Question

假设这是我需要的查询：

SELECT * FROM `products` WHERE `keywords` LIKE "%(SELECT `key` FROM `keywords` WHERE `slug` = '%d8%af%db%8c%d9%88%d8%a7%d8%b1' LIMIT 1)%"

目前它不起作用并且在满足条件的记录时返回零结果我想知道这里有什么问题？

Answer 1

这应该有效：

SELECT * 
FROM products 
WHERE keywords LIKE
    CONCAT('%', (SELECT `key` FROM `keywords` 
        WHERE `slug` = '%d8%af%db%8c%d9%88%d8%a7%d8%b1' LIMIT 1), '%');

Answer 2

尝试这个，不确定它是否会起作用......

SELECT * FROM products
WHERE keywords LIKE "%" + (SELECT key FROM keywords WHERE slug = 'something' LIMIT 1) + "%";

Answer 3

另一个带有like和嵌套SELECT的工作示例：

SELECT COUNT(email) as count FROM table1 t1 
JOIN (
      SELECT company_domains as emailext FROM table2 WHERE company = 'DELL'
     ) t2 
ON t1.email LIKE CONCAT('%', emailext) WHERE t1.event='PC Global Conference";

如果电子邮件扩展名等于多个公司域，则任务是使用过滤器对事件进行计数。如果有人觉得这很有用

如何在LIKE的Mysql中使用嵌套的SELECT？

3 个答案: