Question

我正在试图弄清楚如何检查电子邮件是否已经存在于数据库中，因为我不想复制我尝试使用Google搜索，而且我认为它没有用作旧帖子？

这是我的代码：

$emailcheck = $value2;
$emailcheck = "SELECT email FROM demo WHERE email= '$emailcheck' ";
$result = mysql_query($link, $emailcheck);
$count = mysql_num_rows($result);

if ($count > 0){
    {$_SESSION['email'] = "Email is already used!";}    
}



if (empty($_POST["name"]))
     {$_SESSION['name']= "Name is required";}
   if (empty($_POST["email"]))
     {$_SESSION['email'] = "Email is required";} else {
   if (!filter_var($_POST["email"], FILTER_VALIDATE_EMAIL))
       {$_SESSION['email'] = "Email is incorect";}}



if(empty($_POST["name"]) || empty($_POST["email"]) || (!filter_var($_POST["email"], FILTER_VALIDATE_EMAIL)) || ($count > 0)) 
{   
    header("Location: home.php");
} else {

感谢任何人的帮助！

我几乎完整的代码：

$value = $_POST['name'];
$value2 = $_POST['email'];

$emailcheck = $value2;
$emailcheck = "SELECT email FROM demo WHERE email= '$emailcheck' ";
$result = mysql_query($link, $emailcheck);
$count = mysql_num_rows($result);

if ($count > 0){
    {$_SESSION['email'] = "Email is already used!";}    
}



if (empty($_POST["name"]))
     {$_SESSION['name']= "Name is required";}
   if (empty($_POST["email"]))
     {$_SESSION['email'] = "Email is required";} else {
   if (!filter_var($_POST["email"], FILTER_VALIDATE_EMAIL))
       {$_SESSION['email'] = "Email is incorect";}}



if(empty($_POST["name"]) || empty($_POST["email"]) || (!filter_var($_POST["email"], FILTER_VALIDATE_EMAIL)) || ($count > 0)) 
{   
    header("Location: home.php");
} else {


if (!$link) {
    die('Could not connect: ' . mysql_errno());
}

$db_selected = mysql_select_db(DB_NAME, $link);

if (!$db_selected) {
    die('can\'t use ' . DB_NAME . ':' . mysql_error());
}

$sql = "INSERT INTO demo (name, email) VALUES ('$value', '$value2')";

if (!mysql_query($sql)) {
    die('Error: ' . mysql_error());
}
header("location: https://www.google.co.uk/search?q=Ghostbusters&stick=H4sIAAAAAAAAAONgFuLUz9U3MI43skxR4gIxjUxLcrLTtYQck0vyi0LyQ8Ic81J888syU4sBdwGq3i0AAAA&sa=X&ved=0ahUKEwj0-J6QtuXaAhUQSsAKHaYqB2wQxA0I6gEwHg&biw=1280&bih=699");     
mysql_close();
session_destroy();
}
}
?>

Answer 1

我认为你必须把

die;

之后

header("Location: home.php");

否则代码仍将继续执行

Answer 2

Peshaw和Tin绝对正确;即使您发现重复的电子邮件，也无需改变或停止代码执行。

将知道是否存在重复，但您的代码将不会像当前所写的那样回复。

将结果存储在会话中后，您需要一些东西......

$isDuplicate = false; if ($count > 0){ $_SESSION['email'] = "Email is already used!"; $isDuplicate = true; } if(empty($_POST["name"]) || empty($_POST["email"]) || (!filter_var($_POST["email"], FILTER_VALIDATE_EMAIL)) || ($count > 0) || ($isDuplicate)) { // error handling here header("Location: home.php"); } else { // regular code execution here if (empty($_POST["name"])) {$_SESSION['name']= "Name is required";} if (empty($_POST["email"])) {$_SESSION['email'] = "Email is required";} else { if (!filter_var($_POST["email"], FILTER_VALIDATE_EMAIL)) {$_SESSION['email'] = "Email is incorrect";}} }

...当然，如果没有通常警告将用户输入直接插入查询而不是使用PDO对其进行参数化或至少消毒它们，这个答案就不会完整。请清理您的输入。

Answer 3

您需要将if ($count > 0)之后的其余代码包装在else代码块中，如下所示：

if ($count > 0){
    $_SESSION['email'] = "Email is already used!";

} else {
    if (empty($_POST["name"]))
        {$_SESSION['name']= "Name is required";}
        // the rest of your code
}

检查电子邮件是否在PHP中的Mysql数据库中

3 个答案: