어떻게이 문을 mysqli로 변환 할 수 있습니까? 나는 그것을 시도했지만 작동하지 않습니다. 나는 무엇이 없는가. 로그인 양식에 연결하고 싶습니다.mysqli 변환 가져 오기 쿼리
이것은 변환 한 코드입니다.
<?php
include('admin/dbcon.php');
session_start();
$username = $_POST['username'];
$password = $_POST['password'];
/* teacher */
$query_teacher = mysql_query("SELECT * FROM teacher WHERE username='$username' AND password='$password'")or die(mysql_error());
$num_row_teacher = mysql_num_rows($query_teacher);
$row_teahcer = mysql_fetch_array($query_teacher);
/* admin */
$query_admin = mysql_query("SELECT * FROM users WHERE username='$username' AND password='$password'")or die(mysql_error());
$num_row_admin = mysql_num_rows($query_admin);
$row_admin = mysql_fetch_array($query_admin);
if ($num_row_teacher > 0){
$_SESSION['id']=$row_teahcer['teacher_id'];
echo 'true';
}else if ($num_row_admin > 0){
$_SESSION['id']=$row_admin['user_id'];
echo 'true_admin';
}else{
echo 'false';
}
?>
이것은 변환 된 mysqli이지만 아직 누락 된 항목이 있습니다. 이 문제를 해결하도록 도와 주시면 매우 감사하겠습니다.
<?php
// establishing the MySQLi connection
$con = mysqli_connect("localhost","root","","retreat");
if (mysqli_connect_errno())
{
echo "MySQLi Connection was not established: " . mysqli_connect_error();
}
// checking the user
session_start();
$username = mysqli_real_escape_string($con,$_POST['username'];
$password = mysqli_real_escape_string($con,$_POST['password'];
/* teacher */
$query_teacher = "SELECT * FROM teacher WHERE username='$username' AND password='$password'";
$num_row_teacher = mysqli_query($con,$query_teacher);
$row_teahcer = mysqli_num_rows($num_row_teacher);
/* admin */
$query_admin = "SELECT * FROM users WHERE username='$username' AND password='$password'";
$num_row_admin = mysqli_query($con,$query_admin);
$row_admin = mysqli_num_rows($num_row_admin);
if ($row_teahcer > 0){
//$_SESSION['user_email']=$email;
$_SESSION['']=$row_teacher['teacher_id'];
echo 'true';
}else if ($num_row_admin > 0){
$_SESSION['id']=$row_admin['user_id'];
echo 'true_admin';
}else{
echo 'false';
}
?>
누락; $ results2 = $ num_row_admin-> fetch(); '그 후에'$ _SESSION [ 'id'] = $ results [ 'user_id']' – Shahmee
@Shahmee 그것은 절차 상 OOP가 아닙니다. – Kitson88
은 그것을 $ results = mysqli_fetch_array ($ num_row_teacher)로 바꿉니다; $ results2 = mysqli_fetch_array ($ num_row_admin); ' – Shahmee