Notice: Undefined variable: conn in G:\wamp\www\WWW\WWW\post.php on line 47 = - 第一个错误提示 完全版
$conn 变量未定义 请检查 conn.php 文件定义及加载情况
我自己改了一下 又出现一个错误 Warning: mysql_query(): 4 is not a valid MySQL-Link resource in G:\wamp\www\WWW\WWW\post.php on line 47 require_once("conn.php"); if(mysql_query("insert into name set name='$name',tel='$tel',email='$email',QQ='$QQ',sex='$sex',brandname='$brandname',commoditybrand='$commoditybrand',productID='$productID',ordername='$ordername',shopname='$shopname',shopaddr='$shopaddr',invoicenumber='$invoicenumber',purchasingdate='$purchasingdate'",$con)) { echo "添加成功!"; } else{ echo "添加失败!"; } ?>
if(mysql_query("insert into name set name='$name',tel='$tel',email='$email',QQ='$QQ',sex='$sex',brandname='$brandname',commoditybrand='$commoditybrand',productID='$productID',ordername='$ordername',shopname='$shopname',shopaddr='$shopaddr',invoicenumber='$invoicenumber',purchasingdate='$purchasingdate'",$conn)) { echo "添加成功!"; } else{ echo "添加失败!"; }
请检查 conn.php 文件定义及加载情况
require_once("conn.php");
if(mysql_query("insert into name set name='$name',tel='$tel',email='$email',QQ='$QQ',sex='$sex',brandname='$brandname',commoditybrand='$commoditybrand',productID='$productID',ordername='$ordername',shopname='$shopname',shopaddr='$shopaddr',invoicenumber='$invoicenumber',purchasingdate='$purchasingdate'",$con))
{
echo "添加成功!";
}
else{
echo "添加失败!";
}
?>
<?php
header('Content-Type: text/html; charset=gbk'); $con = mysql_connect("localhost", "root", "");
if (!$con)
{
die('Could not connect: ' . mysql_error());
}$db_selected = mysql_select_db("root", $con);if (!$db_selected)
{
die ("无法连接到数据库 : " . mysql_error());
}mysql_close($con);
?>
conn.php 内容
这能不出错吗?
require_once("conn.php");
if(mysql_query("insert into name set name='$name',tel='$tel',email='$email',QQ='$QQ',sex='$sex',brandname='$brandname',commoditybrand='$commoditybrand',productID='$productID',ordername='$ordername',shopname='$shopname',shopaddr='$shopaddr',invoicenumber='$invoicenumber',purchasingdate='$purchasingdate'",$conn))
{
echo "添加成功!";
}
else{
echo "添加失败!";
}
?><?php
header('Content-Type: text/html; charset=gbk'); $conn = mysql_connect("localhost", "root", "");
if (!$conn)
{
die('Could not connect: ' . mysql_error());
}$db_selected = mysql_select_db("root", $conn);if (!$db_selected)
{
die ("无法连接到数据库 : " . mysql_error());
}mysql_close($conn);
?>
= - 全部重新定义 还是这个报错
Warning: mysql_query(): 4 is not a valid MySQL-Link resource in G:\wamp\www\WWW\WWW\post.php on line 47
那么其后的代码怎么操作数据库呢?
那我把mysql_close($conn); 这个加在equire_once("conn.php");
if(mysql_query("insert into name set name='$name',tel='$tel',email='$email',QQ='$QQ',sex='$sex',brandname='$brandname',commoditybrand='$commoditybrand',productID='$productID',ordername='$ordername',shopname='$shopname',shopaddr='$shopaddr',invoicenumber='$invoicenumber',purchasingdate='$purchasingdate'",$conn))
{
echo "添加成功!";
}
else{
echo "添加失败!";
}
?>这个后面可以吗 = - 还有我都成功了 但是数据库里没有我添加的表单数据