Postgresql select 最小日期值

Question

这些是我的 table：

table: 门票

ticketid：序列号
用户 ID：整数
日期票：日期
timeticket: 时间

table：用户

用户ID：序列号
用户名：varchar
密码：varchar

这些是我的数据：

    userid  username  password
    1        user1         123
    2        user2         123

ticketid  userid  dateticket timeticket
    1        1    2015-05-27 14:47:14
    2        1    2015-05-27 14:47:15
    3        1    2015-05-27 14:47:16
    4        2    2015-05-27 14:47:17
    5        2    2015-05-27 14:47:18
    6        2    2015-05-12 14:47:19

我想显示分配给具有用户 ID 和最小日期的用户的工单数量。

我当前的查询是：

select 
    min(t.dateticket) as mydate,
    (select min(ti.timeticket) as mytime from tickets ti where ti.userid=t.userid),
    t.userid,
    count(t.ticketid) as my_all
from
    tickets t
group by 
    t.userid
order by 
    my_all asc, 
    mydate asc, 
    mytime asc;

The current result is:

        mydate     mytime     userid  my_all
      2015-05-27   14:47:14     1        3
      2015-05-12   14:47:17     2        3


The expected result is:

        mydate     mytime     userid  my_all
      2015-05-12   14:47:19     1        3
      2015-05-27   14:47:14     2        3

Answer 1

我认为这会满足您的要求：

select distinct on (userid) dateticket, timeticket, userid,
       count(*) over (partition by userid) as my_all
from tickets t
order by userid, dateticket, timeticket;

distinct onreturns每组第一条记录，基于order by。诀窍是使用 window 函数获取所有门票的数量。

编辑：

要进行额外排序，请使用子查询：

select t.*
from (select distinct on (userid) dateticket, timeticket, userid,
             count(*) over (partition by userid) as my_all
      from tickets t
      order by userid, dateticket, timeticket
     ) t
order by my_all, dateticket, timeticket.

Postgresql select 最小日期值

Postgresql select min Date value

sql

postgresql

dml

postgresql-9.3