mysql从一个表中选择count行，其中列与另一个表中的值类似

Question

我将尝试用一个例子来更好地解释标题

表1示例

Id  text
1   lorem ipsum doe
2   foo bar lorem ipsum jhon
3   bla bla ipsum tommy

表2示例

Id  fullname  name  surname  keyword
1   jhon doe  jhon  doe      jhon
2   tom asd   tom   asd      tom
3   sam frf   sam   frr      sam

使用like或regexp的预期表结果？

fullname  count(*)
jhon doe  2
tom asd   1
sam frf   0

非常感谢！

Answer 1

最简单的是使用REGEXP。

SELECT fullname, 
       Count(t1.id) 
FROM   table1 t1 
       RIGHT JOIN table2 t2 
               ON t1.text REGEXP t2.keyword 
GROUP  BY fullname 

DEMO

我使用了一个正确的连接，这样你就可以得到零的山姆（否则就会被淘汰）

Answer 2

用我的真实数据进行一些性能测试

t1 => 100,000行并且正在增长

t2 => 207行

测试1

SELECT 
    t2.fullname,
    count(t1.id) AS total
FROM
    table_1 AS t1
        RIGHT JOIN
    table_2 AS t2 ON t1.text REGEXP t2.keyword
GROUP BY t2.fullname
ORDER BY total DESC

212 seconds

测试2

SELECT 
    t2.fullname,
    count(t1.id) AS total
FROM
    table_1 AS t1
        RIGHT JOIN
    table_2 AS t2 ON t1.text LIKE CONCAT('%', t2.keyword, '%')
GROUP BY t2.fullname
ORDER BY total DESC

30 seconds

测试3

SELECT 
    t2.fullname,
    count(t1.id) AS total
FROM
    table_1 AS t1
        RIGHT JOIN
    table_2 AS t2 ON t1.text LIKE lower(CONCAT('%', t2.name, '%')) AND t1.text LIKE lower(CONCAT('%', t2.surname, '%'))
GROUP BY t2.fullname
ORDER BY total DESC

32 seconds

测试4

SELECT 
    t2.fullname,
    count(t1.id) AS total
FROM
    table_1 AS t1
        RIGHT JOIN
    table_2 AS t2 ON t1.text LIKE lower(CONCAT('%', t2.name, '%')) OR t1.text LIKE lower(CONCAT('%', t2.surname, '%'))
GROUP BY t2.fullname
ORDER BY total DESC

40 seconds

测试5

SELECT 
    t2.fullname,
    count(t1.id) as total
FROM
    table_1 as t1
        RIGHT JOIN
    table_2 as t2 ON t1.text LIKE CONCAT('%', t2.keyword, '%') OR (t1.text LIKE lower(CONCAT('%', t2.name, '%')) AND t1.text LIKE lower(CONCAT('%', t2.surname, '%')))
GROUP BY t2.fullname
ORDER BY total DESC

41 seconds

我会选择测试5.最佳折衷结果/性能

还有什么建议吗？

再次感谢你的帮助！