<?php
require 'connectdb.php';
$login_username = mysqli_real_escape_string($dbcon,$_POST['username']);
$login_password = mysqli_real_escape_string($dbcon,$_POST['password']);
$salt = "yingyeenddldk[l[l[ljihhduhudf";
has_login_password = hash_hmac('sha256', $login_password, $salt);
$sql = "SELECT * FROM tb_login WHERE login_username=? AND login_password=?";
$stmt = mysqli_prepare($dbcon,$sql);
mysqli_stmt_bind_param($stmt,"ss",$login_username,$has_login_password);
mysqli_execute($stmt);
$result_user = mysqli_stmt_get_result($stmt);
if ($result_user->num_rows == 1) {
session_start();
$row_user = mysqli_fetch_array($result_user,MYSQLI_ASSOC);
$_SESSION ['login_id'] = $row_user['user_id'];
header ("Location: main.php");
}else{
echo "no no";
}
แล้วมันขึ้นแบบนี้ต้องทำไงคะ แก้มาห้าชั่วโมงแล้ว ไปไม่เป็นคะ
Parse error: syntax error, unexpected '=' in C:\xampp1\htdocs\ying\login.php on line 8
แก้ไขโค้ด login งงมากช่วยหน่อย
require 'connectdb.php';
$login_username = mysqli_real_escape_string($dbcon,$_POST['username']);
$login_password = mysqli_real_escape_string($dbcon,$_POST['password']);
$salt = "yingyeenddldk[l[l[ljihhduhudf";
has_login_password = hash_hmac('sha256', $login_password, $salt);
$sql = "SELECT * FROM tb_login WHERE login_username=? AND login_password=?";
$stmt = mysqli_prepare($dbcon,$sql);
mysqli_stmt_bind_param($stmt,"ss",$login_username,$has_login_password);
mysqli_execute($stmt);
$result_user = mysqli_stmt_get_result($stmt);
if ($result_user->num_rows == 1) {
session_start();
$row_user = mysqli_fetch_array($result_user,MYSQLI_ASSOC);
$_SESSION ['login_id'] = $row_user['user_id'];
header ("Location: main.php");
}else{
echo "no no";
}