Question

我不知道我的代码有什么问题：

<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">
<html xmlns="http://www.w3.org/1999/xhtml">
<head>
<meta http-equiv="Content-Type" content="text/html; charset=utf-8" />
<title>Register</title>
</head>

<body>
<?php

mysql_connect("localhost", "root", "") or die(mysql_error());
mysql_select_db("dressoholic") or die(mysql_error()); 

if($_POST['form_submitted'] == '1'){

$actkey = mt_rand() . mt_rand() . mt_rand() . mt_rand() . mt_rand();
$birthdate = $_POST['year'].'-'.$_POST['month'].'-'.$_POST['day'];

$usrname = mysql_real_escape_string($_POST['uname']);
$pswd = mysql_real_escape_string($_POST['password']);
$emaill = mysql_real_escape_string($_POST['email1']);
# gender moet nog gedaan worden!!!
$sql = "INSERT INTO users(username, fname, lname, email, password, activationkey, bdate, status) VALUES
('$usrname','$_POST[fname]','$_POST[lname]','$emaill','$pswd','$birthdate','$actkey','verify')";

if (!mysql_query($sql))

  {

  die('Error: ' . mysql_error());

  }

  echo "An email has been sent to $_POST[email1] with an activation key. Please check your mail to complete registration.";

##Send activation Email

$to      = $_POST['email1'];

$subject = " Dressoholic.com Registration";

$message = "Welcome to our website!\r\rYou, or someone using your email address, has completed registration at Dressoholic.com. You can complete registration by clicking the following link:\rhttp://localhost/dressoholic/register.php?$activationKey\r\rIf this is an error, ignore this email and you will be removed from our mailing list.\r\rRegards,\ Dressoholic.com Team";

$headers = 'From: blabla@gmail.com' . "\r\n" .

    'Reply-To: blabla@gmail.com' . "\r\n" .

    'X-Mailer: PHP/' . phpversion();

mail($to, $subject, $message, $headers);

} 
else {

##User isn't registering, check verify code and change activation code to null, status to activated on success

$queryString = $_SERVER['QUERY_STRING'];

$query = "SELECT * FROM users"; 

$result = mysql_query($query) or die(mysql_error());

  while($row = mysql_fetch_array($result)){

    if ($queryString == $row["activationkey"]){

       echo "Congratulations!" . $row["username"] . " is now the proud new owner of a Dressoholic.com account.";

       $sql="UPDATE users SET activationkey = '', status='activated' WHERE (id = $row[id])";

       if (!mysql_query($sql))

  {

        die('Error: ' . mysql_error());

  }

 }


}
}


?>

</body>
</html>

这是确切的错误：

Error: Duplicate entry 'arnold-blabla@gmail.c-1987--22' for key 'username'

arnold是我输入的用户名。 blabla@gmail.com是我输入的电子邮件。 1987年出生的那一年和22天！

我必须说我是PHP和MySQL的初学者。感谢。

Answer 1

users列的username表格中有唯一索引。这意味着两个不同的用户不能拥有相同的username。

输入不同的用户名或删除索引，你应该没问题。

Answer 2

您正在尝试在数据库中插入已存在的用户名（username字段）。如果您只是尝试更新用户信息，请尝试选择与arnold-blabla@gmail.c-1987--22不同的名称或执行update句子而不是insert。

Answer 3

这不是PHP问题，这是一个MySQL问题。我的猜测是username是表users上的主键。如果是这种情况，则每个用户名必须是唯一的。 MySQL告诉你有重复，这是不允许的。

在'用户名'上重复输入'bla bla bla bla'？

3 个答案: