SQL Server How to combine 3 queries into one? -

i want condense following 3 queries 1 query , out put totals 3 columns. oh , how make don't have declare date. want "know" current date, month, , year.

declare @mydate datetime set @mydate = '2015-01-1' select sum(amount) 'day total'  [accounting].[dbo].[handpay]  accountingdate>=@mydate , accountingdate<dateadd(day,1,@mydate)  select sum(amount) 'month total'  [accounting].[dbo].[handpay]  accountingdate>=@mydate , accountingdate<dateadd(month,1,@mydate)  select sum(amount) 'day total'  [accounting].[dbo].[handpay]  accountingdate>=@mydate , accountingdate<dateadd(year,1,@mydate) 

what best way this? thanks!

thanks super fast responses! solved.

if i'm understanding problem correctly, ought work:

declare @today date = convert(date, getdate()); select     [day total]   = sum(case when [accountingdate] >= @today , [accountingdate] < dateadd(day,   1, @today) [amount] else 0 end),     [month total] = sum(case when [accountingdate] >= @today , [accountingdate] < dateadd(month, 1, @today) [amount] else 0 end),     [year total]  = sum(case when [accountingdate] >= @today , [accountingdate] < dateadd(year,  1, @today) [amount] else 0 end)     [accounting].[dbo].[handpay]; 

note [month total] , [year total] don't give sums of entries occur within current month/year, rather sum of entries occur within month/a year of today's date. i'm not sure if that's want, seems consistent original queries.

update: suggested d stanley below, can simplify bit since know date ranges compose [day total] , [month total] sums enclosed entirely within date range composes [year total] sum. here's might like:

declare @today date = convert(date, getdate()); select     [day total]   = sum(case when [accountingdate] < dateadd(day,   1, @today) [amount] else 0 end),     [month total] = sum(case when [accountingdate] < dateadd(month, 1, @today) [amount] else 0 end),     [year total]  = sum([amount])     [accounting].[dbo].[handpay]     [accountingdate] >= @today , [accountingdate] < dateadd(year, 1, @today);