Question

鉴于以下数据集，我如何找到那些具有“已接受”决定的大多数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')
;

小提琴：http://sqlfiddle.com/#!9/03bcf2/1

最初，我正在使用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
;

然而，我想到我可以有多个电子邮件地址，这些地址引用了不同的“最接受”的决定（即，可以说是一个平局），那些将被过滤掉（是正确的措辞吗？） LIMIT关键字。

然后，我尝试了对上述查询的变体，将ORDER BY和LIMIT行替换为：

HAVING MAX(CountDecision)

但我意识到这只是半个陈述：MAX(CountDecision)需要与某些东西进行比较。我只是不知道是什么。

任何指针都会非常感激。谢谢！

注意：这是用于家庭作业。

更新：要明确，我正在尝试从Email中找到EmailReferences的值和数量。但是，我只希望行FinalDecision.Decision = 'Accepted'（匹配ApplicantID s）。根据我的数据，结果应该：

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

Answer 1

基本上你需要做两件事......首先，你需要找到maxCount是什么，然后找到最大数量的记录。

现在，您可以在单个嵌套查询中组合这两个步骤，或将结果存储在变量中并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仍然缺乏窗口功能，但是当版本8准备就绪时，这变得更容易了。所以对于fuure参考，或者像Mariadb这样已经具有窗口函数的数据库：

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

email          | CountDecision
:------------- | ------------:
ref10@test.org |             2

dbfiddle here

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;

选择按具有最大聚合的列分组的行

3 个答案: