<select>数据传递的有关问题
Release: 2016-06-13 13:31:06
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数据传递的问题 inquire.html inquire the classroom
CRMS - Inquire the classroom
inquire_classroom.php inquire the classroom
CRMS - Inquire the classroom
// create short variable names $lesson_id=$_POST['lesson_id']; $con = mysql_connect("localhost","root",""); //通过服务器locahost建立连接,用户名为root,无密码 if (!$con) { die('Could not connect: ' . mysql_error()); }//如果不成功,显示错误 mysql_select_db("crms", $con);//选择数据库 if (!mysql_select_db("crms", $con)) die(mysql_error()); $search_lesson_id = "SELECT * FROM use2 WHERE lesson_id = '{$lesson_id}'"; $result = mysql_query($search_lesson_id,$con); if (!$result) { die("The lesson does not exist."); } while($row = mysql_fetch_array($result)) { echo $row['Cno']; echo $row['CID']; echo " "; } mysql_close($con); ?> 下拉框选择了一个值提交后,出现 Notice: Undefined index: lesson_id in D:\xampp\htdocs\inquire_classroom.php on line 9 D301001 的错误。但是这里我不是已经定义了吗?------解决方案--------------------
$lesson_id=$_POST['lesson_id']; 这样其实是OK的。 改一下php.ini的报错级别: error_reporting = E_ALL & ~E_NOTICE 改后重启apache。
------解决方案-------------------- notice并不是error,不影响结果。 想不出现notice,修改下php.ini中的错误提示级别就行
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