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Sat, Mar 14 2009 12:45 PM | Permanent Link |
Uli Becker | Hi,
I have a table like this: Field1: integer Field2: integer MyDate: integer; Since they don't work on all days of the year, I have to find out (as fast as possible) the number of days which has entries in this table. Same thing after that with months. Till now I did it like this: select distinct MyDate from MyTable where MyYear = 2008; and used "recordcount" of this query to find the number of days. Is there a faster and more elegant way to accomplish what I want? Thanks Uli |
Sat, Mar 14 2009 1:42 PM | Permanent Link |
Roy Lambert NLH Associates ![]() | Uli
>I have a table like this: > >Field1: integer >Field2: integer >MyDate: integer; > >Since they don't work on all days of the year, I have to find out (as >fast as possible) the number of days which has entries in this table. >Same thing after that with months. > >Till now I did it like this: > > select distinct MyDate from MyTable where MyYear = 2008; > >and used "recordcount" of this query to find the number of days. > >Is there a faster and more elegant way to accomplish what I want? Apart from I can't see MyYear in the table you could try something like SELECT COUNT(1) AS DaysInYear FROM MyTable GROUP BY MyYear That will give you a count for all years If you want more please give us the full table structure. Roy Lambert [Team Elevate] |
Sat, Mar 14 2009 5:14 PM | Permanent Link |
Ulrich Becker | Roy
> SELECT COUNT(1) AS DaysInYear FROM MyTable GROUP BY MyYear > > That will give you a count for all years I didn't describe it well, sorry. What I want to know is how many different dates have entries in one year. A typical result would be 250 in one year. Uli |
Sat, Mar 14 2009 10:48 PM | Permanent Link |
"James Relyea" | based off your select statement, could something like this work?
select distinct count(MyDate) from MyTable group by MyDate having MyYear = 2008; It should show the # of entries for the year. ![]() jr "Ulrich Becker" <test@test.com> wrote in message news:49BC1E40.7010500@test.com... > Roy > >> SELECT COUNT(1) AS DaysInYear FROM MyTable GROUP BY MyYear >> >> That will give you a count for all years > > I didn't describe it well, sorry. > > What I want to know is how many different dates have entries in one year. > A typical result would be 250 in one year. > > Uli |
Sun, Mar 15 2009 5:40 AM | Permanent Link |
Uli Becker | James,
thank you. > select distinct count(MyDate) from MyTable group by MyDate having MyYear = > 2008; Actually your query shows how many entries per day have been made in one year. What I want is just the number of days in one year, where entries have been made. Your query results in 256 records e.g.. That's the number I need for each year. Uli |
Sun, Mar 15 2009 10:21 AM | Permanent Link |
Roy Lambert NLH Associates ![]() | Ulrich
In that case you need someone a lot better at sql than me. Anything I can figure out is a two stage process much like the one you're currently using. Roy Lambert |
Sun, Mar 15 2009 12:54 PM | Permanent Link |
Uli Becker | Roy,
> In that case you need someone a lot better at sql than me. Anything I can figure out is a two stage process much like the one you're currently using. Yes, I think so. BTW: I know you are quite good at SQL ![]() Uli Becker |
Sun, Mar 15 2009 7:52 PM | Permanent Link |
"John Hay" | Uli
> Field1: integer > Field2: integer > MyDate: integer; > > Since they don't work on all days of the year, I have to find out (as > fast as possible) the number of days which has entries in this table. > Same thing after that with months. > > Till now I did it like this: > > select distinct MyDate from MyTable where MyYear = 2008; > > and used "recordcount" of this query to find the number of days. > > Is there a faster and more elegant way to accomplish what I want? > I don't know about faster but what about SELECT COUNT(*) FROM (SELECT DISTINCT MyDate FROM MyTable WHERE MyYear=2008) t1 If you want totals for individual months (and MyDate is a date field) you could use something like SELECT EXTRACT(YEAR FROM MyDate) AS Yearnum, EXTRACT(MONTH FROM MyDate) AS Monthnum, COUNT(*) FROM (SELECT DISTINCT MyDate FROM MyTable WHERE MyDate between :StartDate and :EndDate) t1 GROUP BY Yearnum,Monthnum ORDER BY Yearnum,Monthnum John |
Sun, Mar 15 2009 9:25 PM | Permanent Link |
"James Relyea" | Try changing your SQL line from:
>select distinct MyDate from MyTable where MyYear = 2008; to 'select distinct count(MyDate) from MyTable where MyYear = 2008; ' I deal with this type of thing all the time but never with ElevateDB. I'm assuming it can be done in 1 statement here too. Could you post a few MyDate values so I have something to play around with? ![]() jr "Uli Becker" <test@test.com> wrote in message news:49BCCCDD.5000303@test.com... > James, > > thank you. > >> select distinct count(MyDate) from MyTable group by MyDate having MyYear >> = 2008; > > Actually your query shows how many entries per day have been made in one > year. > What I want is just the number of days in one year, where entries have > been made. Your query results in 256 records e.g.. That's the number I > need for each year. > > Uli |
Mon, Mar 16 2009 4:03 AM | Permanent Link |
Uli Becker | John,
> I don't know about faster but what about > > SELECT COUNT(*) FROM > (SELECT DISTINCT MyDate FROM MyTable WHERE MyYear=2008) t1 It's not exactly what I want (select some values from MyTable and the number of day with eintries in one year in ONE statement), but actually your solution with a derived table is very fast! It helped. Thank you!. Uli |
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