我有一个表格持续时间：
-------------------------------
| id | from | to |
|----|------------|-----------|
| 1 | 2011-02-22 | NULL |
| 2 | 1999-08-03 |2005-03-30 |
| 3 | 1982-09-03 |1988-01-30 |
| 4 | 1965-12-01 |1980-05-02 |
-------------------------------
我想计算一行中的时差（来自）并将它们加起来。我的疑问是：
SELECT CONCAT(SUM(TIMESTAMPDIFF( YEAR, from, IF(to IS NULL,'2017-10-
18', to))),' Years ',
SUM(TIMESTAMPDIFF( MONTH, from, IF(to IS NULL,'2017-10-18', to))
% 12),' Months ',
SUM(FLOOR( TIMESTAMPDIFF( DAY, from, IF(to IS NULL,'2017-10-18',
to)) % 30.4375 )),' Days ') AS Duration
FROM duration;
我得到了：
------------------------------
| Duration |
|----------------------------|
| 30 Years 23 Months 78 Days |
------------------------------
但我想得到：
-----------------------------
| Duration |
|---------------------------|
| 32 Years 1 Months 18 Days |
-----------------------------
请帮忙！
答案 0 :(得分：1)
我试图在下面提供一系列查询，以显示确定持续时间/月/日的准确方法（尽管这些计算结果 32年1个月22天（可能是因为＆＃34;现在＆＃34; ）的变化而且我还没有添加最终的连接，这是很难实现的。所使用的方法每行准确，但确实依赖在最终计算月份和日期的每月近似天数（30.4375 *）。
TIMESTAMPDIFF(YEAR , from_dt, to_dt )给出每行的年数TIMESTAMPDIFF(MONTH, from_dt + INTERVAL TIMESTAMPDIFF(YEAR , from_dt, to_dt) YEAR , to_dt )将年数添加到开始日期，然后获取该日期与结束日期之间的差异TIMESTAMPDIFF(DAY , from_dt + INTERVAL TIMESTAMPDIFF(MONTH, from_dt, to_dt) MONTH , to_dt )现在将总月数添加到开始日期，然后计算剩余的ays到达结束日期nb：我使用了多个＆＃34;层＆＃34;在下面的查询中，它们更容易阅读（我希望）。
MySQL 5.6架构设置：
CREATE TABLE duration
(
from_dt DATE, to_dt DATE
);
INSERT INTO duration
VALUES ('2011-02-22', NULL )
, ('1999-08-03', '2005-03-30' )
, ('1982-09-03', '1988-01-30' )
, ('1965-12-01', '1980-05-02' )
;
查询1 ：
## adjusted sums( method using timestampdiff )
SELECT
CASE WHEN SUM(d.months) > 12 THEN SUM(d.years) + FLOOR(SUM(d.years/12)) ELSE SUM(d.years) END
as years
, CASE WHEN SUM(d.days) > 30 THEN FLOOR(((SUM(d.months) + FLOOR(SUM(d.days/30.4375)))/12)) ELSE SUM(d.months) END
as months
, CASE WHEN SUM(d.days) > 30 THEN FLOOR(MOD(SUM(d.days), 30)) ELSE SUM(d.months) END
as days
, current_date
FROM (
SELECT
TIMESTAMPDIFF(YEAR , from_dt, to_dt )
as years
, TIMESTAMPDIFF(MONTH, from_dt + INTERVAL TIMESTAMPDIFF(YEAR , from_dt, to_dt) YEAR , to_dt )
as months
, TIMESTAMPDIFF(DAY , from_dt + INTERVAL TIMESTAMPDIFF(MONTH, from_dt, to_dt) MONTH , to_dt )
as days
FROM (
SELECT
from_dt
, COALESCE(to_dt,CURRENT_DATE) to_dt
FROM duration
) d2
) d
<强> Results ：
| years | months | days | current_date |
|-------|--------|------|--------------|
| 32 | 2 | 22 | 2017-10-19 |
查询2 ：
## simple sums( method using timestampdiff )
SELECT
sum(TIMESTAMPDIFF(YEAR , from_dt, to_dt ))
years
, sum(TIMESTAMPDIFF(MONTH, from_dt + INTERVAL TIMESTAMPDIFF(YEAR , from_dt, to_dt) YEAR , to_dt ))
months
, sum(TIMESTAMPDIFF(DAY , from_dt + INTERVAL TIMESTAMPDIFF(MONTH, from_dt, to_dt) MONTH , to_dt ))
days
, current_date
FROM (
SELECT
from_dt
, COALESCE(to_dt,CURRENT_DATE) to_dt
FROM duration
) d
<强> Results ：
| years | months | days | current_date |
|-------|--------|------|--------------|
| 30 | 23 | 82 | 2017-10-19 |
查询3 ：
## method using timestampdiff
SELECT
d.*
, TIMESTAMPDIFF(YEAR , from_dt, to_dt ) AS diff_yr
, TIMESTAMPDIFF(MONTH, from_dt + INTERVAL TIMESTAMPDIFF(YEAR , from_dt, to_dt) YEAR , to_dt ) AS diff_mn
, TIMESTAMPDIFF(DAY , from_dt + INTERVAL TIMESTAMPDIFF(MONTH, from_dt, to_dt) MONTH , to_dt ) AS diff_dy
