當我從表到另一個檢索值時，它插入了空白

Question

我有兩張桌子

1. 使用者
2. 注冊會員

我想確認用戶表中的值，然后將其插入registered_members表中，但問題是..值從用戶表中選擇良好，但插入到具有空白值的registered_members中，為什么？

<?
include('conf.php');

// Passkey that got from link
$passkey=$_GET['passkey'];

$tbl_name1="users";

// Retrieve data from table where row that match this passkey
$sql1="SELECT * FROM $tbl_name1 WHERE confirm_code ='$passkey'";
$result1=mysql_query($sql1) or die ("error1") ;

// If successfully queried
if($result1){

// Count how many row has this passkey
$count=mysql_num_rows($result1);

// if found this passkey in our database, retrieve data from table "users"
if($count==1){

$rows=mysql_fetch_array($result1);
$username=$_POST['uname'];
$fname=$_POST['fname'];
$password=$_POST['password'];
$email=$_POST['email'];
$age=$_POST['age'];
$gender=$_POST['gender'];

$tbl_name2="registered_members";

// Insert data that retrieves from "users" into table "registered_members"
$sql2=("INSERT INTO $tbl_name2 (uname,fname,password,email,age,gender)VALUES('$username','$fname','$password','$email','$age','$gender')");
$result2=mysql_query($sql2) or die ("error insert");
}

// if not found passkey, display message "Wrong Confirmation code"
else {
echo "Wrong Confirmation code";
}

// if successfully moved data from table"users" to table "registered_members" displays message "Your account has been activated" and don't forget to delete confirmation code from table "users"
if($result2){

echo "Your account has been activated";

// Delete information of this user from table "temp_members_db" that has this passkey
$sql3="DELETE FROM $tbl_name1 WHERE confirm_code = '$passkey'";
$result3=mysql_query($sql3);

}

}
?>

Answer 1

因為您使用的是$ _POST而不是#rows

更改

$rows=mysql_fetch_array($result1); 
$username=$_POST['uname']; 
$fname=$_POST['fname']; 
$password=$_POST['password']; 
$email=$_POST['email']; 
$age=$_POST['age']; 
$gender=$_POST['gender'];  

與

$rows=mysql_fetch_array($result1); 
$username=$rows['uname']; 
$fname=$rows['fname']; 
$password=$rows['password']; 
$email=$rows['email']; 
$age=$rows['age']; 
$gender=$rows['gender'];

Answer 2

在所有這些$row中將$_POST更改$_POST $row ：

$username=$_POST['uname'];
$fname=$_POST['fname'];
$password=$_POST['password'];
$email=$_POST['email'];
$age=$_POST['age'];
$gender=$_POST['gender'];

因為您正在以$rows=mysql_fetch_array($result1);在$row中獲取結果$rows=mysql_fetch_array($result1);