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Mysql 具有max(count())的sql返回零行_Mysql_Sql_Count_Max_Having - Fatal编程技术网

Mysql 具有max(count())的sql返回零行

Mysql 具有max(count())的sql返回零行,mysql,sql,count,max,having,Mysql,Sql,Count,Max,Having,我正试图获得课程表重叠的教室,我的桌子:课程: COURSE_ID NAME 11 matematika 22 logika 33 himiya 44 sport 55 algoritmika 66 hedva 77 algebra linearit 附表: ID COURSE_ID ID_ROOM DAY HOUR 1 11

我正试图获得课程表重叠的教室,我的桌子:课程:

COURSE_ID    NAME
11           matematika
22           logika
33           himiya
44           sport
55           algoritmika
66           hedva
77           algebra linearit
附表:

ID  COURSE_ID  ID_ROOM  DAY  HOUR
1   11         105      Mon  10am
2   11         105      Wen  10am
3   11         105      Thu  10am
4   22         105      Mon  10am
5   22         205      Wen  10am
6   22         105      Thu  10am
7   33         305      Mon  11am
8   33         105      Mon  10am
教室:

ID_ROOM  LOCATION  CAPACITY
105      A         20
205      B         10
305      C         30
我的sql是:

select class_room.ID_ROOM as crid, class_room.LOCATION, schedule.DAY as d, schedule.HOUR as h,  count(courses.COURSE_ID) as count 
  from schedule
  natural join class_room
  natural join courses
  group by crid, d, h
  order by count desc;
我得到:

crid  LOCATION  d   h       count
105   A         Mon 10am    3
105   A         Thu 10am    2
305   C         Mon 11am    1
105   A         Wen 10am    1
205   B         Wen 10am    1
但我只需要显示count的所有最大值(目前为1行)。 我试着

但返回的是空表。
怎么了?或者,可能是另一种解决方案的建议,以获得我需要的?

这将返回计数最高的所有行:

select ID_ROOM as crid, DAY as d, HOUR as h,
      count(*) as cnt
from schedule
group by crid, d, h
having
   count(*)
   = ( select count(*) as cnt
       from schedule
       group by ID_ROOM, DAY, HOUR
       order by cnt desc
       limit 1
     )

现在将其加入
教室
以获取
位置

以下内容将返回与最大计数匹配的所有组


您能给出一个用于测试的示例数据集吗?例如,在SQL FIDLE中。我使用replace
max(count)
添加了所有测试数据,并使用
count(courses.COURSE\u ID)=1
。这将返回count=1的行,可能我的问题不清楚,但我需要显示count的最大值!我的意思是,重叠课程是指计数大于1的课程。你能发布你的预期输出吗,我还不清楚。
select ID_ROOM as crid, DAY as d, HOUR as h,
      count(*) as cnt
from schedule
group by crid, d, h
having
   count(*)
   = ( select count(*) as cnt
       from schedule
       group by ID_ROOM, DAY, HOUR
       order by cnt desc
       limit 1
     )
select  class_room.ID_ROOM as crid
        , class_room.LOCATION
        , schedule.DAY as d
        , schedule.HOUR as h
        ,  count(courses.COURSE_ID) as count 
from    schedule
        natural join class_room
        natural join courses
group by 
        crid, d, h
having count(*) = (
                    select  max(count)
                    from    (            
                              select  count(courses.COURSE_ID) as count
                              from    schedule
                                      natural join class_room
                                      natural join courses
                              group by 
                                      id_room, day, hour
                            ) maxcount
                    )