【发布时间】:2011-06-05 19:42:24
【问题描述】:
当我尝试运行此代码时收到上述警告:
$mysqli=new mysqli("localhost", "***", "***","***") or die(mysql_error());
function checklogin($username, $password){
global $mysqli;
$result = $mysqli->prepare("SELECT * FROM users WHERE username = ?");
$result->bind_param("s", $username);
$result->execute();
if($result != false){
$dbArray=mysql_fetch_array($result);
【问题讨论】:
标签: php mysqli prepared-statement