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php和mysql在单击电子邮件激活链接后未激活新用户_Php_Mysql - Fatal编程技术网

php和mysql在单击电子邮件激活链接后未激活新用户

php和mysql在单击电子邮件激活链接后未激活新用户,php,mysql,Php,Mysql,我正在用php创建一个注册/登录系统。我的代码可以通过电子邮件向新用户发送激活链接,但用户点击链接,用户未激活。请任何人帮助,代码如下 users.php function activate($email, $email_code) { $email = mysql_real_escape_string($email); $email_code = mysql_real_escape_string($email_code); if (mysql

我正在用php创建一个注册/登录系统。我的代码可以通过电子邮件向新用户发送激活链接,但用户点击链接,用户未激活。请任何人帮助,代码如下

users.php

function activate($email, $email_code) {
    $email          = mysql_real_escape_string($email);
    $email_code     = mysql_real_escape_string($email_code);

    if (mysql_result(mysql_query("SELECT COUNT(`user_id`) FROM `user` WHERE `email` = '$email' AND `email_code` = '$emal_code' AND `active` = 0"), 0) == 1) {
        if (mysql_query("UPDATE `user` SET `active` = 1 WHERE `email` = '$email'")){
         return true; 
    } else {
         return false;  
    }

}
}

function register_user ($register_data) {
    array_walk($register_data, 'array_sanitize');
    $register_data['password'] = md5($register_data['password']);

    $fields = '`' . implode('`, `', array_keys ($register_data)) . '`';
    $data   = '\'' . implode('\', \'', $register_data) . '\'';

    mysql_query("INSERT INTO `user` ($fields) VALUES ($data)");
    email($register_data['email'], 'Activate your account', "Hello " .      $register_data['username'] . ",\n\nYou need to activate your account by clicking the link below:\n\nhttp://fredhosting.com/real/activate.php?email=" . $register_data['email'] . "&email_code=" . $register_data['email_code'] . "\n\n - Fredhosting.com");
}
Activate.php

if (isset($_GET['success']) === true && empty($_GET['success']) === true) {
?>
    <h2>Thanks, we have activated your account...</h2>
    <p>You can now Log in!</p>
<?php
} else if (isset($_GET['email'], $_GET['email_code']) === true) {

    $email       = trim($_GET['email']);
    $email_code  = trim($_GET['email_code']);

    if (email_exists($email) === false) {
        $errors[] = 'Oops, something went wrong and we could not find that email address.';
    } else if (activate($email, $email_code) === false) {
        $errors[] = 'We had some issues activating your account.';
    }

    if (empty($errors) === false) {
    ?>
        <h2>Oops...</h2>
    <?php
        echo output_errors($errors);
    } else {
        header('Location: activate.php?success');
        exit(); 
    }

}
if(isset($\u GET['success'])==true&&empty($\u GET['success'])==true){
?>
谢谢,我们已激活您的帐户。。。
你现在可以登录了

哎呀。。。 错误:

不正确的变量$emal\u代码

但是,在发布代码之前,拼写错误通常会被重新阅读几次

似乎您正在处理关闭的PHP错误,请激活并推荐它们

ini_set('display_errors', '1');
error_reporting(E_ALL);

您的脚本是否产生错误或其他问题?您是否进行过任何类型的调试以缩小问题的范围?我的猜测是,如果您在代码中添加一些简单的
var\u dump()
语句进行调试,您将立即发现问题。
activate()中的另一个指针
function绝对没有理由首先进行选择。如果您正确处理了所有结果/错误条件(无论如何都应该这样做),您可以直接转到更新并消除多次查询的需要.大家好…谢谢你们的回复…它实际上根本没有产生任何错误,只是没有激活新用户。@user2983998为什么它没有激活用户?在代码执行的什么时候它没有达到您期望的效果?您是否尝试过在MySQL中处理错误,看看是否有问题?正确的代码不应该只是激活检查结果,看看你是否得到了预期的结果。此外,你甚至不应该使用
mysql.*
函数编写新代码,因为它们已被弃用。建议使用
mysqli
PDO
。哈哈哈,我认为dat是我的错误。。。我太棒了上帝保佑你无论你在哪里谢谢
ini_set('display_errors', '1');
error_reporting(E_ALL);