Question

我的注册表格有问题。它正在完美地注册新用户，尽管验证导致了我的问题。当具有相同用户名的用户尝试注册验证时，完美地选择了这一点，但我仍然会弹出确认用户已成功注册的diagloue框。当发生这种情况时我检查数据库，并且用户在执行验证时没有重新编译。我该如何阻止这种情况发生？

基本上我想要实现的是当新用户尝试输入已经存在的用户名时，消息停止说'成功'，因为验证正在提取这是我的代码

                <?php


    if ($_POST["register"]=="yes") {


//Declares the variables that will be posted to the database    
$username = $_POST['username'];
$first_name = $_POST['first_name'];
$last_name = $_POST['last_name'];
$email = $_POST['email'];
$password = md5($_POST['password']);



//Checks the database to see if username exists already
$query = "SELECT * FROM users WHERE username = '$username'";
$result = mysqli_query($db_connection, $query);
$nums = mysqli_num_rows($result);

//Checks the database to see if email address exists already
$query2 = "SELECT * FROM users WHERE email = '$email'";
$result2 = mysqli_query($db_connection, $query2);
$nums2 = mysqli_num_rows($result2);

    if ($nums >= 1)
    //informs user if username already exists
     echo "Username already exists, click <a href = 'register1.php'>HERE </a> to try again";
    else if ($nums2 >=1)

    //informs user if email already exists
    echo "Email Address already exists, click <a href = 'register1.php'>HERE </a> to try again";

    else
    {

$insert = 'INSERT INTO users(username, first_name, last_name, email, password,role_id, disable, activate) VALUES("'.$username.'","'.$first_name.'","'.$last_name.'","'.$email.'","'.$password.'", 2,0,0)';

mysqli_query($db_connection, $insert);
  mysqli_close($db_connection);
           ("Location: mainpage.php");



  }


{
  $i = 1;

if ($i > 0) {
    die;
}
else 

      "<script>alert (\"Registration Successful, your account has been sent for approval, and will be activated within 24 hours. \");</script>";
       echo "<meta http-equiv='refresh' content='0'>";
     }




     }



?>



<!-- Registration form -->
<form name="register" id="register" method="post" action="register1.php">
<input type='hidden' name='register' value='yes'>
<table border = '0'>
<tr><td><strong> First Name:</strong></td><td><input type = "text" name = "first_name" size="30" value=""></td></tr> 
<tr><td><strong> Last Name:</strong></td><td><input type = "text" name = "last_name" size="30" value=""></td></tr> 
<tr><td><strong> Username:</strong></td><td> <input type = "text" name = "username" size="30" value=""> </td></tr>
<tr><td><strong> Email: </strong></td><td><input type = "text" name = "email" size="30" value=""> </td></tr>
<tr><td><strong> Password:</strong></td><td><input type = "password" name = "password" size="30" value=""> </td></tr>
<tr><td> <strong>Verify Password:</strong></td><td><input type = "password" name = "password2" size="30" value=""> </td></tr>
</table>
<p>
<input type ="button" value="Register" input class='button' onclick="verifyForm()"/>
</p>
</form>

Answer 1

您的条件语句DO打印所需的结果，但不会停止程序执行。

例如：

$i = 1;

if ($i > 0) {
    echo 'I want to stop the program, you have entered invalid $i';
}
Register();

但是，您将打印该消息，但Register()也会执行。因为if()块不一定意味着停止执行。

您必须在函数/方法中使用die()，exit或return

E.g：

$i = 1;

if ($i > 0) {
    die('I want to stop the program, you have entered invalid $i');
}
Register();

在PHP中注册表单无法正常工作

1 个答案: