Extracting Day/Hour/Minute - sql

Question

I have the following code which is working, however how do I get the terminal minutes to show the output as Day, Hour, Minute? If this cannot be done, is it possible to add a +1 on the time which would indicate it's the following day?
The problem I am having is that when our orders run past 23.59 PM, the system is not displaying the correct format because of the 24 hour time period.
I am stumped and hope I am not confusing matters.
SELECT FOLIO_NUMBER, TERMINAL_NAME,
format((START_LOAD_TIME - ORDER_ENTRY_TIME), 'HH:mm') AS STAGING_MINUTES,
format((TERM_END_LOAD_TIME - START_LOAD_TIME), 'HH:mm') AS LOADING_MINUTES,
format((TERM_END_LOAD_TIME - ORDER_ENTRY_TIME), 'HH:mm') AS TERMINAL_MINUTES
FROM ORDERS
JOIN TERMINAL_OWNER ON ORDERS.LOADING_TERMINAL_ID = TERMINAL_OWNER.TERMINAL_ID

Answer 1

DECLARE #INT INT
SET #INT = DATEDIFF(SECOND,GETDATE(),GETDATE()+1)
select
convert(varchar(10), (#INT/86400)) + ':' +
convert(varchar(10), ((#INT%86400)/3600)) + ':'+
convert(varchar(10), (((#INT%86400)%3600)/60)) + ':'+
convert(varchar(10), (((#INT%86400)%3600)%60)) as 'DD:HH:MM:SS'
Courtesy of Nat-MS. See here

Answer 2

I think you are looking for something like this
Declare #theMinutes Varchar(10)
Set #theMinutes = '19:25'
declare #totMintute int
Select
#totMintute = (Cast(
Cast(left(#theMinutes,charindex(':',#theMinutes)-1) as Int) * 60
+ Cast(substring(#theMinutes,charindex(',',#theMinutes)+4,len(#theMinutes)) as Int)
as Int ) * 60) / 60
--For 12 hour 1 days
Select #totMintute / 720 as NoDays -- 720 minutes per day
, (#totMintute % 720) / 60 as NoHours -- modulo 720
, (#totMintute % 60) as NoMinutes -- modulo 60
--For 24 hour 1 days
Select #totMintute / 1440 as NoDays -- 1440 minutes per day
, (#totMintute % 1440) / 60 as NoHours -- modulo 1440
, (#totMintute % 60) as NoMinutes -- modulo 60
The output will look like as shown below.
You can convert this query data source table as shown below.
Create table #Temp (MinValue Varchar(8))
insert into #Temp Values ('19:25')
Select TotMinute / 720 as NoDays -- 1440 minutes per day
, (TotMinute % 720) / 60 as NoHours -- modulo 1440
, (TotMinute % 60) as NoMinutes -- modulo 60
from(
select
(Cast(
Cast(left(MinValue,charindex(':',MinValue)-1) as Int) * 60
+ Cast(substring(MinValue,charindex(',',MinValue)+4,len(MinValue)) as Int)
as Int ) * 60) / 60 as TotMinute
from #Temp
)a
You can find the live demo here.

Answer 3

I would say that you should remove the need to identify days/hours in the SQL Output and just get the difference in minutes, which you can then work with in your application layer.
Take this sample code:
create table #orders (
FOLIO_NUMBER int,
START_LOAD_TIME datetime,
ORDER_ENTRY_TIME datetime,
TERM_END_LOAD_TIME datetime
)
insert into #orders (FOLIO_NUMBER,START_LOAD_TIME,ORDER_ENTRY_TIME,TERM_END_LOAD_TIME)
values (1, getdate(),getdate() - 1,getdate() + 1)
select *,
datediff(mi, ORDER_ENTRY_TIME, START_LOAD_TIME) AS STAGING_MINUTES,
datediff(mi, START_LOAD_TIME, TERM_END_LOAD_TIME) AS LOADING_MINUTES,
datediff(mi, ORDER_ENTRY_TIME, TERM_END_LOAD_TIME) AS TERMINAL_MINUTES
from #orders
drop table #orders
This will output the minutes difference between the events:
FOLIO_NUMBER STAGING_MINUTES LOADING_MINUTES TERMINAL_MINUTES
1 1440 1440 2880
You can then perform some simple maths with these values to extract, days, hours and minutes.