日期一次又一次显示，我如何使群组由

Question

使用MS Access 2003

我的查询。

select a.personid, a.cardeventdate as newday, a.intime as intime, 
    max(b.cardeventdate) as prevday, b.outtime as outtime 
from (
    select personid, cardeventdate, min(cardeventtime) as Intime, 
     max(cardeventtime) as outtime 
    from (
     SELECT T_PERSON.cardno, T_PERSON.NAME, T_TITLE.TITLENAME, 
      T_CARDEVENT.personid, T_CARDEVENT.CARDEVENTDATE, 
      T_CARDEVENT.CARDEVENTTIME 
     FROM (
      T_TITLE INNER JOIN T_PERSON 
       ON T_TITLE.TITLECODE = T_PERSON.TITLECODE) 
      INNER JOIN T_CARDEVENT ON T_PERSON.PERSONID = T_CARDEVENT.PERSONID 
     ORDER BY T_PERSON.TITLECODE) 
    group by personid, cardeventdate 
    ) as a INNER JOIN (
    select personid, cardeventdate, min(cardeventtime) as Intime, 
     max(cardeventtime) as outtime 
    from (
     SELECT T_PERSON.cardno, T_PERSON.NAME, 
      T_TITLE.TITLENAME, T_CARDEVENT.personid, 
      T_CARDEVENT.CARDEVENTDATE, T_CARDEVENT.CARDEVENTTIME 
     FROM (T_TITLE INNER JOIN T_PERSON 
      ON T_TITLE.TITLECODE = T_PERSON.TITLECODE) INNER JOIN T_CARDEVENT ON 
      T_PERSON.PERSONID = T_CARDEVENT.PERSONID 
     ORDER BY T_PERSON.TITLECODE) 
    group by personid, cardeventdate) as b ON a.PERSONID = b.PERSONID 
AND a.CARDEVENTDATE > b.CARDEVENTDATE  
GROUP BY a.PERSONID, a.CARDEVENTDATE, a.intime, b.outtime 

表格上面的查询，输出是

personid newday  intime prevday  outtime 

127   20081112 073540 20081111 073024 
127   20081113 073420 20081111 073024 
127   20081113 073420 20081112 122737 
127   20081117 073710 20081111 073024 
127   20081117 073710 20081113 073420 
127   20081117 073710 20081112 122737 
1496  20080819 081935 20080818 110555 
1496  20080903 064211 20080819 081935 
1496  20080903 064211 20080818 110555 
1496  20080904 124129 20080903 064254 
1496  20080904 124129 20080819 081935 
1496  20080904 124129 20080818 110555 
1496  20080908 134345 20080903 064254 
1496  20080908 134345 20080819 081935 
1496  20080908 134345 20080818 110555 
1496  20080908 134345 20080904 130936 
1606  20080831 132538 20080824 102744 
1696  20080825 143758 20080824 182058 
1696  20080831 121407 20080825 153204 
1696  20080831 121407 20080824 182058 
1696  20080901 110704 20080831 121407 
1696  20080901 110704 20080825 153204 
1696  20080901 110704 20080824 182058 
1696  20080902 103342 20080831 121407 

等等...

日期被一次又一次地显示，所以显示的日期只有一次

的预期输出

personid newday intime prevday outtime 

127   20081112 073540 20081111 073024 
127   20081113 073420 20081112 122737 
127   20081117 073710 20081113 073024 

等...

所以从上面的查询我怎么可以做出分组。

查询帮助？

Answer 1

将DISTINCT放入您的第一个选择

例如，

select distinct a.personid, a.cardeventdate as newday,...