mysql 如何实现 like in?
https://blog.csdn.net/qq_36800514/article/details/115380100
-- 阅读权限 全239 部分9
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, '%')



 
                    
                     
                    
                 
                    
                
 
                
            
         
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浙公网安备 33010602011771号