SQL Server selecting one record for a day -

i have query selects [orderdate] sites in last week. want able select 1 record given site in day. solution follows:

select       sitename      , orderdate        table1         orderdate >= dateadd(dd,(datediff(dd,-53690,getdate()-1)/7)*7,-53690) 

,but when @ results, notice date 21st site2 gives 2 records. need select one.

sitename..................orderdate  site1....................2014-04-21 16:00:37.650  site2....................2014-04-21 16:00:39.697  site2....................2014-04-21 16:00:39.697  site3....................2014-04-21 16:00:35.180     site1....................2014-04-22 16:00:46.113  site2....................2014-04-22 16:00:50.817  site3....................2014-04-22 16:00:53.163    site1....................2014-04-23 16:00:50.993  site2....................2014-04-23 16:00:53.193  site3....................2014-04-23 16:00:55.727 

** editing question **

hi, sorry complicated inserting distinct. included below entire query, included part of simplicity sake:

select distinct         sitename         ,sum(case              when                 orderdate >= dateadd(dd,(datediff(dd,-53690,getdate()-1)/7)*7,-53690)                  1             else                  0              end           ) completed         table1         client    in ('site1','site2','site3','site4','site5','site6','site7')  ,        sitename != '(site8)'  group        sitename  order        sitename  

i select 7 sites avoiding site 8 result

sitename..................completed

site1..........................3

site2..........................4

site3..........................3

site4..........................3

site5..........................3

site6..........................2

sit7...........................2

for site 2 count should 3 instead of 4 despite using distinct query still adding 2 records date

select distinct sitename , orderdate table1   orderdate >= dateadd(dd,(datediff(dd,-53690,getdate()-1)/7)*7,-53690)