Sql 在其他数据库管理系统中。这是一个很好的解决方案。如果你喜欢，请检查答案。如果你喜欢，请检查答案 event_date id ---------- --- 2015-11-18 x1 2015-11-18 x2 2015

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在其他数据库管理系统中。这是一个很好的解决方案。如果你喜欢，请检查答案。如果你喜欢，请检查答案

event_date    id     
----------    ---  
2015-11-18    x1
2015-11-18    x2
2015-11-18    x3
2015-11-18    x4
2015-11-18    x5
2015-11-19    x1
2015-11-19    x2
2015-11-19    y1
2015-11-19    y2
2015-11-19    y3
2015-11-20    x1
2015-11-20    y1
2015-11-20    z1
2015-11-20    z2

event_date    count(id)
-----------   ---------
2015-11-18      5
2015-11-19      3
2015-11-20      2

SELECT   event_date, COUNT(DISTINCT id)
FROM     mytable
GROUP BY event_date

SELECT
    mytable.event_date
    event_date_count.id,
    event_date_count.event_date_count
FROM
    mytable
    INNER JOIN
        (
        SELECT
            id,
            event_date,
            COUNT(event_date) as event_date_count
        FROM
            mytable
        GROUP BY
            id,
           event_date
        ) event_date_count
    ON event_date_count.event_date = mytable.event_date

select t.event_date,
       count(1)
from ( 
      -- Record first occurrence of each id along with the earliest date occurred 
      select id,
              min(event_date) as event_date
       from
       mytable
       group by id
      ) t
group by t.event_date;

SELECT   EVENT_DATE,COUNT (DISTINCT ID)
FROM     MYTABLE 
WHERE NOT EXISTS 
      (SELECT * FROM MYTABLE T2 WHERE     
       T2.EVENT_DATE<MYTABLE.EVENT_DATE AND T2.ID=MYTABLE.ID)
GROUP BY EVENT_DATE

select 
a.event_date,
count(a.id) cnt_id
from
table_name a
left outer join 
(
    select x.id, min(x.event_date) min_event_date from table_name x
) b on 
    a.id = b.id AND 
    a.event_date = b.min_event_date
GROUP BY
a.event_date