我有这个PHP代码 - :
$q = "SET @session = '1', @buddys = '12,7,10', @rejects = 'post_0'; SELECT f.* FROM feed as f"; $r = mysqli_num_rows($q);.
这会导致此错误 - :You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near 'SELECT f.* FROM feed as f' at line 3
。
但令人惊讶的是,当在phpmyadmin中运行相同的mysql查询时,它会根据需要运行。
是什么事?谢谢你的帮助...
答案 0 :(得分:0)
在选择
之前,您的问题是分号;
尝试用逗号,
$q = "SET @session = '1', @buddys = '12,7,10', @rejects = 'post_0' , SELECT f.* FROM feed as f";
$r = mysqli_num_rows($q);
或者你也应该像你那样分开你的查询
$q = "SET @session = '1', @buddys = '12,7,10', @rejects = 'post_0' ";
$q .= "SELECT f.* FROM feed as f";