如何获取 mysql 中 IN 子句中不存在的数据

Question

我在 'email' table

上尝试了以下查询

SELECT email_id,count(*) as count FROM email 
where  email_id in ('a@a.com','b@b.com','c@c.com') group by email_id

结果

email_id   count

a@a.com     3

c@c.com     2

但我想要如下序列的结果

email_id     count

a@a.com       3

b@b.com       0

c@c.com       2

供您参考 b@b.com 未出现在电子邮件 table

中

Answer 1

您可以使用：

SELECT e1.email_id, COUNT(e2.email_id) AS count 
FROM (SELECT 'a@a.com' AS email_id 
      UNION 
      SELECT 'b@b.com'
      UNION
      SELECT 'c@c.com'
 ) AS e1
LEFT JOIN email AS e2 ON e1.email_id = e2.email_id
GROUP BY e1.email_id

Demo here

Answer 2

架构

create table users
(   userId int auto_increment primary key,
    userName varchar(100) not null,
    emailAddr varchar(200) not null
);

create table emails
(   id int auto_increment primary key,
    userId int not null
-- etc
);

insert users(userName,emailAddr) values ('Sally','sally10@gmail.com'),('Henry','h@h.com'),('JT','abcde@xxx.org');

insert emails(userId) values (1),(1),(1),(3),(3);

查询

select u.userId,u.emailAddr,count(e.id) as theCount
from users u
left join emails e
on e.userId=u.userId
group by u.userId,u.emailAddr
+--------+-------------------+----------+
| userId | emailAddr         | theCount |
+--------+-------------------+----------+
|      1 | sally10@gmail.com |        3 |
|      2 | h@h.com           |        0 |
|      3 | abcde@xxx.org     |        2 |
+--------+-------------------+----------+

Left join 捡起丢失的电子邮件，可以这么说。

Answer 3

大概是这样的：

SELECT t.email_id,count(e.email_id) as count 
FROM 
(
select 'a@a.com' as email_id union all
select 'b@b.com' union all
select 'c@c.com'
) t
left join email e on t.email_id = e.email_id
group by t.email_id

Answer 4

试试这个：

SELECT 'a@a.com' as email_id,count(1)  as `count` FROM email where email_id='a@a.com'
union
SELECT 'b@b.com' as email_id,count(1) as `count`  FROM email where email_id='b@b.com'
union
SELECT 'c@c.com' as email_id,count(1) as `count`  FROM email where email_id='c@c.com'

如何获取 mysql 中 IN 子句中不存在的数据

How to get data which is not present in IN clause in mysql

mysql

group-by

count

in-clause

架构

查询