0

抱歉，如果这看起来很愚蠢但很简单，但我似乎无法弄清楚为什么会出现这些错误：

警告：mysql_result() 期望参数 1 是资源，在第 25 行 C:\Program Files (x86)\EasyPHP-5.3.9\www\Image Upload\func\user.func.php 中给出的布尔值

注意：未定义索引：第 37 行 C:\Program Files (x86)\EasyPHP-5.3.9\www\Image Upload\register.php 中的 user_id

这是每个文件的代码

用户.func.php：

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

这是register.php：

if (isset($_POST['register_email'], $_POST['register_name'], $_POST['register_password'])){
$register_email = $_POST['register_email'];
$register_name = $_POST['register_name'];
$register_password = $_POST['register_password'];

$errors = array();

if(empty($register_email) || empty($register_name) || empty($register_password)){
    $errors[] = 'All fields must be filled out';
} 
else{
        if(filter_var($register_email, FILTER_VALIDATE_EMAIL) === false){
            $errors[] = 'Email address not valid';
        }
        if(strlen($register_email) > 255 || strlen($register_name) > 35 || strlen($register_password) > 35){
            $errors[] = 'One or more fields contains too many characters';
        }
        if(user_exists($register_email) === true){
            $errors[] = 'That email has already been registered to another user';
        }
    }

if(!empty($errors)){
    foreach ($errors as $error){
        echo $error, '<br />';
    }
} else {
    $register = user_register($register_email, $register_name, $register_password);
    $SESSION['user_id'] = $register;
    echo $_SESSION['user_id'];
}

}

谢谢你的帮助！-TechGuy24

4

2 回答 2

1

查询失败..应该是email = '$email'（而不是用反引号包围第二封电子邮件）。

另请查阅准备好的陈述和 PDO。

mysql_queryFALSE失败时将返回（布尔值），成功时将返回您正在寻找的“资源”。

于 2012-08-29T00:22:14.243 回答
0

You're using backticks "`" on a value, so your query is failing, use single quotes '

$query = mysql_query("SELECT COUNT(`user_id`) FROM `users` WHERE `email` = '$email'");
于 2012-08-29T00:22:36.740 回答