SQL 多个 table JOINS、GROUP BY 和 HAVING

Question

我有一个 table 结构有点类似于此：

CREATE TABLE `user`
 (`id` int, `name` varchar(7));
CREATE TABLE `email`
  (`id` int, `email_address` varchar(50), `verified_flag` tinyint(1),`user_id` int);
CREATE TABLE `social`
 (`id` int,`user_id` int);

INSERT INTO `user`
 (`id`, `name`)
VALUES
 (1,'alex'),
 (2,'jon'),
 (3,'arya'),
 (4,'sansa'),
 (5,'hodor')
;
INSERT INTO `email`
     (`id`,`email_address`,`verified_flag`,`user_id`)
VALUES
 (1,'alex@gmail.com','1',1),
 (2,'jon@gmail.com','0',1),
 (3,'arya@gmail.com','0',3),
 (4,'sansa@gmail.com','1',4),
 (5,'reek@gmail.com','0',3),
 (6,'hodor@gmail.com','0',5),
 (7,'tyrion@gmail.com','0',1)
;
INSERT INTO `social`
     (`id`,`user_id`)
VALUES
 (1,4),
 (2,4),
 (3,5),
 (4,4),
 (5,4)
;

我要获取的是所有邮件：

未验证
属于没有（即 0）验证电子邮件的用户
属于没有（即 0）社交记录的用户

通过以下查询，我可以应用第一个和第三个条件，但不能应用第二个条件：

SELECT *
FROM `email`
INNER JOIN `user` ON `user`.`id` = `email`.`user_id`
LEFT JOIN `social` ON `user`.`id` = `social`.`user_id`
WHERE `email`.`verified_flag` = 0
GROUP BY `email`.`user_id`,`email`.`email_address`
HAVING COUNT(`social`.`id`) = 0

我怎样才能达到这个结果？还有sqlfiddle

Answer 1

您可以使用以下查询：

SELECT e.`id`, e.`email_address`, e.`verified_flag`, e.`user_id`
FROM (
   SELECT `id`,`email_address`,`verified_flag`,`user_id`
   FROM `email`
   WHERE `verified_flag` = 0) AS e
INNER JOIN (
   SELECT `id`, `name`
   FROM  `user` AS t1
   WHERE NOT EXISTS (SELECT 1
                     FROM  `email` AS t2
                     WHERE `verified_flag` = 1 AND t1.`id` = t2.`user_id`)

         AND 

         NOT EXISTS (SELECT 1
                     FROM  `social` AS t3
                     WHERE t1.`id` = t3.`user_id`)
) AS u ON u.`id` = e.`user_id`;

此查询使用两个派生表：

e 实现第一个条件，即 returns 所有未验证的电子邮件
u 实现第二个和第三个条件，即它 returns 一组没有经过验证的电子邮件的所有用户和没有社交记录。

在 e 和 u 之间执行 INNER JOIN returns 所有满足条件号的电子邮件。 1 属于满足条件的用户2 和 3.

Demo here

您也可以使用此查询：

SELECT *
FROM `email`
WHERE `user_id` IN (
   SELECT `email`.`user_id`
   FROM `email`
   INNER JOIN `user` ON `user`.`id` = `email`.`user_id`
   LEFT JOIN `social` ON `user`.`id` = `social`.`user_id`
   GROUP BY `email`.`user_id`
   HAVING COUNT(`social`.`id`) = 0 AND 
          COUNT(CASE WHEN `email`.`verified_flag` = 1 THEN 1 END) = 0 )

使用子查询是为了select全部user_id满足条件号。 2 和 3. 条件编号1 是多余的，因为如果用户没有经过验证的电子邮件，那么经过验证的电子邮件就不可能与该用户相关。

Demo here

Answer 2

有趣且棘手的一个。

我看到你那里发生了一些事情。但是当你的表变大时，有和子查询就变成了一个非常的坏主意。

方法见下文。不要忘记设置索引！

SELECT * from email
LEFT JOIN social on email.user_id = social.user_id

 -- tricky ... i'm going back to email table to pick verified emails PER user
LEFT JOIN email email2 on email2.user_id = email.user_id AND email2.verified_flag = 1
WHERE
     -- you got this one going already :)
    email.verified_flag = 0

     -- user does not have any social record
    AND social.id is null

     -- email2 comes in handy here ... we limit resultset to include only users that DOES NOT have a verified email
    AND email2.id is null
ORDER BY email.user_id asc;

Answer 3

简单地运行一个联合查询：

SELECT `user_id`, `email_address`, `verified_flag`, 'No Email' as `Type` 
FROM `email` RIGHT JOIN `user` ON `user`.`id` = `email`.`user_id` 
WHERE `email`.`user_id` IS NULL

UNION

SELECT `user_id`, `email_address`, `verified_flag`, 'Not Verified' as `Type` 
FROM `email` INNER JOIN `user` ON `user`.`id` = `email`.`user_id` 
WHERE `email`.`verified_flag` = 0

UNION

SELECT `user_id`, `email_address`, `verified_flag`, 'No Social' as `Type` 
FROM `email` INNER JOIN `user` ON `user`.`id` = `email`.`user_id`
LEFT JOIN `social` ON `user`.`id` = `social`.`user_id`  
GROUP BY `user_id`, `email_address`, `verified_flag` 
HAVING COUNT(IFNULL(`social`.`id`, 0)) = 0;

Answer 4

    SELECT
      u.id AS u_id
    , u.name AS u_name
    , e.email_address AS e_email
    , e.verified_flag AS e_verify
    , e.user_id AS e_uid
    , s.id AS s_id
    , s.user_id AS u_id
    , COALESCE(ver_e.ver_email_count,0) as ver_email_count
FROM
    email as e
LEFT OUTER JOIN
    user as u
        ON u.id = e.user_id
LEFT OUTER JOIN
    social AS s
        ON u.id = s.user_id
LEFT OUTER JOIN
    (
        SELECT
            COUNT(email_address) AS ver_email_count
            , user_id 
        FROM
            email
    ) AS ver_e
        ON u.id = ver_e.user_id
GROUP BY
    e.user_id
HAVING e.verified_flag = 0
AND
ver_email_count = 0
AND
ISNULL(s.id)

使用一个派生的 table 来获取每个用户已获得的已验证电子邮件地址的数量

SQL 多个 table JOINS、GROUP BY 和 HAVING

SQL Multiple table JOINS, GROUP BY and HAVING

mysql

sql

join

group-by

having