获取sql中的几个月和几天中两个日期之间的差异

时间:2012-07-16 07:49:09

标签: sql oracle date-arithmetic

我需要得到两个日期之间的差异,如果差异是84天,我应该有2个月和14天的输出,我刚刚给出的总代码。这是代码

SELECT Months_between(To_date('20120325', 'YYYYMMDD'),
       To_date('20120101', 'YYYYMMDD'))
       num_months,
       ( To_date('20120325', 'YYYYMMDD') - To_date('20120101', 'YYYYMMDD') )
       diff_in_days
FROM   dual; 

输出是:

NUM_MONTHS    DIFF_IN_DAYS
2.774193548       84

我需要例如此查询的输出最差为2个月和14天,否则我不介意我是否可以确定几个月之后的确切天数,因为那些日子实际上不是14天因为所有月份没有30天。

10 个答案:

答案 0 :(得分:19)

select 
  dt1, dt2,
  trunc( months_between(dt2,dt1) ) mths, 
  dt2 - add_months( dt1, trunc(months_between(dt2,dt1)) ) days
from
(
    select date '2012-01-01' dt1, date '2012-03-25' dt2 from dual union all
    select date '2012-01-01' dt1, date '2013-01-01' dt2 from dual union all
    select date '2012-01-01' dt1, date '2012-01-01' dt2 from dual union all
    select date '2012-02-28' dt1, date '2012-03-01' dt2 from dual union all
    select date '2013-02-28' dt1, date '2013-03-01' dt2 from dual union all
    select date '2013-02-28' dt1, date '2013-04-01' dt2 from dual union all
    select trunc(sysdate-1)  dt1, sysdate               from dual
) sample_data

结果:

|                        DT1 |                       DT2 | MTHS |     DAYS |
----------------------------------------------------------------------------
|  January, 01 2012 00:00:00 |   March, 25 2012 00:00:00 |    2 |       24 |
|  January, 01 2012 00:00:00 | January, 01 2013 00:00:00 |   12 |        0 |
|  January, 01 2012 00:00:00 | January, 01 2012 00:00:00 |    0 |        0 |
| February, 28 2012 00:00:00 |   March, 01 2012 00:00:00 |    0 |        2 |
| February, 28 2013 00:00:00 |   March, 01 2013 00:00:00 |    0 |        1 |
| February, 28 2013 00:00:00 |   April, 01 2013 00:00:00 |    1 |        1 |
|   August, 14 2013 00:00:00 |  August, 15 2013 05:47:26 |    0 | 1.241273 |

链接到测试:SQLFiddle

答案 1 :(得分:2)

更新了正确性。最初由@jen回答。

with DATES as (
   select TO_DATE('20120101', 'YYYYMMDD') as Date1,
          TO_DATE('20120325', 'YYYYMMDD') as Date2
   from DUAL union all
   select TO_DATE('20120101', 'YYYYMMDD') as Date1,
          TO_DATE('20130101', 'YYYYMMDD') as Date2
   from DUAL union all
   select TO_DATE('20120101', 'YYYYMMDD') as Date1,
          TO_DATE('20120101', 'YYYYMMDD') as Date2
   from DUAL union all
   select TO_DATE('20130228', 'YYYYMMDD') as Date1,
          TO_DATE('20130301', 'YYYYMMDD') as Date2
   from DUAL union all
   select TO_DATE('20130228', 'YYYYMMDD') as Date1,
          TO_DATE('20130401', 'YYYYMMDD') as Date2
   from DUAL
), MONTHS_BTW as (
   select Date1, Date2,
          MONTHS_BETWEEN(Date2, Date1) as NumOfMonths
   from DATES
)
select TO_CHAR(Date1, 'MON DD YYYY') as Date_1,
       TO_CHAR(Date2, 'MON DD YYYY') as Date_2,
       NumOfMonths as Num_Of_Months,
       TRUNC(NumOfMonths) as "Month(s)",
       ADD_MONTHS(Date2, - TRUNC(NumOfMonths)) - Date1 as "Day(s)"
from MONTHS_BTW;

SQLFiddle 演示:

    +--------------+--------------+-----------------+-----------+--------+
    |   DATE_1     |   DATE_2     | NUM_OF_MONTHS   | MONTH(S)  | DAY(S) |
    +--------------+--------------+-----------------+-----------+--------+
    | JAN 01 2012  | MAR 25 2012  | 2.774193548387  |        2  |     24 |
    | JAN 01 2012  | JAN 01 2013  | 12              |       12  |      0 |
    | JAN 01 2012  | JAN 01 2012  | 0               |        0  |      0 |
    | FEB 28 2013  | MAR 01 2013  | 0.129032258065  |        0  |      1 |
    | FEB 28 2013  | APR 01 2013  | 1.129032258065  |        1  |      1 |
    +--------------+--------------+-----------------+-----------+--------+

注意,对于最后两个日期,Oracle会错误地报告月份的小数部分(给出天数)。 0.1290对应于Oracle考虑一个月内4天(3月和4月)的确切31天。

答案 2 :(得分:2)

我认为你的问题定义不够好,原因如下。

依赖于months_between的答案必须处理以下问题:该功能在2013-02-28和2013-03-31之间以及2013-01-28和2013-02-28之间正好报告一个月,以及在2013-01-31和2013-02-28之间(我怀疑有些回答者在实践中没有使用过这些功能,或者现在需要检查一些生产代码!)