FROM (
SELECT
from_dt
, COALESCE(to_dt,CURRENT_DATE) to_dt
, current_date
FROM duration
) d
<强> Results ：
| from_dt | to_dt | current_date | diff_yr | diff_mn | diff_dy |
|------------|------------|--------------|---------|---------|---------|
| 2011-02-22 | 2017-10-19 | 2017-10-19 | 6 | 7 | 27 |
| 1999-08-03 | 2005-03-30 | 2017-10-19 | 5 | 7 | 27 |
| 1982-09-03 | 1988-01-30 | 2017-10-19 | 5 | 4 | 27 |
| 1965-12-01 | 1980-05-02 | 2017-10-19 | 14 | 5 | 1 |
查询4 ：
## testing (method using timestampdiff)
## is able to reproduce to_dt ?
select
from_dt
, to_dt
, from_dt + INTERVAL diff_yr YEAR
+ INTERVAL diff_mn MONTH
+ INTERVAL diff_dy DAY
recalculated_to_dt
, current_date
FROM (
SELECT
d.*
, TIMESTAMPDIFF(YEAR , from_dt, to_dt ) AS diff_yr
, TIMESTAMPDIFF(MONTH, from_dt + INTERVAL TIMESTAMPDIFF(YEAR , from_dt, to_dt) YEAR , to_dt ) AS diff_mn
, TIMESTAMPDIFF(DAY , from_dt + INTERVAL TIMESTAMPDIFF(MONTH, from_dt, to_dt) MONTH , to_dt ) AS diff_dy
FROM (
SELECT
from_dt
, COALESCE(to_dt,CURRENT_DATE) to_dt
FROM duration
) d
) d2
<强> Results ：
| from_dt | to_dt | recalculated_to_dt | current_date |
|------------|------------|--------------------|--------------|
| 2011-02-22 | 2017-10-19 | 2017-10-19 | 2017-10-19 |
| 1999-08-03 | 2005-03-30 | 2005-03-30 | 2017-10-19 |
| 1982-09-03 | 1988-01-30 | 1988-01-30 | 2017-10-19 |
| 1965-12-01 | 1980-05-02 | 1980-05-02 | 2017-10-19 |
答案 1 :(得分：0)
我chaane table nane和两个naness nanes：
find . -type f | git check-ignore --stdin
看来这就是你想要的......
答案 2 :(得分：0)
SELECT CONCAT(y, ' Years ', m, ' Months ', d, ' Days') as Duration
FROM(
SELECT
IF(SUM(Month) > 12, SUM(d.Year) + FLOOR(SUM(d.Year/12)), SUM(d.Year)) as y,
IF(SUM(d.DAY) > 30, FLOOR(MOD(((SUM(d.Month) + FLOOR(SUM(d.Day/30.4375)))), 12)), SUM(d.Month)) as m,
IF(SUM(d.DAY) > 30, FLOOR(MOD(SUM(d.Day), 30)), SUM(d.Month)) as d
FROM (
SELECT
TIMESTAMPDIFF( YEAR, a, b ) Year,
TIMESTAMPDIFF( MONTH, a, b ) % 12 Month,
FLOOR( TIMESTAMPDIFF( DAY, a, b ) % 30.4375 ) Day
from dates
) as d
) as c
答案 3 :(得分：0)
正确的SQL是
选择 求和（d.months）> 12 THEN SUM（d.years）+地板（Sum（d.months / 12））ELSE Sum（d.years）的情况 AS年 ，当SUM（d.days）> 30 THEN FLOOR（MOD（（SUM（d.months）+ FLOOR（SUM（d.days / 30.4375））），12））ELSE FLOOR（MOD（SUM（d.months） ），12））结束 AS个月 ，当SUM（d.days）> 30 THEN FLOOR（MOD（SUM（d.days），30））ELSE FLOOR（MOD（SUM（d.days），30））结束时 AS天
从（
选择
TIMESTAMPDIFF（YEAR，from_date，to_date）
AS年
，TIMESTAMPDIFF（MONTH，from_date + INTERVAL TIMESTAMPDIFF（YEAR，from_date，to_date）YEAR，to_date）
AS个月
，TIMESTAMPDIFF（DAY，from_date + INTERVAL TIMESTAMPDIFF（MONTH，from_date，to_date）MONTH，to_date）
AS天
来自（
选择
从日期
，COALESCE（to_date，CURRENT_DATE）到_date
来自user_id ='1'的经验
d2
）d