SQL DISTINCT TOP 30

Question

我需要在列＆＃34; productid＆＃34;上设置DISTINCT。来自我的查询：

SELECT TOP 30 s.fid, s.productid, s.partsreplaced, s.labor, p.part_number, p.name, p.img_small, p.discontinued 
FROM frequentrepairs s 
INNER JOIN products p ON s.productid = p.id     
ORDER BY p.part_number ASC

我试过了：

SELECT TOP 30 s.fid, s.productid, s.partsreplaced, s.labor, p.part_number, p.name, p.img_small, p.discontinued 
FROM (SELECT DISTINCT productid FROM frequentrepairs) frequentrepairs s 
INNER JOIN products p ON s.productid = p.id     
ORDER BY p.part_number ASC

很抱歉没有提供数据样本。这是我的表FrequentRepairs的样子：

所以基本上，我不想显示相同的＆＃34; productid＆＃34;两次。

Answer 1

重新开始： 我遇到的问题是我还不了解问题/问题。

假设：

1. Products.ID是主键。
2. FrequentRepairs.ProductID是Products.ID的外键
3. 产品与FrequentRepairs之间的关系是1对多
4. 您要实现的目标的总体目标是大多数维修中涉及的前30个部分的列表？

SELECT TOP 30 
       max(s.fid) as max_fid
     , p.productid
     , max(s.partsreplaced) as partsReplaced --(just picking 1)
       --if it needs to be coorlated to the max_FID then we would have 
       --to dos something different 
     , sum(s.labor) as Sum_labor
     , p.part_number
     , p.name
     , p.img_small
     , p.discontinued 
FROM frequentrepairs s 
INNER JOIN products p 
  ON s.productid = p.id     
GROUP BY p.Part_number, P.Name, p.img_small, p.Discontinued, 
ORDER BY count(p.part_number) DESC, part_number ASC