Select all fields with group by clause in Ms. Access -

i have table contain checktime, userid, , checktype. checktime type datetime. want these data without duplicate checktime date. example data this:

userid | checktime           | checktype 1      | 2016-05-01 06:58:00 | 1      | 2016-05-01 06:59:00 | 1      | 2016-05-02 06:58:00 | 1      | 2016-05-02 06:58:10 | 1      | 2016-05-02 06:59:00 | 

does know how remove duplicate date? so, above data, can result this:

userid | checktime           | checktype 1      | 2016-05-01 06:58:00 | 1      | 2016-05-02 06:58:00 | 

i tried use distinct, doesn't works, because have different time each checktime. , use group clause, can give me result 1 field.

my query right this:

select userid, checktype, checktime, day(checktime) checkinout userid = 259  , month(checktime)='5'  , year(checktime)='2016' , checktype = 'i'  group day(checktime) order day(checktime) asc 

and cames error aggregate function.

do need time?

select distinct  userid, format([checktime],"yyyy/mm/dd") checkdate, checktype checkinout; 

results in 2 records without time

userid | checktime  | checktype 1      | 2016-05-01 | 1      | 2016-05-02 | 

or looking first time each unique day? it

edit: added order date clause

select distinct userid, checktype,      format([checktime],"yyyy/mm/dd") checkdate,      min(checkinout.checktime) firstchecktime checkinout (month([checktime])=5) , (year([checktime])=2016) group userid, checktype, format([checktime],"yyyy/mm/dd") having ((userid=1) , (checktype="i")) order format([checktime],"yyyy/mm/dd"); 

looks this:

userid  checktype   checkdate   firstchecktime 1               2016/05/01    2016-05-01 06:58:00 1               2016/05/02    2016-05-02 06:58:00