ето това ми е кода в view.php
, но ми излиза следната грешка. Не знам от какво е
ето и демо
http://87.252.174.185/obqvi/index.php
Код:
<head>
<meta http-equiv="Content-Type" content="text/html; charset=windows-1251">
</head>
<?php
include("config.php");
echo "
Покажи обяви само от тип: </br>
<a href='$self?izberi=buy'>Купувам</a><br/>
<a href='$self?izberi=sell'>Продавам</a><br/>";
echo "Под категории:</br>
<a href='$self?category='avto_i_moto'>Авто И Мото</a><br/>
<a href='$self?category='nauka'>Наука</a><br/>
<a href='$self?category='shoping'>Шопинг</a><br/>
<a href='$self?category='antichnost'>Античност</a><br/>
<a href='$self?category='parceli'>Парцели</a><br/>";
$vip = $_GET['vip'];
$izberi = $_GET['izberi'];
$izberi = htmlspecialchars($izberi);
$izberi = strip_tags($izberi);
$izberi = get($izberi);
$category = $_GET['category'];
$category = htmlspecialchars($category);
$category = strip_tags($category);
$category = get($category);
//nachalo na stranicioniraneto
$redove = ($pageNum - 1) * $broinastranica;
$broinastranica = 4;
$pageNum = 1;
if(isset($_GET['page']))
{
$pageNum = $_GET['page'];
$pageNum = htmlspecialchars($pageNum);
$pageNum = strip_tags($pageNum);
$pageNum = get($pageNum);
}
$redove = ($pageNum - 1) * $broinastranica;
$maxPage = ceil($numrows/$broinastranica);
$nomernastranici = '';
for($page = 1; $page <= $maxPage; $page++)
{
if($page == $pageNum)
{
$nomernastranici .= $page;
}
else
{
$cat = (isset($_GET["category"])) ? "&category=" . $_GET["category"] : "";
$izberi = (isset($_GET["izberi"])) ? "&izberi=" . $_GET["izberi"] : "";
$nomeranastranici .= " <a href=\"$self?page=$page$cat$izberi\">$page</a> ";
}
}
if(isset($_GET['izberi']))
{
$zapisi = "SELECT * FROM obqvi WHERE izberi='$izberi' ORDER BY id DESC LIMIT $redove, $broinastranica";
}
elseif(isset($_GET['category']))
{
$zapisi = "SELECT * FROM obqvi WHERE category='$category ORDER BY id DESC LIMIT $redove, $broinastranica";
}
elseif(!isset($_GET['izberi']) & (!isset($_GET['category'])))
{
$zapisi = "SELECT * FROM obqvi ORDER BY id DESC LIMIT $redove, $broinastranici";
}
$result = mysql_query($zapisi) or die(mysql_error());
while($row = mysql_fetch_array($result))
{
$ime = $row['ime'];
$email = $row['email'];
$cena = $row['cena'];
$opisanie = $row['opisanie'];
$gsm = $row['gsm'];
$izberi = $row['izberi'];
$snimka = $row['snimka'];
$date = $row['date'];
$category = $row['category'];
$vip = $row['vip'];
echo "
<table align='center' valign='top' border=1 cellspacing=0 cellpadding=0>
<tr>
<td>
Име: $ime
</td>
</tr>
<tr>
<td>
Email: $email
</td>
</tr>
<tr>
<td>
Цена: $cena
</td>
</tr>
<tr>
<td>
Описание: $opisanie
</td>
</tr>
<tr>
<td>
Избери: $izberi
</td>
</tr>
<tr>
<td>
Снимка: $snimka
</td>
</tr>
<tr>
<td>
Дата на добавяне: $date
</td>
</tr>
<tr>
<td>
Категория: $category
</td>
</tr>
<tr>
<td>
V.I.P: $vip
</td>
</tr>";
}
?>, но ми излиза следната грешка. Не знам от какво е
Код:
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 '' at line 1ето и демо
http://87.252.174.185/obqvi/index.php