Question

我已经成功地允许用户通过向他们发送带有唯一邀请码的电子邮件来邀请朋友，

但是我正在尝试添加能力来检查它是否是有效的电子邮件地址，以及该电子邮件是否已在另一个表格中注册用户＆＃39; （相同的数据库）因为已经注册接收邀请电子邮件的人会很头疼。

我试图通过编写此脚本来检查有效的电子邮件是否存在：

  function email_registered($email) {
  $email = sanitize($email);
  return (mysql_result(mysql_query("SELECT COUNT(`user_id`) FROM `users` WHERE `email` =   '$email'"), 0)  ==1) ? true : false;
}      


    if (filter_var($_POST['email'], FILTER_VALIDATE_EMAIL) === false) {
        $errors[] = 'A valid email address is required';
    }
    if (email_registered($_POST['email']) === true) {
        $errors[] = 'Sorry, the email \'' . $_POST['email'] . '\' is already in use';
    }

这在注册用户时检查电子邮件地址是成功的，但注册帐户与已注册的帐户在同一个表中。我不确定如何在邀请代码中使用相同的脚本，因为我正在尝试检查已注册的单独表格中的电子邮件。

目前它不检查它是否是有效的电子邮件或是否存在。

完整的PHP：

include 'config.php';

function email_registered($email) {
$email = sanitize($email);
return (mysqli_result(mysqli_query("SELECT mysqli_num_rows()(`user_id`) FROM `users` WHERE `email` = '$email'"), 0)  ==1) ? true : false;
} 


if(!empty($_POST)){
if(!empty($_POST['email'])){

    $email = mysqli_real_escape_string($conn,$_POST['email']);         

    $length = 10;
    $inviteCode = "";
    $characters = "0123456789abcdefghijklmnopqrstuvwxyz";

    for ($p = 0; $p < $length; $p++) {
    $inviteCode .= $characters[mt_rand(10, strlen($characters))];
    }

function email_registered($email)
{
    if (!empty($email)) {
        $ret_val = false;

        $query = sprintf(
            "SELECT id FROM users WHERE email = '%s'",
            mysqli_real_escape_string($email)
        );

        $result = mysqli_query($query);
        $num_rows = mysqli_num_rows($result);
        if ($num_rows > 0) {
            //email exists
            ?> 
            <p>User Exists</p>
            <?php 

            $ret_val = true;
        } else {
            $query = sprintf(
                "SELECT id FROM invites WHERE email = '%s'",
                mysqli_real_escape_string($email)
            );

            $result = mysqli_query($query);
            $num_rows = mysqli_num_rows($result);
            if ($num_rows > 0) {
                //email exists
                ?>
                <p>User Exists</p>
                <?php

                $ret_val = true;
            }
        }
        return $ret_val;
    }
 }

else { 

     $query = mysqli_query($conn, "INSERT INTO `referrals` (`email`, `inviteCode`) VALUES   ('$email', '$inviteCode') "); }
     //you might want to consider checking more here such as $query == true as it can return other statuses that you may not want
     if($query){ 

  include 'userinvite.php';

     ?>
         <p> "Thank you for inviting your friends!"</p>
     <?php 

    }
    else{
    ?>
       <p>Sorry there must have been a problem</p>
    <?php
        die('Error querying database. ' . mysqli_error($conn));
    }
}
else {
?>
    <p>Please enter an email</p>
<?php
}
}
?>

我只想检查电子邮件是否已在＆＃39;用户＆＃39;中注册。表格，如果输入的电子邮件是有效的电子邮件。

Answer 1

我认为您的主要问题是如何检查另一个表中是否存在电子邮件。如果错了，请告诉我，我可以更新我的答案：P这是您应该能够使用的函数的草稿。

我假设您有这两个表：

表1：用户

||id||email||name||

表2：邀请

||id||email||inviter_user_id||

您可以使用此功能检查表格中是否存在电子邮件

<?php
    /**
     * Check if the given email already exists in the DB
     *
     * @param $email string the email to check
     */
    function email_exists($email)
    {
        if (!empty($email)) {
            $ret_val = false;

            $query = sprintf(
                "SELECT id FROM users WHERE email = '%s'",
                mysqli_real_escape_string($email)
            );

            $result = mysqli_query($query);
            $num_rows = mysqli_num_rows($result);
            if ($num_rows > 0) {
                //email exists
                $ret_val = true;
            } else {
                $query = sprintf(
                    "SELECT id FROM invites WHERE email = '%s'",
                    mysqli_real_escape_string($email)
                );

                $result = mysqli_query($query);
                $num_rows = mysqli_num_rows($result);
                if ($num_rows > 0) {
                    //email exists
                    $ret_val = true;
                }
            }
            return $ret_val;
        }
    }
?>

检查有效的电子邮件及其是否存在

1 个答案: