Question

我只需要选择一行中计数最多的一行。我怎么做？

这是我目前的代码：

 select firstname, lastname, count(*) as total
 from trans 
 join work 
 on trans.workid = work.workid
 join artist
 on work.artistid = artist.artistid
 where datesold is not null
 group by firstname, lastname;

当前示例：

FIRSTNAME |  LASTNAME |  TOTAL
------------------------------
Tom       |   Cruise  |   3
Angelina  |   Jolie   |   9
Britney   |   Spears  |   5
Ellie     |  Goulding |   4

我需要它才能选择这个：

FIRSTNAME  |  LASTNAME  |  TOTAL
--------------------------------
Angelina   |  Jolie     |   9

Answer 1

您可以添加order by total desc和fetch first 1 row only（仅限Oracle 12c r1 ，否则您应该将结果用作临时表并使用select来使用rownum = 1子句中的where限制，以防您total对于不同的组不能相同。另一种方法是添加此having子句，以便列出所有人员total的最大值：

having count(*) = (select max(total) from (select count(*) as total from <your_query>) tmp)

或那：

having count(*) = (select count(*) as total from <your_query> order by total desc fetch first 1 row only)

Answer 2

在Oracle 12中，您可以这样做：

 select firstname, lastname, count(*) as total
 from trans join
      work 
      on trans.workid = work.workid join
      artist
      on work.artistid = artist.artistid
 where datesold is not null
 group by firstname, lastname
 order by count(*) desc
 fetch first 1 row only;

在旧版本中，您可以使用子查询执行此操作：

select twa.*
from (select firstname, lastname, count(*) as total
      from trans join
           work 
           on trans.workid = work.workid join
           artist
           on work.artistid = artist.artistid
      where datesold is not null
      group by firstname, lastname
      order by count(*) desc
     ) twa
where rownum = 1;

Answer 3

这将适用于sql server 2012 ..

＆＃13;

with CTECount (firstname, lastname,total)
as
(
	select firstname, lastname, count(1) as total
	 from trans 
	 join work 
	 on trans.workid = work.workid
	 join artist
	 on work.artistid = artist.artistid
	 where datesold is not null
	 group by firstname, lastname
 )

 select top(1) with ties from CTECount 
 order by total desc

＆＃13;

由于

Answer 4

你可以这样做，非常简单：

选择 TOP 1 firstname，lastname，count（*）作为trans的总数加入trans.workid = work.workid的工作在work.artistid = artist.artistid上加入艺术家其中dateold不为null 按名字，姓氏分组 按总DESC排序;

只选择一个在sql中具有最高计数的行

4 个答案: