选择按汇总最大的列分组的行

Question

给定以下数据集，我如何找到作为大多数具有“已接受”决策的ApplicationID的引用的电子邮件地址？

CREATE TABLE IF NOT EXISTS `EmailReferences` (
  `ApplicationID` INT NOT NULL,
  `Email` VARCHAR(45) NOT NULL,
  PRIMARY KEY (`ApplicationID`, `Email`)
);
INSERT INTO EmailReferences (ApplicationID, Email)
VALUES
(1, 'ref10@test.org'), (1, 'ref11@test.org'), (1, 'ref12@test.org'),
(2, 'ref20@test.org'), (2, 'ref21@test.org'), (2, 'ref22@test.org'),
(3, 'ref11@test.org'), (3, 'ref31@test.org'), (3, 'ref32@test.org'),
(4, 'ref40@test.org'), (4, 'ref41@test.org'), (4, 'ref42@test.org'),
(5, 'ref50@test.org'), (5, 'ref51@test.org'), (5, 'ref52@test.org'),
(6, 'ref60@test.org'), (6, 'ref11@test.org'), (6, 'ref62@test.org'),
(7, 'ref70@test.org'), (7, 'ref71@test.org'), (7, 'ref72@test.org'),
(8, 'ref10@test.org'), (8, 'ref81@test.org'), (8, 'ref82@test.org')
;

CREATE TABLE IF NOT EXISTS `FinalDecision` (
  `ApplicationID` INT NOT NULL,
  `Decision` ENUM('Accepted', 'Denied') NOT NULL,
  PRIMARY KEY (`ApplicationID`)
);
INSERT INTO FinalDecision (ApplicationID, Decision)
VALUES
(1, 'Accepted'), (2, 'Denied'),
(3, 'Accepted'), (4, 'Denied'),
(5, 'Denied'),   (6, 'Denied'),
(7, 'Denied'),   (8, 'Accepted')
;

相同的小提琴： http ://sqlfiddle.com/#!9/03bcf2/1

最初，我使用LIMIT 1和ORDER BY CountDecision DESC ，如下所示：

SELECT  er.email, COUNT(fd.Decision) AS CountDecision
FROM    EmailReferences AS er
JOIN    FinalDecision AS fd ON er.ApplicationID = fd.ApplicationID
WHERE   fd.Decision = 'Accepted'
GROUP   BY er.email
ORDER   BY CountDecision DESC
LIMIT   1
;

但是，我想到我可以有多个电子邮件地址引用不同的“最受接受”决策（例如，可以说是平局），并且可以使用LIMIT关键字将其过滤掉（是正确的措词吗？）。 。

然后，我尝试对上述查询进行变体，将ORDER BY和LIMIT行替换为：

HAVING MAX(CountDecision)

但是我意识到那只是陈述的一半：需要将MAX(CountDecision)与某种事物进行比较。 我就是不知道

任何指针将不胜感激。 谢谢！

注意：这是用于家庭作业。

更新：明确地说，我正在尝试从EmailReferences查找Email的值和计数。 但是，我只希望具有FinalDecision.Decision = 'Accepted' （在匹配的ApplicantID ）。 根据我的数据，结果应为：

Email          | CountDecision
---------------+--------------
ref10@test.org | 2
ref11@test.org | 2

Answer 1

基本上，您需要做两件事...首先，您需要找到什么是maxCount，然后找到具有最大计数的记录。

现在，您可以将这两个步骤合并到一个嵌套查询中，或者将结果存储在变量中，然后在第二个查询中使用它。 我个人试图避免内部查询，因为它们会导致性能问题并且读取起来更复杂，因此我在这里使用变量选项：

-- Find out what max count is and store it in a variable
SELECT  @maxcount := COUNT(fd.Decision) AS CountDecision
FROM    EmailReferences AS er
JOIN    FinalDecision AS fd ON er.ApplicationID = fd.ApplicationID
WHERE   fd.Decision = 'Accepted'
GROUP   BY er.email
ORDER BY CountDecision desc
Limit 1;

-- get emails with @maxcount
SELECT  er.Email, COUNT(fd.Decision) AS CountDecision
FROM    EmailReferences AS er
JOIN    FinalDecision AS fd ON er.ApplicationID = fd.ApplicationID
WHERE   fd.Decision = 'Accepted'
GROUP   BY er.email
HAVING  COUNT(fd.Decision) = @maxcount;

Answer 2

MySQL仍然缺少窗口功能，但是当版本8可以投入生产时，这将变得更加容易。 因此，对于将来的参考，或者对于那些已经具有窗口功能的数据库（例如Mariadb）：

 CREATE TABLE IF NOT EXISTS `EmailReferences` ( `ApplicationID` INT NOT NULL, `Email` VARCHAR(45) NOT NULL, PRIMARY KEY (`ApplicationID`, `Email`) ); 

 INSERT INTO EmailReferences (ApplicationID, Email) VALUES (1, 'ref10@test.org'), (1, 'ref11@test.org'), (1, 'ref12@test.org'), (2, 'ref20@test.org'), (2, 'ref21@test.org'), (2, 'ref22@test.org'), (3, 'ref30@test.org'), (3, 'ref31@test.org'), (3, 'ref32@test.org'), (4, 'ref40@test.org'), (4, 'ref41@test.org'), (4, 'ref42@test.org'), (5, 'ref50@test.org'), (5, 'ref51@test.org'), (5, 'ref52@test.org'), (6, 'ref60@test.org'), (6, 'ref11@test.org'), (6, 'ref62@test.org'), (7, 'ref70@test.org'), (7, 'ref71@test.org'), (7, 'ref72@test.org'), (8, 'ref10@test.org'), (8, 'ref81@test.org'), (8, 'ref82@test.org') ; 

 CREATE TABLE IF NOT EXISTS `FinalDecision` ( `ApplicationID` INT NOT NULL, `Decision` ENUM('Accepted', 'Denied') NOT NULL, PRIMARY KEY (`ApplicationID`) ); 

 INSERT INTO FinalDecision (ApplicationID, Decision) VALUES (1, 'Accepted'), (2, 'Denied'), (3, 'Accepted'), (4, 'Denied'), (5, 'Denied'), (6, 'Denied'), (7, 'Denied'), (8, 'Accepted') ; 

 select email, CountDecision from ( SELECT er.email, COUNT(fd.Decision) AS CountDecision , max(COUNT(fd.Decision)) over() maxCountDecision FROM EmailReferences AS er JOIN FinalDecision AS fd ON er.ApplicationID = fd.ApplicationID WHERE fd.Decision = 'Accepted' GROUP BY er.email ) d where CountDecision = maxCountDecision 

\n 电邮|  计数决定\n ：------------- |  ------------：\n ref10@test.org |  2\n

dbfiddle 在这里

Answer 3

例如...

SELECT a.*
  FROM 
     ( SELECT x.email
            , COUNT(*) total
         FROM emailreferences x
         JOIN finaldecision y
           ON y.applicationid = x.applicationid
        WHERE y.decision = 'accepted'
        GROUP
           BY x.email
     ) a
  JOIN
     ( SELECT COUNT(*) total
         FROM emailreferences x
         JOIN finaldecision y
           ON y.applicationid = x.applicationid
        WHERE y.decision = 'accepted'
        GROUP
           BY x.email
        ORDER 
           BY total DESC 
        LIMIT 1
     ) b
    ON b.total = a.total;