SQL Query for midnight transactions - sql

Question

What query do I use to get transactions including midnight (past midnight). For instance:
On 1st January 2018
Our Department Stores opens daily from 7am until 2am(next day)
So 2am transactions as seen as the sale for '1 January 2018'
This is my Query
WHERE Date BETWEEN '2018-01-01' AND '2018-01-02'
and Time >= '1754-01-01 07:00:00.000' and Time <='1754-01-01 01:59:00.000'
Currently my query only takes transactions until 11:59:59
But our store opens past midnight,
I want to get transaction everyday until 2am

Answer 1

Ideally you should combine the Date and Time column inside a single DATETIME column. Having said that, the following should work:
WITH testdata(Date, Time) AS (
SELECT '2018-01-01', '1754-01-01 06:00:00' UNION
SELECT '2018-01-01', '1754-01-01 07:00:00' UNION
SELECT '2018-01-01', '1754-01-01 08:00:00' UNION
SELECT '2018-01-01', '1754-01-01 23:00:00' UNION
SELECT '2018-01-02', '1754-01-01 00:00:00' UNION
SELECT '2018-01-02', '1754-01-01 01:00:00' UNION
SELECT '2018-01-02', '1754-01-01 06:00:00' UNION
SELECT '2018-01-02', '1754-01-01 07:00:00' UNION
SELECT '2018-01-02', '1754-01-01 08:00:00' UNION
SELECT CAST(NULL AS DATE), CAST(NULL AS DATETIME) -- type cast
)
SELECT *
FROM testdata
WHERE Date BETWEEN '2018-01-01' AND '2018-01-02'
AND (
Date = '2018-01-01' AND Time >= '1754-01-01 07:00:00' OR
Date = '2018-01-02' AND Time < '1754-01-01 07:00:00'
)
For completeness, I am including any transaction before 7:00 AM to previous day. Result:
| Date | Time |
|------------|-------------------------|
| 2018-01-01 | 1754-01-01 07:00:00.000 |
| 2018-01-01 | 1754-01-01 08:00:00.000 |
| 2018-01-01 | 1754-01-01 23:00:00.000 |
| 2018-01-02 | 1754-01-01 00:00:00.000 |
| 2018-01-02 | 1754-01-01 01:00:00.000 |
| 2018-01-02 | 1754-01-01 06:00:00.000 |

Answer 2

if date column has time then just use
WHERE Date>='2018-01-01 07:00:00.000' AND Date<='2018-01-02 01:59:00.000'

Answer 3

As your day is offset by 2 hours into the following day, I would just pull your date back by 2 hours.
Here is some example / test code. The select subtracts 2 hours from the input #testdate datetime and then truncates it to the day, chopping off the hours
declare #testdate as datetime = '2018-01-01 10:15'
select cast(dateadd(hour,-2,#testdate) as date)
set #testdate = '2018-01-02 01:15'
select cast(dateadd(hour,-2,#testdate) as date)
Both select statements return
2018-01-01
If you store date and time separately additionally run the following sql to combine them
cast(Date as datetime) + cast(Time as datetime)
e.g
select cast(dateadd(hour,-2,cast(Date as datetime) + cast(Time as datetime)) as date)

Answer 4

If you just want to get transactions for the given day, you could declare the date in a variable and then have the query search for records that are between (#date + 7 hours: 7 am) and (#date + 26 hours: 2 am the next day), which you can do by using the dateadd function. For simplicity, I've declared the startdate and enddate in variables as well.
Declare #date datetime = '2018-01-01';
Declare #startdate datetime = dateadd(hour, 7, #date);
Declare #enddate datetime = dateadd(hour, 26, #date);
select *
from table
WHERE Date BETWEEN #startdate AND #enddate
This will only work if your date column is of datetime datatype, though.
If you're storing your date and time in separate columns, you can combine them. Like so:
Declare #date datetime = '2018-01-01';
Declare #startdate datetime = dateadd(hour, 7, #date);
Declare #enddate datetime = dateadd(hour, 26, #date);
select *
from table
WHERE (cast(date as datetime) + cast(cast(time as time) as datetime))
BETWEEN #startdate AND #enddate