Question

set search_path = 'project';

select g.firstname,g.lastname, g.rating, sh.showname
from guest g, shows sh, timeslot ts
where sh.shownumber = ts.shownumber
and ts.guestnumber = g.guestnumber 
order by sh.showname

这是结果

"Charlie ";"Sheen";7.2;"Cooking Show"
"Charlie";"Sheen";7.2;"Cooking Show"
"Carl";"Makky";7.0;"Cooking Show"
"Barack";"Obama";6.7;"Fitness Mania"
"Vladimir";"Putin";8.2;"Fitness Mania"
"kim";"jung";5.3;"Fitness Mania"
"kim";"jung";5.3;"Gamers Heaven"
"Justin";"Trudeau";8.5;"Kids Play Time"
"Charlie";"Sheen";7.2;"Kids Play Time"
"ellen";"page";9.2;"Weather News"

我想要的是评分最高的嘉宾PER SHOW，所以它应该有5条记录，每条可用的节目中有一条最高评价的客人记录

Answer 1

这是row_number()函数的一个很好的应用：

select firstname, lastname, rating, showname
from (select g.firstname, g.lastname, g.rating, sh.showname,
             row_number() over (partition by sh.showname order by g.rating desc) as seqnum
      from guest g shows sh join
           timeslot ts
           on sh.shownumber = ts.shownumber join
           guest g
           on ts.guestnumber = g.guestnumber  
     ) gr
where seqnum = 1
order by showname;

我还修改了代码以使用显式join语法。

如果您希望每个节目有五位嘉宾，请使用seqnum <= 5。

Answer 2

如果每场演出只需要一位嘉宾 - 您可以使用DISTINCT ON：

select distinct on (sh.showname) g.firstname, g.lastname, g.rating, sh.showname
from guest g, shows sh, timeslot ts
where sh.shownumber = ts.shownumber
and ts.guestnumber = g.guestnumber 
order by sh.showname, g.rating desc

最大和不同的SQL查询

2 个答案: