[英]Need output by combining DATEDIFF(hh,StartTime,EndTime) + 'Minutes' in SQL Server
[英]DateDiff to output hours and minutes
我的代碼以小時為單位給出了 TOTAL HOURS,但我正在嘗試 output 之類的
TotalHours
8:36
其中 8 代表小時部分,36 代表分鍾部分,表示一個人一天在辦公室工作的總小時數。
with times as (
SELECT t1.EmplID
, t3.EmplName
, min(t1.RecTime) AS InTime
, max(t2.RecTime) AS [TimeOut]
, t1.RecDate AS [DateVisited]
FROM AtdRecord t1
INNER JOIN
AtdRecord t2
ON t1.EmplID = t2.EmplID
AND t1.RecDate = t2.RecDate
AND t1.RecTime < t2.RecTime
inner join
HrEmployee t3
ON t3.EmplID = t1.EmplID
group by
t1.EmplID
, t3.EmplName
, t1.RecDate
)
SELECT EmplID
, EmplName
, InTime
, [TimeOut]
, [DateVisited]
, DATEDIFF(Hour,InTime, [TimeOut]) TotalHours
from times
Order By EmplID, DateVisited
很簡單:
CONVERT(TIME,Date2 - Date1)
例如:
Declare @Date2 DATETIME = '2016-01-01 10:01:10.022'
Declare @Date1 DATETIME = '2016-01-01 10:00:00.000'
Select CONVERT(TIME,@Date2 - @Date1) as ElapsedTime
產量:
ElapsedTime
----------------
00:01:10.0233333
(1 row(s) affected)
試試這個查詢
select
*,
Days = datediff(dd,0,DateDif),
Hours = datepart(hour,DateDif),
Minutes = datepart(minute,DateDif),
Seconds = datepart(second,DateDif),
MS = datepart(ms,DateDif)
from
(select
DateDif = EndDate-StartDate,
aa.*
from
( -- Test Data
Select
StartDate = convert(datetime,'20090213 02:44:37.923'),
EndDate = convert(datetime,'20090715 13:24:45.837')) aa
) a
輸出
DateDif StartDate EndDate Days Hours Minutes Seconds MS
----------------------- ----------------------- ----------------------- ---- ----- ------- ------- ---
1900-06-02 10:40:07.913 2009-02-13 02:44:37.923 2009-07-15 13:24:45.837 152 10 40 7 913
(1 row(s) affected)
可以做這樣的小改動
SELECT EmplID
, EmplName
, InTime
, [TimeOut]
, [DateVisited]
, CASE WHEN minpart=0
THEN CAST(hourpart as nvarchar(200))+':00'
ELSE CAST((hourpart-1) as nvarchar(200))+':'+ CAST(minpart as nvarchar(200))END as 'total time'
FROM
(
SELECT EmplID, EmplName, InTime, [TimeOut], [DateVisited],
DATEDIFF(Hour,InTime, [TimeOut]) as hourpart,
DATEDIFF(minute,InTime, [TimeOut])%60 as minpart
from times) source
我會將您的最終選擇設為:
SELECT EmplID
, EmplName
, InTime
, [TimeOut]
, [DateVisited]
, CONVERT(varchar(3),DATEDIFF(minute,InTime, TimeOut)/60) + ':' +
RIGHT('0' + CONVERT(varchar(2),DATEDIFF(minute,InTime,TimeOut)%60),2)
as TotalHours
from times
Order By EmplID, DateVisited
任何嘗試使用DATEDIFF(hour,...
的解決方案都會很復雜(如果它是正確的),因為DATEDIFF
計算轉換 - DATEDIFF(hour,...09:59',...10:01')
將返回 1因為從 9 點到 10 點的時間轉換。所以我只是在分鍾上使用DATEDIFF
。
如果涉及秒數,上述內容仍然可能存在細微錯誤(它可能會稍微多計,因為它的計數分鍾轉換)所以如果您需要秒或毫秒精度,您需要調整DATEDIFF
以使用這些單位,然后應用合適的除法常數(根據上一小時)只返回小時和分鍾。
只需更改
DATEDIFF(Hour,InTime, [TimeOut]) TotalHours
部分到
CONCAT((DATEDIFF(Minute,InTime,[TimeOut])/60),':',
(DATEDIFF(Minute,InTime,[TimeOut])%60)) TotalHours
/60 給你小時,%60 給你剩余的分鍾,而 CONCAT 讓你在它們之間加上一個冒號。
我知道這是一個老問題,但我遇到了它,並認為如果其他人遇到它可能會有所幫助。
將 MS 中的Datediff
除以一天中的毫秒數,轉換為Datetime
,然后轉換為時間:
Declare @D1 datetime = '2015-10-21 14:06:22.780', @D2 datetime = '2015-10-21 14:16:16.893'
Select Convert(time,Convert(Datetime, Datediff(ms,@d1, @d2) / 86400000.0))
如果你想要 08:30 ( HH:MM) 格式然后試試這個,
SELECT EmplID
, EmplName
, InTime
, [TimeOut]
, [DateVisited]
, RIGHT('0' + CONVERT(varchar(3),DATEDIFF(minute,InTime, TimeOut)/60),2) + ':' +
RIGHT('0' + CONVERT(varchar(2),DATEDIFF(minute,InTime,TimeOut)%60),2)
as TotalHours from times Order By EmplID, DateVisited
請輸入您的相關值並嘗試以下操作:
declare @x int, @y varchar(200),
@dt1 smalldatetime = '2014-01-21 10:00:00',
@dt2 smalldatetime = getdate()
set @x = datediff (HOUR, @dt1, @dt2)
