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 为资源, bool 值在 C:\Program Files (x86)\EasyPHP-5.3.9\www\Image Upload\func\user.func.php 第 25 行给出
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 $_SESSION['user_id'];
}
}
感谢您的帮助! -TechGuy24
最佳答案
查询失败..它应该是email = '$email'
(而不是用反引号包围第二封电子邮件)。
另请查找准备好的语句和 PDO。
mysql_query
当失败时将返回 FALSE
( bool 值),当成功时将返回您正在寻找的“资源”。
关于php - 为什么我会收到所有这些错误?,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/12169312/