mysql從一個表中選擇count行，其中列與另一個表中的值類似

Question

我將嘗試用一個例子來更好地解釋標題

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的預期表結果？

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

非常感謝！

Answer 1

最簡單的是使用REGEXP。

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

DEMO

我使用了一個正確的連接，這樣你就可以得到零的山姆（否則就會被淘汰）

Answer 2

用我的真實數據進行一些性能測試

t1 => 100,000行並且正在增長

t2 => 207行

測試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

測試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

測試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

測試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

測試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

我會選擇測試5.最佳折衷結果/性能

還有什么建議嗎？

再次感謝你的幫助！