Mã:
<?php
if (!defined('RK_MEDIA')) die("You does have access to this!");
$connect = mysql_connect(SERVER_HOST, DATABASE_USER, DATABASE_PASS, true);
$dataconnect = mysql_select_db(DATABASE_NAME, $connect);
if (!$dataconnect) die('Error: ' . mysql_error());
// Class kết nối cơ sở dữ liệu
class MySql {
public static function dbselect($item, $table, $con) {
global $connect;
$table = trim($table);
$arr = null;
$i = (float) 0;
$sql = "SELECT $item FROM " . DATABASE_NAME . '.' . DATABASE_FX . $table;
if ($con != "") $sql .= " WHERE $con";
$result = mysql_query($sql, $connect);
if ($result) {
while ($myrow = mysql_fetch_row($result)) {
$count = mysql_num_fields($result);
for ($j = (float) 0; $j < $count; $j++)
$arr[$i][$j] = $myrow[$j];
$i++;
}
mysql_free_result($result);
return $arr;
} else return false;
}
public static function dbdelete($table, $item) {
mysql_query('DELETE FROM ' . DATABASE_FX . $table . " WHERE $item");
}
public static function dbupdate($table, $name, $item) {
mysql_query('UPDATE ' . DATABASE_FX . $table . " SET $name WHERE $item");
}
public static function dbinsert($table, $name, $item) {
mysql_query('INSERT INTO ' . DATABASE_FX . $table . " ($name) VALUES ($item)");
}
}
?>
Mình đã viết lại SQL cái code phimle gì đó share đã config các kiểu, chỉnh lại domain, nhưng nó báo lỗi: Fatal error: Uncaught Error: Call to undefined function mysql_connect() in C:\xampp\htdocs\include\lib\mysql.php:3
Mong được trợ giúp sớm! Xin cảm ơn =)))