其他函数，例如FIND_IN_SET，它返回逗号分隔列表中可用的字符串数

Question

As per official documentation, FIND_IN_SET returns the position of the first occurrence of string available from a comma-separated list. But I want to get the count of available string in a comma-separated list.

For example :

SELECT id, name, FIND_IN_SET(id, '1,1,2,3,4,4,5,6,7,7,7') as num FROM users;

It returns the following result

+----+---------------+
| id | name    | num |
+----+---------------+
|  1 | Jack    |   1 |
|  2 | Alex    |   3 |
|  3 | John    |   4 |
|  4 | Brett   |   5 |
|  5 | Keith   |   7 |
|  6 | Richard |   8 |
|  7 | Kent    |   9 |
+----+---------------+

SQL Fiddle: http://sqlfiddle.com/#!9/9eeacb/2

But I want the count of ids which matches the id of the table. So, the expected output should be like below.

SELECT id, name, SOME_FUNCTION(id, '1,1,2,3,4,4,5,6,7,7,7') as num FROM users;

    +----+---------------+
    | id | name    | num |
    +----+---------------+
    |  1 | Jack    |   2 |
    |  2 | Alex    |   1 |
    |  3 | John    |   1 |
    |  4 | Brett   |   2 |
    |  5 | Keith   |   1 |
    |  6 | Richard |   1 |
    |  7 | Kent    |   3 |
    +----+---------------+

Answer 1

The best answer is to move away from the CSV paradigm, and instead get the set of id s to match into its own table:

CREATE TABLE ids (id int);
INSERT INTO ids (id)
VALUES (1),(1),(2),(3),(4),(4),(5),(6),(7),(7),(7);

Then use the following simple query:

SELECT
    u.id,
    u.name,
    COUNT(i.id) AS num
FROM users u
LEFT JOIN ids i
    ON u.id = i.id
GROUP BY
    u.id,
    u.name;

Answer 2

Use string functions tricks:

set @s = '1,1,2,3,4,4,5,6,7,7,7';
select id, name,
  ceiling((length(@s) - length(trim(both ',' from replace(
    replace(concat(',', replace(@s, ',', ',,'), ','), concat(',', id, ',' ), ''),
    ',,', ','
  )))) / (length(id) + 1)) as num
from users

See the demo (with other possible cases).
Results:

id | name    | num
 1 | Jack    | 2
 2 | Alex    | 1
 3 | John    | 1
 4 | Brett   | 2
 5 | Keith   | 1
 6 | Richard | 1
 7 | Kent    | 3

Answer 3

You are passing in the comparison values as a string. You can do some string tricks to count the number of values within the string and then add that up:

SELECT u.id, u.name,
       COUNT(*),
       SUM( (LENGTH(REPLACE(CONCAT(',', ids, ','), u.id, CONCAT(u.id, 'X'))) -
             LENGTH(CONCAT(',', ids, ','))
            ) 
          )
FROM users u CROSS JOIN
     (SELECT '1,1,2,3,4,4,5,6,7,7,7' as ids) x
WHERE FIND_IN_SET(u.id, x.ids) > 0
GROUP BY u.id, u.name;

Note: I created the subquery x as a convenience, so I can refer to ids . You can pass in the value as a parameter instead:

SELECT u.id, u.name,
       COUNT(*),
       SUM( (LENGTH(REPLACE(CONCAT(',', ?, ','), u.id, CONCAT(u.id, 'X'))) -
             LENGTH(CONCAT(',', ?, ','))
            ) 
          )
FROM users u 
WHERE FIND_IN_SET(u.id, ?) > 0
GROUP BY u.id, u.name;