问题：在 JOIN 中跨多行应用条件来检索匹配的记录。
目标：选择同时拥有这两者的用户'tag1' 和 'tag2' 标签。
解决这个问题主要有两种方法：
A.存在：
SELECT * FROM users WHERE EXISTS (SELECT * FROM tags WHERE user_id = users.id AND name ='tag1') AND EXISTS (SELECT * FROM tags WHERE user_id = users.id AND name ='tag2')
B.子查询：
SELECT * FROM users WHERE id IN (SELECT user_id FROM tags WHERE name ='tag1') AND id IN (SELECT user_id FROM tags WHERE name ='tag2')
C.连接：
SELECT u.* FROM users u INNER JOIN tags t1 ON u.id = t1.user_id INNER JOIN tags t2 ON u.id = t2.user_id WHERE t1.name = 'tag1' AND t2.name = 'tag2'
A.计数：
SELECT users.id, users.user_name
FROM users
INNER JOIN tags ON users.id = tags.user_id
WHERE tags.name IN ('tag1', 'tag2')
GROUP BY users.id, users.user_name
HAVING COUNT(*) = 2B.字符串处理：
SELECT user.id, users.user_name, GROUP_CONCAT(tags.name) as all_tags
FROM users INNER JOIN tags ON users.id = tags.user_id
GROUP BY users.id, users.user_name
HAVING FIND_IN_SET('tag1', all_tags) > 0 AND FIND_IN_SET('tag2', all_tags) > 0 以上是如何使用 JOIN 在 SQL 中选择同时具有'tag1”和'tag2”的用户？的详细内容。更多信息请关注PHP中文网其他相关文章！
