Comments on: SQL: Virtual Grouping?

Question: SQL: Virtual Grouping?

odinsdream — Tue, 13 Dec 2005 11:42:16 -0800

SQL: Virtual Grouping?

I'd like to get a count of transactions organized by the hour, so that I can determine which time period in the day is the busiest. I can get the count for a single hour with this:

Select Count(Distinct TimeStamp) From Transactions Where TimeStamp Like '00:%'

Our timestamps go from 00:00:00 to 23:60:60.

So, what I want to do is consolidate a long list of these queries into something more efficient and versatile, if possible a single query:

Select Count(Distinct TimeStamp) From Transactions Where TimeStamp Like '00:%'
Select Count(Distinct TimeStamp) From Transactions Where TimeStamp Like '01:%'
...
Select Count(Distinct TimeStamp) From Transactions Where TimeStamp Like '22:%'
Select Count(Distinct TimeStamp) From Transactions Where TimeStamp Like '23:%'

What are some good ways to do this? Many thanks!

By: odinsdream

odinsdream — Tue, 13 Dec 2005 11:42:54 -0800

Er, the above should be Count(Distinct RecordID) in each line, sorry about that.

By: ldenneau

ldenneau — Tue, 13 Dec 2005 11:53:16 -0800



select 

  count(left(timestamp, 2)), left(timestamp, 2) 

from transactions 

group by left(timestamp, 2)

By: ltdan

ltdan — Tue, 13 Dec 2005 13:03:50 -0800

select datepart(hh,timestamp), count(distinct recordid)
from transactions
group by datepart(hh)

or better yet, count(1) rather than count(distinct recordid) if you're just counting the number of records.

By: ltdan

ltdan — Tue, 13 Dec 2005 13:04:39 -0800

whoops...messed up the 'group by'. should be 'group by datepart(hh,timestamp).

By: SNACKeR

SNACKeR — Tue, 13 Dec 2005 13:32:07 -0800

select datepart(hh, timestamp) as Hour, count(*) as Count
from transactions
group by datepart(hh, timestamp)
order by datepart(hh, timestamp)

By: odinsdream

odinsdream — Tue, 13 Dec 2005 13:40:57 -0800

itdan, there could be several transactions with the same RecordID, which should be treated as one "count" in the total. At least, that's what I was trying to accomidate with Count(Distinct RecordID).

Thanks everyone, I'll try these out in the morning.

By: blue mustard

blue mustard — Tue, 13 Dec 2005 14:10:10 -0800

It looks like the timestamp column isn't a date datatype, so I don't think datepart() will work. ldenneau's answer looks right, except for counting the duplicate records... your count(distinct RecordID) should fix it....

select count(distinct recordid), left(ts, 2)

from transactions

group by left(ts, 2)

By: blue mustard

blue mustard — Tue, 13 Dec 2005 14:18:55 -0800

Oops, looks like I'm wrong about the datepart(). It's perfectly happy with expressions like '00:00:00'.

By: andrew cooke

andrew cooke — Tue, 13 Dec 2005 14:29:15 -0800

it probably does a cast.