这是记录在案的行为,其中各自月份中的最后日期或同月的日期被判定为相隔几个月的整数。

因此,在将2013-02-28与2013-01-28或2013-01-31进行比较时,您会得到与“1”相同的结果,但将其与2013-01-29或2013-01-30进行比较分别给出0.967741935484和0.935483870968 - 因此当一个日期接近另一个日期时,此函数报告的差异可能会增加。

如果这不是一个可接受的情况,那么你将不得不编写一个更复杂的函数,或者只依赖于假设每月30(例如)天的计算。在后一种情况下,您将如何处理2013-02-28和2013-03-31?

答案 3 :(得分:1)

这就是你的意思吗?

select trunc(months_between(To_date('20120325', 'YYYYMMDD'),to_date('20120101','YYYYMMDD'))) months,
             round(To_date('20120325', 'YYYYMMDD')-add_months(to_date('20120101','YYYYMMDD'),
                           trunc(months_between(To_date('20120325', 'YYYYMMDD'),to_date('20120101','YYYYMMDD'))))) days
        from dual;

答案 4 :(得分:1)

在这里,我只是在今天和表格中的CREATED_DATE DATE字段之间做了区别,这显然是过去的日期:

SELECT  
((FLOOR(ABS(MONTHS_BETWEEN(CREATED_DATE, SYSDATE))) / 12) * 12) || ' months, '  AS MONTHS,
-- we take total days - years(as days) - months(as days) to get remaining days
FLOOR((SYSDATE - CREATED_DATE) -      -- total days
(FLOOR((SYSDATE - CREATED_DATE)/365)*12)*(365/12) -      -- years, as days
-- this is total months - years (as months), to get number of months, 
-- then multiplied by 30.416667 to get months as days (and remove it from total days)
FLOOR(((SYSDATE - CREATED_DATE)/365)*12 - (FLOOR((SYSDATE - CREATED_DATE)/365)*12)) * (365/12))
|| ' days ' AS DAYS 
FROM MyTable

我使用(365/12)或30.416667作为我的转换因子,因为我使用总天数并删除年和月(以天为单位)来获取剩余天数。无论如何,这对我的目的来说已经足够了。

答案 5 :(得分:0)

我发布的解决方案将考虑一个月30天

  select CONCAT (CONCAT (num_months,' MONTHS '), CONCAT ((days-(num_months)*30),' DAYS '))
  from ( 
  SELECT floor(Months_between(To_date('20120325', 'YYYYMMDD'),
   To_date('20120101', 'YYYYMMDD')))
   num_months,
   ( To_date('20120325', 'YYYYMMDD') - To_date('20120101', 'YYYYMMDD') )
   days
  FROM   dual);

答案 6 :(得分:-2)

MsSql语法: DATEDIFF(datepart,startdate,enddate)

Oracle :这将返回天数

    select
  round(Second_date - First_date)  as Diff_InDays,round ((Second_date - First_date) / (30),1)  as Diff_InMonths,round ((Second_date - First_date) * (60*24),2)  as TimeIn_Minitues
from
  (
  select
    to_date('01/01/2012 01:30:00 PM','mm/dd/yyyy hh:mi:ss am') as First_date
   ,to_date('05/02/2012 01:35:00 PM','mm/dd/yyyy HH:MI:SS AM') as Second_date
  from
    dual
  ) result;

演示:http://sqlfiddle.com/#!4/c26e8/36

答案 7 :(得分:-2)

SELECT   (MONTHS_BETWEEN(date2,date1) +  (datediff(day,date2,date1))/30) as num_months,
datediff(day,date2,date1) as diff_in_days  FROM  dual;

// You should replace date2 with TO_DATE('2012/03/25', 'YYYY/MM/DD')
// You should replace date1 with TO_DATE('2012/01/01', 'YYYY/MM/DD')
// To get you results

答案 8 :(得分:-2)

找出Oracle Sql中两年之间的年 - 月 - 日

select 
trunc(trunc(months_between(To_date('20120101', 'YYYYMMDD'),to_date('19910228','YYYYMMDD')))/12) years ,
trunc(months_between(To_date('20120101', 'YYYYMMDD'),to_date('19910228','YYYYMMDD'))) 
-
(trunc(trunc(months_between(To_date('20120101', 'YYYYMMDD'),to_date('19910228','YYYYMMDD')))/12))*12
months,
             round(To_date('20120101', 'YYYYMMDD')-add_months(to_date('19910228','YYYYMMDD'),
                           trunc(months_between(To_date('20120101', 'YYYYMMDD'),to_date('19910228','YYYYMMDD'))))) days
        from dual;

答案 9 :(得分:-4)

请参阅下面的查询(假设@ dt1> = @ dt2);

Declare @dt1 datetime = '2013-7-3'
Declare @dt2 datetime = '2013-5-2'

select abs(DATEDIFF(DD, @dt2, @dt1)) Days,
case when @dt1 >= @dt2
    then case when DAY(@dt2)<=DAY(@dt1)
        then Convert(varchar, DATEDIFF(MONTH, @dt2, @dt1)) + CONVERT(varchar, ' Month(s) ') + Convert(varchar, DAY(@dt1)-DAY(@dt2)) + CONVERT(varchar, 'Day(s).')
        else Convert(varchar, DATEDIFF(MONTH, @dt2, @dt1)-1) + CONVERT(varchar, ' Month(s) ') + convert(varchar, abs(DATEDIFF(DD, @dt1, DateAdd(Month, -1, @dt1))) - (DAY(@dt2)-DAY(@dt1))) + CONVERT(varchar, 'Day(s).')
    end
    else 'See asumption: @dt1 must be >= @dt2'
end In_Months_Days

返回:

Days | In_Months_Days

62   |   2 Month(s) 1Day(s).