一段php添加代码，错了，就是不知道拿错了

Question

$email=$_POST['Email_name'];
$pwd=$_POST['pwd'];
$Fname=$_POST['F_name'];
$Lname=$_POST['L_name'];
$companyname=$_POST['company_name'];
$streetaddress=$_POST['street_address'];
$postcode=$_POST['post_code'];
$cityname=$_POST['city_name'];
$Continentname=$_POST['Continent_name'];
$countryname=$_POST['country_name'];
$telephone=$_POST['Telephone_num'];
$faxnumber=$_POST['Fax_num'];
$Question=$_POST['Question'];
$Answer=$_POST['Answer'];
$sql_into="insert into tabcustomer(Email,Pwd,Fname,lname,companyname,address,postcode,City,Continent,Countryid,Tele,Fax,Questions,Answer,Regtime)
values ('".$email."',".$pwd.",'".$Fname."','".$Lname."','".$companyname."','".$streetaddress."','".$postcode."','".$cityname."','".$Continentname."','".$countryname."','".$telephone."','".$faxnumber."','".$Question."','".$Answer."','".$email."',".date("Y-m-d H:i:s").")";
$result = mysql_query($sql_into);
if($result==true){
echo "<script>alert('留言成功！');window.location.href='adduser.php';</script>";
}else{
echo "shibai";
}运行结果是 shibai，说明没有添加成功，页面也没有提示错误在多少行，江湖求救啊，

Answer 1

$sql_into="insert into tabcustomer(Email,Pwd,Fname,lname,companyname,address,postcode,City,Continent,Countryid,Tele,Fax,Questions,Answer,Regtime)
values ('".$email."',".$pwd.",'".$Fname."','".$Lname."','".$companyname."','".$streetaddress."','".$postcode."','".$cityname."','".$Continentname."','".$countryname."','".$telephone."','".$faxnumber."','".$Question."','".$Answer."','".$email."',".date("Y-m-d H:i:s").")";改为$sql_into="insert into tabcustomer(Email,Pwd,Fname,lname,companyname,address,postcode,City,Continent,Countryid,Tele,Fax,Questions,Answer,Regtime)
values ('$email',$pwd,'$Fname','$Lname','$companyname','$streetaddress','$postcode','$cityname','$Continentname','$countryname','$telephone','$faxnumber','$Question','$Answer','$email',date(Y-m-d H:i:s))";

Answer 2

//if($result==true){//有这种判断方法吗？
直接
if($result){
echo " <script>alert('留言成功！');window.location.href='adduser.php'; </script>";
}else{
echo "shibai";
} 

Answer 3

我也建议用一楼的方法！PHP中，单引和双引是不一样的。

Answer 4

Warning: mysql_query(): supplied argument is not a valid MySQL-Link resource in D:\Debug\php\phphuanjing\htdocs\agoodic\admin\adduser_sql.php on line 28还是不行啊，出现如下错误