为什么这个SQL给出了语法错误？

想看看视频类“me_id”和“you_id”双双与观看：为什么这个SQL给出了语法错误？

SELECT c.title, COUNT(*) AS popularity 
FROM video v 
JOIN user u ON v.user_id = u.id 
JOIN v_cat vc ON c.id = vc.vid_id 
JOIN cat c ON c.id = vc.cat_id 
JOIN u_cat uc ON uc.cat_id = c.id 
WHERE uc.user_id = '$me_id' 

INTSERSECT 

SELECT c.title, COUNT(*) AS popularity 
FROM video v 
JOIN user u ON v.user_id = u.id 
JOIN v_cat vc ON c.id = vc.vid_id 
JOIN cat c ON c.id = vc.cat_id 
JOIN u_cat uc ON uc.cat_id = c.id 
WHERE uc.user_id = '$you_id' 

GROUP BY c.title 
ORDER BY uc.id DESC LIMIT 0, 10

我用PHP/MySQL的工作有什么想法？

来源

2012-02-21 algorithmicCoder

什么是错误信息？ – zad 2012-02-21 06:39:14

打印查询并向我们显示，以便我们观察并发现错误 – 2012-02-21 06:40:11