MYSQL通过比较值使用一个查询的行数来限制一秒钟的结果

Question

How do you compare the returned results on table1 with the quota on table2 using a subquery. The output should be each row that doesn't meet the quota from table2.

table1
--------------------------
id | master | info | here
--------------------------
0  | a      | 1234 | abcd
1  | b      | 2345 | bcde
2  | a      | 3456 | cdef
3  | c      | 4567 | defg
4  | b      | 5678 | efgh
5  | b      | 6789 | fghi

table2
--------------------------
id | master | quota | info
--------------------------
0  | a      | 2     | abcd
1  | b      | 4     | bcde
2  | c      | 5     | cdef

The output should return table2 rows 1 & 2; row 0 is satisfied because the quota is 2, and there are two rows with master a. Row 1 just missed the quota by 1, therefore it is supposed to be returned along with row 2.

Here is my train of thought:

(mysql influenced psuedo - code, needless to say it doesn't work!) -

SELECT * FROM table2 
WHERE quota > COUNT(
    SELECT id FROM table1 
    WHERE table1.master = table2.master
)

Answer 1

Try using HAVING with a join

SELECT * FROM table2
INNER JOIN table1 USING (master)
GROUP BY master HAVING COUNT(*) < quota

Answer 2

You could INNER JOIN from table2 to a count subquery of table one and return the cases where quota > count

SELECT
    t2.id,
    t2.master,
    c.master_count,
    t2.quota,
    t2.info
FROM table2 AS t2
INNER JOIN (
    SELECT master, COUNT(id) AS master_count
    FROM table1
    GROUP BY master
    ) AS c
    ON t2.master = c.master
WHERE
    c.master_count >= t2.quota