这个命令有什么问题?我收到错误,不知道出了什么问题
$qrys = mysql_query("
SELECT *
FROM users
LEFT JOIN follows ON follows.followbyid != '$userid'
WHERE users.active = 1 AND users.followex = 1 AND users.credits > 0 AND users.twid != '$userid' AND users.twimg != ''
ORDER BY users.featured DESC, users.buyer DESC, users.retweets DESC
LIMIT 49
") or die (mysql_error());
答案 0 :(得分:0)
这不应该是:
$qrys = mysql_query("
SELECT *
FROM users
LEFT JOIN follows ON **[A JOIN CRITERIA]**
WHERE users.active = 1 AND users.followex = 1 AND users.credits > 0 AND users.twid != '$userid' AND users.twimg != '' **AND follows.followbyid != '$userid'**
ORDER BY users.featured DESC, users.buyer DESC, users.retweets DESC
LIMIT 49
") or die (mysql_error());
也停止使用mysql_query并至少使用mysqli。