SQL DISTINCT COUNT IN DATE RANGE

Question

Dear All
Thanks for your attention.
A table is storing the task summary of each order
e.g.
Date                            | Order Number    | Task Number
2015-01-01 00:00:01   | ABC123456        | JOB001
2015-01-01 00:01:02   | ABC123456        | JOB002
2015-01-01 00:10:02   | ABC123444        | JOB001
2015-01-01 10:12:59   | ABC123456        | JOB002   (Since the job002 is not done correctly, need to work again)
2015-01-01 18:20:05   | ABC888888        | JOB001
2015-01-01 20:22:42   | ABC789456        | JOB001
2015-01-01 21:02:11   | BBB121212        | JOB002
I would like to write a single query that get the distinct count of order number in sepcific date range
result as expected in three columns:
2015-01-01
task_number            | AM | PM
JOB001         | 2    | 2
JOB002         | 1    | 1
explain the figures here:
JOB001 AM = 2 (ABC123456, ABC123444)
JOB002 AM = 1  (two records of ABC123456 count as 1 since we would like to know the number of orders)
JOB001 PM = 2  (ABC888888, ABC789456)
JOB002 PM = 1  (BBB121212)
I wrote a query get similar result but the count of order is not distinct
SELECT task_number,
AM=SUM(CASE WHEN date >= ('2015-01-01 00:00:00') AND date < ('2015-01-01 12:00:00') THEN 1 ELSE 0 END),
PM=SUM(CASE WHEN date >= ('2015-01-01 12:00:00') AND date < ('2015-01-02 00:00:00') THEN 1 ELSE 0 END)
FROM MyTable
WHERE task_number = 'JOB001'
Could anyone advise how to enhance this query to let the result become disintct of order number?
Many Thanks,
swivan

Answer 1

Try
select task_number,
count(distinct case when datepart(hour, [date]) >=0 and datepart(hour, [date]) < 12 then [OrderNumbers] END) as AM,
count(distinct case when datepart(hour, [date]) >=12 and datepart(hour, [date]) < 24 then [OrderNumbers] END) as PM FROM myTable
GROUP BY Task_Number
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