這句sql哪裡出錯了呀。

$sql="select * from student where studentid={$studentid} and likeclassid={$classid}";

PS:likeclassid 是一個完整的欄位名

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 'and likeclassid=1176' at line 1

回複內容：

$sql="select * from student where studentid={$studentid} and likeclassid={$classid}";

PS:likeclassid 是一個完整的欄位名

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 'and likeclassid=1176' at line 1

把大括弧去掉 而且like後面少空格 匹配格式也不對

文法上應為 classid like {$classid}
但既然是like了就要模糊比對，取決於你想匹配的內容使用%,*。

