mysql 如何实现 like in?

发布时间 2023-10-08 11:47:35作者: 20211103

https://blog.csdn.net/qq_36800514/article/details/115380100

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select * from hljtxeip_institution where state = 1 and deleteState = 0 and permissionType = 1

 

SELECT
*
FROM
hljtxeip_institution as a
JOIN (
SELECT
substring_index( substring_index( '12,13,14,112,772,280,1115', ',', help_topic_id + 1 ), ',',- 1 ) AS sub_name
FROM
mysql.help_topic
WHERE
help_topic_id <(
length( '12,13,14,112,772,280,1115' )- length(
REPLACE ( '12,13,14,112,772,280,1115', ',', '' ))+ 1 )
)as b
on a.post like concat(b.sub_name, '%')