DateDiff 到 output 小时和分钟

Question

my code gives TOTAL HOURS in hours, but i am trying to output something like

TotalHours 
  8:36

where 8 represents hour part and 36 represents minutes part mean totalHours a person has worked in a single day at office.

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 

Answer 1

Very simply:

CONVERT(TIME,Date2 - Date1)

For example:

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

Yelds:

ElapsedTime
----------------
00:01:10.0233333

(1 row(s) affected)

Answer 2

Try this query

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

Output

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)

Answer 3

Small change like this can be done

  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

Answer 4

I would make your final select as:

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 

Any solution trying to use DATEDIFF(hour,... is bound to be complicated (if it's correct) because DATEDIFF counts transitions - DATEDIFF(hour,...09:59',...10:01') will return 1 because of the transition of the hour from 9 to 10. So I'm just using DATEDIFF on minutes.

The above can still be subtly wrong if seconds are involved (it can slightly overcount because its counting minute transitions) so if you need second or millisecond accuracy you need to adjust the DATEDIFF to use those units and then apply suitable division constants (as per the hours one above) to just return hours and minutes.

Answer 5

Just change the

DATEDIFF(Hour,InTime, [TimeOut]) TotalHours

part to

CONCAT((DATEDIFF(Minute,InTime,[TimeOut])/60),':',
       (DATEDIFF(Minute,InTime,[TimeOut])%60)) TotalHours 

The /60 gives you hours, the %60 gives you the remaining minutes, and CONCAT lets you put a colon between them.

I know it's an old question, but I came across it and thought it might help if someone else comes across it.

Answer 6

Divide the Datediff in MS by the number of ms in a day, cast to Datetime , and then to time:

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))

Answer 7

If you want 08:30 ( HH:MM) format then try this,

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

Answer 8

Please put your related value and try this :

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

Answer 9

[hh:mm:ss] 中的两个时间差

select FORMAT((CONVERT(datetime,'2021-12-01 19:24:40') - CONVERT(datetime,'2021-12-01 17:00:00')),'hh:mm:ss')DffTime

Answer 10

this would hep you

 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 

Answer 11

Since any DateTime can be cast to a float, and the decimal part of the number represent the time itself:

DECLARE @date DATETIME = GETDATE()

SELECT CAST(CAST(@date AS FLOAT) - FLOOR(CAST(@date AS FLOAT)) AS DATETIME

This will result a datetime like '1900-01-01 hour of the day' you can cast it as time, timestamp or even use convert to get the formatted time.

I guess this works in any version of SQL since cast a datetime to float is compatible since version 2005.

Hope it helps.

Answer 12

In case someone is still searching for a query to display the difference in hr min and sec format: (This will display the difference in this format: 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 can be in the format as in the question:

CAST(DATEDIFF(minute, StartDateTime, EndDateTime)/ 60 as nvarchar(20)) + ':' + CAST(DATEDIFF(second, StartDateTime, EndDateTime)/60 as nvarchar(20))

Answer 13

Sharing a variant that works for more than 24 hours.

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'
    )

Output:

3d 2h 10m 5s

Answer 14

For people that has MySql version < 5.6 as me they don't have TIMESTAMPDIFF so, I wrote MYTSDIFF a function that accepts %s (%m or %i) for minutes %h flags to get the difference in seconds, minutes and hours between 2 timestamps.

Enjoy

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 ;

Answer 15

No need to jump through hoops. Subtracting Start from End essentially gives you the timespan (combining Vignesh Kumar's and Carl Nitzsche's answers) :

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

Ouput

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

See the full cast and convert options here: https://msdn.microsoft.com/en-us/library/ms187928.aspx