如果这看起来很愚蠢但很简单,我很抱歉,但我似乎不明白为什么会出现这些错误:
警告:mysql_result()要求参数1为resource,布尔值在第25行的C:''Program Files(x86)''EasyHP-5.3.9''www''Image Upload''func''user.func.php中给定
注意:第37行C:''Program Files(x86)''EasyHP-5.3.9''www''Image Upload''register.php中的未定义索引:user_id
这是每个文件的代码
user.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($query, 0) == 1) ? true : false;
}
这里是register.php:
if (isset($_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
查询失败。。它应该是CCD_ 1(而不是用反勾号包围第二封电子邮件)。
还请查阅准备好的报表和PDO。
mysql_query
将在失败时返回FALSE
(布尔值),并在成功时返回您正在查找的"资源"。
您对值使用了backticks"`;",因此查询失败,请使用单引号'
$query = mysql_query("SELECT COUNT(`user_id`) FROM `users` WHERE `email` = '$email'");