SELECT * FROM `image_appreciations`
WHERE `image_id` IN(SELECT `id` FROM `images` WHERE `user_id` = '1')
是我当前的查询,它返回零结果
SELECT `id` FROM `images` WHERE `user_id` = '1'
作为子查询返回大约 8 个 id,其中两个位于
SELECT * FROM `image_appreciations`
WHERE `image_id` IN(77,89)
这很好用。但总的来说它失败了。我究竟做错了什么?