$sql="select * from student where studentid={$studentid} and likeclassid={$classid}";
PS:likeclassid 是一个完整的字段名
You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near 'and likeclassid=1176' at line 1
语法上应为 classid like {$classid}
但既然是like了就要模糊匹配，取决于你想匹配的内容使用%,*。
把大括号去掉 而且like后面少空格 匹配格式也不对