在SQL中获取两个日期之间的差异（以月为单位）和以天为单位

小编典典

在SQL中获取两个日期之间的差异（以月为单位）和以天为单位

sql

我需要得到两个日期之间的差额，例如，如果差额是84天，我可能应该将结果输出为2个月和14天，我刚才的代码给出了总计。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;

输出为：

NUM_MONTHS    DIFF_IN_DAYS
2.774193548       84

例如，我需要将此查询的输出在最坏的情况下设为2个月和14天，否则我不介意是否可以在月份数字之后得到确切的天数，因为那几天实际上不是14天，因为所有月份都没有30天。

阅读 191

2021-03-17

共1个答案

小编典典

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 |

链接测试：SQLFiddle

2021-03-17