爲什麼這個SQL給出了語法錯誤？

想看看視頻類「me_id」和「you_id」雙雙與觀看：爲什麼這個SQL給出了語法錯誤？

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的工作有什麼想法？

來源

2012-02-21 algorithmicCoder

什麼是錯誤信息？ – zad 2012-02-21 06:39:14

打印查詢並向我們顯示，以便我們觀察並發現錯誤 – 2012-02-21 06:40:11