Mysql 如何选择最后记录sql的第n行？

我遇到了这样一种情况，我必须显示前9行的域名电子邮件和电子邮件的出现次数，第10行作为剩余域的总和，并显示为其他域

我所做的一切

我已经使用下面的查询成功地获取了不同的域及其出现的情况

SELECT (SUBSTRING_INDEX(SUBSTR(email, INSTR(email, '@') + 1),'.',1)) as domain,
       COUNT(*) as C
FROM newsletter_recipient
where LENGTH(email) > 0
GROUP BY (SUBSTRING_INDEX(SUBSTR(email, INSTR(email, '@') + 1),'.',1))
ORDER BY COUNT(*) DESC

当前结果

domain           C
------------------------    
gmail           12  
dddd            2   
mmmmm           2   
dsf             2   
aaaa            1   
bbbb            1   
ccc             1   
yopmail         1   
yahoo           1   
dde             1   
rfg             1   
eedd            1   
dfdg            1   
sad             1   
dfdf            1   
sfd             1   
web             1   

预期结果

domain           C
------------------------    
gmail           12  
dddd            2   
mmmmm           2   
dsf             2   
aaaa            1   
bbbb            1   
ccc             1   
yopmail         1   
yahoo           1
others          8

希望有人能帮忙

表达时使用一个case-

---

使用两个聚合级别：

SELECT (CASE WHEN seqnum < 10 THEN domain ELSE 'Others' END) as domain,
       SUM(c)
FROM (SELECT SUBSTRING_INDEX(SUBSTR(email, INSTR(email, '@') + 1), '.', 1) as domain,
             COUNT(*) as C,
             ROW_NUMBER() OVER (ORDER BY COUNT(*) DESC) as seqnum
      FROM newsletter_recipient
      WHERE LENGTH(email) > 0
      GROUP BY domain
     ) d
GROUP BY (CASE WHEN seqnum < 10 THEN domain ELSE 'Others' END)
ORDER BY SUM(c) DESC;

---

你可以试试下面的方法

WITH cte AS (
     SELECT (SUBSTRING_INDEX(SUBSTR(email, INSTR(email, '@') + 1),'.',1)) as domain,
     COUNT(*) as C
     FROM newsletter_recipient
     WHERE LENGTH(email) > 0
     GROUP BY (SUBSTRING_INDEX(SUBSTR(email, INSTR(email, '@') + 1),'.',1))
     ORDER BY COUNT(*) DESC
)
SELECT 
     IF(rn < 10, domain, 'others') AS domain_name,
     SUM(c) AS count
FROM (
     SELECT domain, c, ROW_NUMBER() OVER () AS rn FROM cte
) T
GROUP BY domain_name

---

我会先通过10+个工会来解决这个问题：

SELECT * from (

SELECT 
    (SUBSTRING_INDEX(SUBSTR(email, INSTR(email, '@') + 1),
            '.',
            1)) AS domain,
    COUNT(*) AS C
FROM
     newsletter_recipient
WHERE
    LENGTH(email) > 0
GROUP BY (SUBSTRING_INDEX(SUBSTR(email, INSTR(email, '@') + 1),
        '.',
        1))
ORDER BY COUNT(*) DESC
Limit 10) as FIRST_10

UNION 

SELECT 'others', sum(ALL_OTHERS.C) from (

SELECT 
    COUNT(*) AS C
FROM
    newsletter_recipient
WHERE
    LENGTH(email) > 0
GROUP BY (SUBSTRING_INDEX(SUBSTR(email, INSTR(email, '@') + 1),
        '.',
        1))
ORDER BY COUNT(*) DESC
Limit 11,18446744073709551615
) as ALL_OTHERS

---

18446744073709551615:足够大，可以跳过前10个并确保检索到所有其他邮件，请参阅：

ohh..我不能在not in中硬编码邮件客户端，因为它们是动态的：-

SELECT * from (

SELECT 
    (SUBSTRING_INDEX(SUBSTR(email, INSTR(email, '@') + 1),
            '.',
            1)) AS domain,
    COUNT(*) AS C
FROM
     newsletter_recipient
WHERE
    LENGTH(email) > 0
GROUP BY (SUBSTRING_INDEX(SUBSTR(email, INSTR(email, '@') + 1),
        '.',
        1))
ORDER BY COUNT(*) DESC
Limit 10) as FIRST_10

UNION 

SELECT 'others', sum(ALL_OTHERS.C) from (

SELECT 
    COUNT(*) AS C
FROM
    newsletter_recipient
WHERE
    LENGTH(email) > 0
GROUP BY (SUBSTRING_INDEX(SUBSTR(email, INSTR(email, '@') + 1),
        '.',
        1))
ORDER BY COUNT(*) DESC
Limit 11,18446744073709551615
) as ALL_OTHERS