Question

我的目标是获得一张能够计算游戏正确答案的表格。

例如我想要这个

| G.Name | E.Action 
| game 1 | correctAnswer 
| game 1 | correctAnswer 
| game 2 | correctAnswer 
| game 3 | correctAnswer 
| game 3 | correctAnswer

成为这个

| G.Name | Count(*) 
| game 1 | 2 
| game 2 | 1 
| game 3 | 2

问题是我得到了这个：

| G.Name | Count(*)
| game 1 | 5 
| game 2 | 5 
| game 3 | 5

其中5是2 + 1 + 2的总和

这是我的查询

  SELECT G.Name, Count(*)
    FROM enduser EU
         INNER JOIN prescription P 
                    ON EU.UserRefID = P.EndUserRefID
         INNER JOIN prescriptiongame PG 
                    ON PG.PrescriptionRefID = P.PrescriptionID
         INNER JOIN games G 
                    ON G.GameID = PG.GameRefID
         INNER JOIN session S 
                    ON S.PrescriptionRefID = P.PrescriptionID
         INNER JOIN entries E 
                    ON E.SessionGameRefID = S.Session
   WHERE P.EndUserRefID = 889
         AND E.Action =  'correctAnswer'
GROUP BY G.Name
ORDER BY G.Name

我怎么能解决这个问题，当我用谷歌搜索人们使用相同的方法，但他们得到了良好的结果

提前致谢

Answer 1

我敢打赌，你的一个联盟正在夸大结果。将查询删除至关键表：prescription，session和entries。你还得到过多的结果吗？如果没有，请尝试一次添加一个连接，直到您计算过多为止。

Answer 2

您的联接以某种方式乘以行数。当您沿着彼此不相关的不同维度连接表时会发生这种情况。这是一个常见问题，但我不了解您的数据结构，因此以下是最佳猜测。

您可以使用count(distinct)修复您的示例。所以试试这个：

SELECT G.Name, Count(distinct SessionGameRefID)

session和game之间是否存在关系？

我想知道你的查询的简化是否会解决问题（它会删除前两个表）：

  SELECT G.Name, Count(*)
    FROM prescriptiongame PG 
         INNER JOIN games G 
                    ON G.GameID = PG.GameRefID
         INNER JOIN session S 
                    ON S.PrescriptionRefID = PG.PrescriptionRefID
         INNER JOIN entries E 
                    ON E.SessionGameRefID = S.Session
   WHERE PG.EndUserRefID = 889
         AND E.Action =  'correctAnswer'
  GROUP BY G.Name
  ORDER BY G.Name;

Answer 3

假设需要所有连接，请尝试以下查询：

  SELECT G.Name, Count(G.Name)
    FROM enduser EU
         INNER JOIN prescription P 
                    ON EU.UserRefID = P.EndUserRefID
         INNER JOIN prescriptiongame PG 
                    ON PG.PrescriptionRefID = P.PrescriptionID
         INNER JOIN games G 
                    ON G.GameID = PG.GameRefID
         INNER JOIN session S 
                    ON S.PrescriptionRefID = P.PrescriptionID
         INNER JOIN entries E 
                    ON E.SessionGameRefID = S.Session
   WHERE P.EndUserRefID = 889
         AND E.Action =  'correctAnswer'
GROUP BY G.Name
ORDER BY G.Name

Answer 4

perl -v

This is perl 5, version 16, subversion 3 (v5.16.3) built for x86_64-linux
file `which perl`
/sv/app/perx/third-party/bin/perl: ELF 64-bit LSB executable, x86-64, version 1 (SYSV), dynamically linked (uses shared libs), for GNU/Linux 2.6.18, not stripped

 2.198696207 - 2.134326286: 0.0643699210000004
 2.198696207 - 2.134326286: 0.047901175

按查询分组的SQL查询计数不按预期工作

4 个答案: