= - 求大神 指导下 非常感谢 ~
1 | <br /><?php<br /><br /> echo "姓名:" . $_POST [ 'name' ]. "<br/>" ;<br />
|
Copier après la connexion
错误报告 如下 :
Notice: Undefined variable: con in D:\wamp\www\WWW\post.php on line 55
Call Stack
( ! ) Warning: mysql_close() expects parameter 1 to be resource, null given in D:\wamp\www\WWW\post.php on line 55
= - 求助
------解决思路----------------------mysql_close($con);改成mysql_close($conn);
------解决思路----------------------$con 写错了,应该是$conn
你上面的程序有写的。
mysql_query("insert into name set
name='$name',tel='$tel',email='$email',QQ='$QQ',sex='$sex',brandname='$brandname',commodity
brand='$commoditybrand',productID='$productID',ordername='$ordername',shopname='$shopname',
shopaddr='$shopaddr',invoicenumber='$invoicenumber',purchasingdate='$purchasingdate'",
$conn))