set @y = @x * 60 - DATEDIFF(minute,@dt1, @dt2)
set @y = cast(@x as varchar(200)) + ':' + @y
Select @y
[hh:mm:ss] 中的兩個時間差
select FORMAT((CONVERT(datetime,'2021-12-01 19:24:40') - CONVERT(datetime,'2021-12-01 17:00:00')),'hh:mm:ss')DffTime
這會幫助你
DECLARE @DATE1 datetime = '2014-01-22 9:07:58.923'
DECLARE @DATE2 datetime = '2014-01-22 10:20:58.923'
SELECT DATEDIFF(HOUR, @DATE1,@DATE2) ,
DATEDIFF(MINUTE, @DATE1,@DATE2) - (DATEDIFF(HOUR,@DATE1,@DATE2)*60)
SELECT CAST(DATEDIFF(HOUR, @DATE1,@DATE2) AS nvarchar(200)) +
':'+ CAST(DATEDIFF(MINUTE, @DATE1,@DATE2) -
(DATEDIFF(HOUR,@DATE1,@DATE2)*60) AS nvarchar(200))
As TotalHours
由於任何 DateTime 都可以轉換為浮點數,並且數字的小數部分表示時間本身:
DECLARE @date DATETIME = GETDATE()
SELECT CAST(CAST(@date AS FLOAT) - FLOOR(CAST(@date AS FLOAT)) AS DATETIME
這將產生一個日期時間,如“一天中的 1900-01-01 小時”,您可以將其轉換為時間、時間戳,甚至使用轉換來獲取格式化的時間。
我想這適用於任何版本的 SQL,因為自 2005 版以來將日期時間轉換為浮點是兼容的。
希望能幫助到你。
如果有人仍在搜索查詢以顯示 hr min 和 sec 格式的差異:(這將以這種格式顯示差異:2 hr 20 min 22 secs)
SELECT
CAST(DATEDIFF(minute, StartDateTime, EndDateTime)/ 60 as nvarchar(20)) + ' hrs ' + CAST(DATEDIFF(second, StartDateTime, EndDateTime)/60 as nvarchar(20)) + ' mins' + CAST(DATEDIFF(second, StartDateTime, EndDateTime)% 60 as nvarchar(20)) + ' secs'
OR 可以是問題中的格式:
CAST(DATEDIFF(minute, StartDateTime, EndDateTime)/ 60 as nvarchar(20)) + ':' + CAST(DATEDIFF(second, StartDateTime, EndDateTime)/60 as nvarchar(20))
共享一個工作時間超過 24 小時的變體。
DECLARE @sd DATETIME = CONVERT(DATETIME, '12/07/2022 11:10:00', 103)
DECLARE @ed DATETIME = CONVERT(DATETIME, '15/07/2022 13:20:05', 103)
Select Concat (
DATEDIFF(DAY, @sd, @ed), 'd ',
DATEPART(Hour, CONVERT(Time,@ed - @sd)), 'h ',
DATEPART(Minute, CONVERT(Time,@ed - @sd)), 'm ',
DATEPART(Second, CONVERT(Time,@ed - @sd)), 's'
)
輸出:
3d 2h 10m 5s
對於像我一樣擁有 MySql 版本 < 5.6 的人,他們沒有 TIMESTAMPDIFF 所以,我寫了 MYTSDIFF 一個 function,它接受 %s (%m 或 %i)分鍾 %h 標志以獲得秒、分和小時之間的差異2個時間戳。
享受
DROP FUNCTION IF EXISTS MYTSDIFF;
DELIMITER $$
CREATE FUNCTION `MYTSDIFF`( date1 timestamp, date2 timestamp, fmt varchar(20))
returns varchar(20) DETERMINISTIC
BEGIN
declare secs smallint(2);
declare mins smallint(2);
declare hours int;
declare total real default 0;
declare str_total varchar(20);
select cast( time_format( timediff(date1, date2), '%s') as signed) into secs;
select cast( time_format( timediff(date1, date2), '%i') as signed) into mins;
select cast( time_format( timediff(date1, date2), '%H') as signed) into hours;
set total = hours * 3600 + mins * 60 + secs;
set fmt = LOWER( fmt);
if fmt = '%m' or fmt = '%i' then
set total = total / 60;
elseif fmt = '%h' then
set total = total / 3600;
else
/* Do nothing, %s is the default: */
set total = total + 0;
end if;
select cast( total as char(20)) into str_total;
return str_total;
END$$
DELIMITER ;
無需跳過箍。 從結束減去開始基本上會給你時間跨度(結合 Vignesh Kumar 和 Carl Nitzsche 的答案):
SELECT *,
--as a time object
TotalHours = CONVERT(time, EndDate - StartDate),
--as a formatted string
TotalHoursText = CONVERT(varchar(20), EndDate - StartDate, 114)
FROM (
--some test values (across days, but OP only cares about the time, not date)
SELECT
StartDate = CONVERT(datetime,'20090213 02:44:37.923'),
EndDate = CONVERT(datetime,'20090715 13:24:45.837')
) t
輸出
StartDate EndDate TotalHours TotalHoursText
----------------------- ----------------------- ---------------- --------------------
2009-02-13 02:44:37.923 2009-07-15 13:24:45.837 10:40:07.9130000 10:40:07:913
在此處查看完整的演員表和轉換選項: https ://msdn.microsoft.com/en-us/library/ms187928.aspx
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