通过向用户发送包含唯一邀请码的电子邮件,我已经成功地允许用户邀请朋友,
但是我试图增加一种功能,以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“用户”(相同的数据库)中注册,因为这对于已经注册接收邀请电子邮件的人来说会很头疼。
我试图通过编写以下脚本来检查电子邮件是否有效:
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 \'' . $_POST['email'] . '\' is already in use';
}
注册用户时成功检查电子邮件地址,但是注册帐户与已注册帐户位于同一表中。我不确定如何在邀请代码中使用相同的脚本,因为我试图检查已注册的单独表中的电子邮件。
当前,它不检查它是否为有效电子邮件或是否存在。
完整的PHP:
include 'config.php';
function email_registered($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
}
}
?>
我只是想检查电子邮件是否已在“用户”表中注册,以及输入的电子邮件是否为有效电子邮件。
最佳答案
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。如果那是错误的,请告诉我,我可以更新我的答案:P这是您应该可以使用的函数的粗略草稿。
我假设您有以下两个表:
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任一表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;
}
}
?>
关于php - 检查有效的电子邮件以及是否存在,我们在Stack Overflow上找到一个类似的问题:https://stackoverflow.com/questions/26047155/