您好,我正在执行登录系统 我遇到了这个错误
致命错误:未捕获错误:在第6行的C:\ wamp64 \ www \ website project \ Processer.php中调用未定义的函数mysql_connect()
现在从网上可以找到的东西中,我需要去php.ini
是他们使用wamp64而不用此即时消息的任何其他方式 如果没有,请逐步帮助我,我真的希望这段代码能正常工作
我在其他文件中的所有其他代码就可以正常工作
$username =$_POST['user'];
$password=$_POST['pass'];
$con=mysql_connect("localhost","root","");
mysql_select_db($con,"bcs350");
$result = mysql_query("select * from users where username = '$username' and Password='$password' ");
$row=mysql_fetch_array($result);
if($row['username'] == $username && $row['Password'] == $password ){
echo "Login success !! welcome ".$row['username'];
}
else{
echo "failed to login";
}
?>
所以我做了一些编辑和更改,使它可以正常工作,但现在它以fetch数组结尾,并没有任何建议
错误:mysqli_fetch_array()期望参数1为mysqli_result
$username=$_POST['username'];
$Password=$_POST['password'];
session_start();
$con=mysqli_connect("localhost","","bcs350");
$result = mysqli_query($con,"SELECT * FROM 'forumtable' WHERE 'username' =
'$username' && 'Password'='$Password'");
$row=mysqli_fetch_array($result);
if($row['username'] == $username && $row['Password'] == $Password ){
echo "Login success !! welcome ".$row['username'];
}
else{
echo "sorry no account exist with that name try again or make an account";
}