HTML код в переменную и вывод на странице - PHP - Ответ 5094935

if (!empty($_POST['preparate'])){
$preparate = strip_tags($_POST['preparate']);
$preparate = htmlspecialchars($preparate);}
$city = intval($_POST['city']);
$Category = !empty($_POST['Category']) ? $_POST['Category'] : array(); // только нужно этот отфильтровать от хакерских внедрений
if (( sizeof($Category)>0 ) && (!in_array ('6', $_POST['Category']))) {
$variable = $row['Prep'];
 $strSQL = "SELECT Tovar.idPreparat,
Tovar.idForma,
Tovar.idDozirovka,
Tovar.idManufacturer,
Tovar.idCountry,Preparat.Prep,
Forma.Frm,Dozirovka.doz,
Manufacturer.Manf,
Country.CountryName,
Preparat.Reactant,
Preparat.Recipe,
min(Stock.Price) min_price,
max(Stock.Price) max_price,
count(Apteki.id) apt
FROM  ((Stock INNER JOIN
(Preparat INNER JOIN 
(Manufacturer INNER JOIN 
(Forma INNER JOIN 
(Dozirovka INNER JOIN
(Country INNER JOIN Tovar ON Country.id = Tovar.idCountry)ON Dozirovka.id=Tovar.idDozirovka) 
ON Forma.id = Tovar.idForma) 
ON Manufacturer.id = Tovar.idManufacturer) 
ON Preparat.id = Tovar.idPreparat) 
ON Stock.idTovar = Tovar.id) INNER JOIN Apteki 
ON Stock.idApteki = Apteki.id) INNER JOIN City 
ON Apteki.idCity = City.id WHERE Prep Like '%$preparate%' AND City.id='".$city."' AND Preparat.Category IN ('".join("','",$Category)."') Group by Tovar.idPreparat,Tovar.idForma,Tovar.idDozirovka,Tovar.idManufacturer,Tovar.idCountry";
}