-1993

最早的是03-apr-1992

感谢任何帮助（ - ：

我正在使用Oracle SQL Developer

日期采用日期格式

Answer 1

如果您使用的是SQL Server，请尝试：

SELECT TOP(1) date FROM table ORDER BY Month(date), Day(date)

对于MySQL，这应该可以解决问题：

SELECT date FROM table ORDER BY Month(date), Day(date) LIMIT 1;

对于Oracle：

SELECT date FROM table ORDER BY Month(date), Day(date) WHERE ROWNUM <= 1;

Answer 2

尝试以下查询：

SELECT column_name,column_name FROM table_name ORDER BY column_name,column_name ASC|DESC;

来源w3school.

Answer 3

这样的事情怎么样......

create table #temp1
(
 ID INT , 
 CreatedDate datetime
)

INSERT #temp1 (Id, CreatedDate )values(1, '06-apr-1990')
INSERT #temp1 (Id, CreatedDate )values(2, '07-may-1991')
INSERT #temp1 (Id, CreatedDate )values(3, '03-apr-1992')
INSERT #temp1 (Id, CreatedDate )values(4, '07-jun-1993')

SELECT A.CREATEDDATE FROM
(
    SELECT TOP 1 CREATEDDATE, CAST(YEAR(CreatedDate) as char(4)) [YEAR], DATENAME(MONTH,CreatedDate)   [MONTH]
    , DATENAME(DAY,CreatedDate) [DAY]
    FROM #temp1
    GROUP BY CreatedDate
    ORDER BY [DAY], [MONTH], [YEAR]
 ) A

Answer 4

ANSI上支持

Oracle 12 SQL：

select * 
from tbl 
order by extract(month from the_date), extract(day from the_date)
fetch first 1 row only;

对于较低版本的Oracle，您可以使用：

with a as
(
  select 

      row_number() 
          over(order by 
               extract(month from the_date), extract(day from the_date)) as rn, 

      tbl.the_date

  from tbl 
)
select the_date 
from a
where rn = 1;

直播代码：http://sqlfiddle.com/#!4/e6a75/20

多年来一年中最早的日期

4 个答案: