MySQL：获得每个类别的前100条记录

Question

我有一个表称为：consultant_project_score它具有以下字段id,project_id,final_score。我想要生成一个查询，显示每个project_id的前100个final_score。我正在使用MySQL workbench 6.3，我无法使用任何create/insert陈述。有什么办法可以做到这一点？

Answer 1

没有测试，但它应该work.The关键是产生一个行号为每个项目

select id, project_id, final_score 
from 
(select id, project_id, final_score, 
     @rn:= if(@project_id= project_id, @rn+ 1, 1) as rn, 
     @project_id:= project_id as pi 
    from t,(SELECT @rn:=0) b 
    order by project_id, final_score desc) x 
where x.rn<= 100; 

Answer 2

这会为你工作。内部查询将通过project_id生成等级组，然后外部查询将过滤前100个结果。

注意：如果您有类似的project_id和分数组合，它将不起作用。创造这个小提琴得到2个等级。

http://sqlfiddle.com/#!9/9e94d5/1

SELECT id1, 
     project_id, 
     final_score 
FROM 
    (SELECT max(t1.id) AS id1, 
      t1.project_id, 
      t1.final_score, 
      count(*) AS rnk 
    FROM consultant_project_score t1 
    INNER JOIN consultant_project_score t2 ON t1.project_id=t2.project_id 
    WHERE t1.final_score<=t2.final_score 
    GROUP BY t1.project_id, 
      t1.final_score) t 
WHERE rnk<=100 

Answer 3

这一个是没有Group BY并且使用self join概念。这应该给你正确的输出

SQL

SELECT cps2.id, cps2.project_id, cps2.final_score FROM 
        ( SELECT DISTINCT project_id FROM consultant_project_score) cps1 
        JOIN consultant_project_score cps2 
              ON cps2.id IN (SELECT cps3.id FROM consultant_project_score cps3 
          WHERE cps3.project_id = cps1.projectid 
          ORDER BY cps3.final_score DESC LIMIT 100) 
        ORDER BY cps2.project_id ,cps2.final_score DESC