Здравейте! Написах си код, но ми изписва: You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near 'from, about, to, message) VALUES ('', 'Покана в гилдията ff', 'user', 'Нашата ги' at line 1. Ето и кода:
Код:
<?php
require_once("../connection.php");
mysql_query("SET NAMES cp1251");
$name=$_GET['guild'];
$select=mysql_query("SELECT * FROM `guilds` WHERE `name`='$name'") or die(mysql_error());
$s=mysql_fetch_assoc($select);
$members=$s['members'];
echo "Име: <strong style='color: #1db1ee;'>".$s['name']."</strong>";
echo "<br/>";
echo "Основател: <strong style='color: #1db1ee;'>".$s['founder']."</strong>";
echo "<br/>";
echo "Девиз: <strong style='color: #1db1ee;'>".$s['motto']."</strong>";
echo "<br/>";
echo "Членове: <strong style='color: #1db1ee;'>".$s['members']."</strong>";
echo "<br/>";
echo "Пари: <strong style='color: #1db1ee;'>".$s['money']."</strong>";
echo "<form method='post' action='' >";
echo "<input type='text' name='invitep'>";
echo "<input type='submit' name='invite'>";
echo "</form>";
$useri=$_POST['invitep'];
$check_user=mysql_query("SELECT * FROM `users` WHERE `user` = '$useri'");
$user_exist=mysql_num_rows($check_user);
if(isset($_POST['invite']) && ($user_exist > 0)) {
$from=$logged;
$to=$useri;
$about="Това е ".$name."";
$message="Чрез това: <a href='confirm.php?accept=".$name."'>Тук</a>";
$ins=mysql_query("INSERT INTO `messages` (from, about, to, message) VALUES ('$from', '$about', '$to', '$message')") or die(mysql_error());
if($ins) {
echo "Вие успешно изпратихте покана до:<strong style='color: #1db1ee;'>".$to."</strong>";
}else{
echo "Възникна грешка при изпращането на поканата ви!";
}
}
?>