Reputation: 47
ApptStart
2005-02-18 10:00:00.000
2005-02-18 13:00:00.000
2005-02-18 11:00:00.000
2005-02-21 09:00:00.000
2005-02-18 15:30:00.000
2005-02-18 14:30:00.000
.
.
.
I have a column in our database that looks like the above. I want to count appointments for the month for a given doc. On Fridays most of them do a half day. So I do not want to count Fridays with appointments only in the morning. If the appointment is in the afternoon, after 12:00:00.000 I want to include that day in the distinct count.
So far I have:
SELECT
ScheduleDays = count(distinct CONVERT(datetime, convert(char(12), a.ApptStart, 1)))
FROM Appointments a
WHERE
ApptKind = 1 AND
--filter on current month
a.ApptStart >= ISNULL(DATEADD(month, DATEDIFF(month, 0, GETDATE()), 0),'1/1/1900') AND
a.ApptStart < ISNULL(DATEADD(month, DATEDIFF(month, 0, GETDATE())+1, 0),'1/1/3000') AND
--Filter on doctor
a.ResourceID in (201)
This worked when I counted every day with appointments on it, but like I said I need to exclude those half days.So I was thinking about only looking at the last right chars of the ApptStart
and comparing it x > noon in a case inside the distinct count...
I tried the following, but it did not work:
ScheduleDays = count(distinct case when (Right(a.ApptStart, 12)) > '12:00:00:000' then 1 else 0 END)
Thanks in advance!
EDIT I tried:
SELECT
ScheduleDays=COUNT(DISTINCT(CAST(datediff(d,0,a.ApptStart) as datetime)))
FROM Appointments a
WHERE
ApptKind = 1 AND
--filter on current month
a.ApptStart >= ISNULL(DATEADD(month, DATEDIFF(month, 0, GETDATE()), 0),'1/1/1900') AND
a.ApptStart < ISNULL(DATEADD(month, DATEDIFF(month, 0, GETDATE())+1, 0),'1/1/3000')AND
--filter all days that aren't Friday, and then give you all Fridays with an hour > 12.
DATENAME(weekday, a.ApptStart) <> 'Friday' OR DATEPART(hour, a.ApptStart) > 12 AND
--Filter on doctor
a.ResourceID in (201)
for 1808 as the count
Upvotes: 1
Views: 472
Reputation: 2032
You probably want to treat dates as dates, rather than strings. You can determine whether a particular timestamp is on a Friday -- or the hour of a timestamp -- using DATEPART
, without having to CONVERT it into a CHAR:
datename(weekday, timestamp_value)
-- returns Friday
datepart(weekday, timestamp_value)
-- returns either 5 or 6, depending on the value of SET DATEFIRST
. (Get day of week in SQL 2005/2008)
datepart(hour, timestamp_value)
-- returns hour part
Using these, you can test whether a timestamp is on Friday at or after noon by checking if datepart(weekday, timestamp_value)
= 6 and datepart(hour, timestamp_value)
>= 12.
bendataclear pointed out that you're using distinct on a case statement which can only ever return 0 or 1, so your total will only ever be 0, 1, or 2. If you're trying to determine which days the doctors worked more than half a day, you'll need to select distinct dates --
SELECT COUNT(DISTINCT(CAST(datediff(d,0,timestamp_value) as datetime)))
FROM table_name
WHERE DATENAME(weekday, timestamp_value) <> 'Friday' OR DATEPART(hour, timestamp_value) > 12
AND the rest of your filters here
The WHERE clause there will give you all days that aren't Friday, and then give you all Fridays that have an hour > 12.
Upvotes: 1