= - 求大神 指导下 非常感谢 ~解决办法

= - 求大神 指导下 非常感谢 ~

<br /><?php<br /><br />echo "姓名：".$_POST['name']."<br/>";<br />// echo $_POST['price']."<br/>";	<br />// echo $_POST['type']."<br/>";	<br />// echo $_POST['name']."<br/>";	<br />echo "电话：". $_POST['tel']."<br/>";<br />echo "电邮：".$_POST['email']."<br/>";	<br />echo "QQ：".$_POST['QQ']."<br/>";<br />echo "地址：".$_POST['addr']."<br/>";<br />echo "性别：".$_POST['sex']."<br/>";<br />echo "商品名称：".$_POST['brandname']."<br/>";<br />echo "商品品牌：".$_POST['commoditybrand']."<br/>";<br />echo "商品编号:".$_POST['productID']."<br/>"; 	<br />echo "商品序号:".$_POST['ordername']."<br/>";<br />echo "商铺名称：".$_POST['shopname']."<br/>";<br />echo "商铺地址:".$_POST['shopaddr']."<br/>";<br />echo "发票编号:".$_POST['invoicenumber']."<br/>";<br />echo "购买日期:".$_POST['purchasingdate']."<br/>";<br /><br /><br />$dingdanhao = date("Y-m-dH-i-s");<br />$dingdanhao = str_replace("-","",$dingdanhao);<br />$dingdanhao .= rand(1000,2000);<br /><br />echo "保用编号:".  $dingdanhao;<br /><br /> <br />$name=$_POST['name'];<br />$tel=$_POST['tel'];<br />$email=$_POST['email'];<br />$QQ=$_POST['QQ'];<br />$sex=$_POST['sex'];<br />$brandname=$_POST['brandname'];<br />$commoditybrand=$_POST['commoditybrand'];<br />$productID=$_POST['productID']; <br />$ordername=$_POST['ordername'];<br />$shopname=$_POST['shopname'];<br />$shopaddr=$_POST['shopaddr'];<br />$invoicenumber=$_POST['invoicenumber'];<br />$purchasingdate=$_POST['purchasingdate'];<br /><br /><br />require_once("conn.php");<br />if(mysql_query("insert into name set <br /><br />name='$name',tel='$tel',email='$email',QQ='$QQ',sex='$sex',brandname='$brandname',commodity<br /><br />brand='$commoditybrand',productID='$productID',ordername='$ordername',shopname='$shopname',<br /><br />shopaddr='$shopaddr',invoicenumber='$invoicenumber',purchasingdate='$purchasingdate'",<br /><br />$conn)){<br />echo "添加成功！<a href='testmysql.php'>查看数据库</a>";<br />}<br /> else{<br />	echo '添加失败！';<br />}<br /><br />mysql_close($con);<br />?><br />

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))