mysqli/mysql查询里面的函数不能正常工作

Question

我正在尝试为我的网站构建一些函数,其中一些函数包括从mysql数据库中获取数据.当我测试函数之外的代码时,它似乎正常工作.所以在这里,第一页:

require('db.php');
require('functions.php');

$email = 'sample@gmail.com';

if (user_exists($email) == true){
 echo "Good news, this exists";
}

现在db.php:

$db = new MySQLi("localhost","test","test","test");
if ($db->connect_errno){
    echo "$db->connect_errno";
}

和functions.php文件:

function sanitize ($data){
    $db->mysqli_real_escape_string($data);
}
function user_exists($usermail){
    $usermail = sanitize($usermail);
    $query = $db->query("SELECT COUNT(userId) FROM users WHERE userEmail= '$usermail' ");
    $check = $query->num_rows;
    return ($check == 1) ? true : false;
}

我访问第一个文件时得到的错误是:

Notice: Undefined variable: db in C:\xampp\htdocs\auctior\inc\functions.php on line 6

Fatal error: Call to a member function query() on a non-object in C:\xampp\htdocs\auctior\inc\functions.php on line 6

所以我需要/包括db.php,其中$ db是mysqli连接.在同一个文件(第一个文件)中,我调用了functions.php中的函数

提前谢谢你,我很感激你的帮助,因为这让我很生气......

Answer 1

您可能需要使用global关键字,否则$db在本地范围内被视为var.

function sanitize ($data){
    global $db;
    $db->mysqli_real_escape_string($data);
}

function user_exists($usermail){
    global $db;
    $usermail = sanitize($usermail);
    $query = $db->query("SELECT COUNT(userId) FROM users WHERE userEmail= '$usermail' ");
    $check = $query->num_rows;
    return ($check == 1) ? true : false;
}