問題:在 JOIN 中跨多行應用條件來擷取符合的記錄。
目標:選擇同時擁有這兩者的使用者'tag1' 和 'tag2' 標籤。
解決這個問題主要有兩種方法:
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.子查詢:
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.連接:
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'
C.連接:
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(*) = 2
C.連接:
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 中選擇同時具有「tag1」和「tag2」的使用者?的詳細內容。更多資訊請關注PHP中文網其他相關文章!