Question

我有一个会计表，从这张表我监控用户的活动。我有以下3个查询：第一

select count(distinct username),
       DATE_FORMAT(StartTime,'%H %d/%m/%y')
from acct
where date(starttime) = date '2014-10-02'
group by DATE_FORMAT(StartTime,'%H %d/%m/%y')

第二

select count(distinct username),
       DATE_FORMAT(UpdateTime,'%H %d/%m/%y')
from acct
where date(updatetime) = date '2014-10-02'
group by DATE_FORMAT(UpdateTime,'%H %d/%m/%y')

第三

select count(distinct username),
       DATE_FORMAT(StopTime,'%H %d/%m/%y')
from acct
where date(stoptime) = date '2014-10-02'
group by DATE_FORMAT(StopTime,'%H %d/%m/%y')

我有点困惑，有可能让它们成为一个查询吗？我想检查当天每小时有多少用户在线。你能指导我在哪里解决我的问题吗？感谢

字段starttime，updatetime和stoptime是datetime。

Answer 1

如果您在所有选择查询中获取相同数量的字段，则可以使用UNION。但在这种情况下，您可能只想使用OR：

select COUNT(DISTINCT username) AS count_distinct_users,
       CASE WHEN date(StartTime) = date '2014-10-02' 
            THEN DATE_FORMAT(StartTime,'%H %d/%m/%y')
            WHEN date(UpdateTime) = date '2014-10-02' 
            THEN DATE_FORMAT(UpdateTime,'%H %d/%m/%y')
            WHEN date(StopTime) = date '2014-10-02' 
            THEN DATE_FORMAT(StopTime,'%H %d/%m/%y')
       END AS date
from acct
where date(StartTime) = date '2014-10-02' 
or date(UpdateTime) = date '2014-10-02'
or date(StopTime) = date '2014-10-02'
group by date

<强> sqlFiddle demo

Answer 2

使用UNION：

尝试此操作

select count(distinct username),
       DATE_FORMAT(StartTime,'%H %d/%m/%y')
from acct
where date(starttime) = date '2014-10-02'
group by DATE_FORMAT(StartTime,'%H %d/%m/%y')
UNION
select count(distinct username),
       DATE_FORMAT(UpdateTime,'%H %d/%m/%y')
from acct
where date(updatetime) = date '2014-10-02'
group by DATE_FORMAT(UpdateTime,'%H %d/%m/%y')
UNION
select count(distinct username),
       DATE_FORMAT(StopTime,'%H %d/%m/%y')
from acct
where date(stoptime) = date '2014-10-02'
group by DATE_FORMAT(StopTime,'%H %d/%m/%y')

如何在3中组合3个查询

2 个答案: