Question

有人可以帮助解决下面的代码。我正在尝试创建一个注册查询，但是当它被提交时，我收到以下行的错误：

$insert_query = "insert into members (First_name, last_name, Address_1, Address_2, Postcode, Email, Membership_Number, Password) values('$fname','$lname','$address1','$address2','$postcode','$email','$member','$password')";

这仅影响first_name，因为其他字段名称已成功提交。

非常感谢您的帮助!!

    <?php
$con = mysql_connect("localhost","root","") or die(mysql_error());
$select_db = mysql_select_db("thistlehc",$con);
if(isset($_POST['register']))

 $fname = mysql_real_escape_string($_POST['fname']);
 $lname = mysql_real_escape_string($_POST['lname']);
 $address1 = mysql_real_escape_string($_POST['address1']);
 $address2 = mysql_real_escape_string($_POST['address2']);
 $postcode = mysql_real_escape_string($_POST['postcode']);
 $email = mysql_real_escape_string($_POST['email']);
 $member = mysql_real_escape_string($_POST['member']);
 $password = mysql_real_escape_string($_POST['password']);

 $query = "select membership_number from members where membership_number='$member'";
 $link = mysql_query($query)or die(mysql_error());
     $num = mysql_num_rows($link);

     if ($num>0){
      echo 'Membership Number already exists'; //Membership number already taken
     }

     else {
     $insert_query = "insert into members (First_name, last_name, Address_1, Address_2, Postcode, Email, Membership_Number, Password) values('$fname','$lname','$address1','$address2','$postcode','$email','$member','$password')";
     $result = mysql_query($insert_query)or die(mysql_error());
     echo "Registered Successfully!";
     }

?>

Answer 1

请注意，就像你忘了封装if声明的内容一样。

if(isset($_POST['register']))

因为在要执行的代码周围没有大括号，所以只执行紧跟在后面的第一行。在您的情况下，if语句似乎返回false，并且定义$fname的行是而不是执行，因此是未定义的变量。

你想使用类似的东西 -

if(isset($_POST['register'])){
 $fname = mysql_real_escape_string($_POST['fname']);
 $lname = mysql_real_escape_string($_POST['lname']);
 $address1 = mysql_real_escape_string($_POST['address1']);
 ...
}

SQL第一个条目不添加到数据库

1 个答案: