Mysql query problem

I had this php+mysql db done for me long time ago.
Before, a property was either "house" or "apartment" or "new development". Selected from a dropdown..

mysql field name was "type". so properties.php?type=house or properties.php?type=new-development or properties.php?type=apartment was working fine. Even properties.php?type=house&type2=new-development would bring both the "houses" and "new-developments"

Now I am trying to make some properties "house" and "new development" at same time.

So I have altered my property edit screen to be ;
*house
*apartment
*new-development
*etc etc..

Altered my db to be "type1 - house" , "type2 - apartment" , "type3 - new-development" etc etc..

But I can not get the mysql query line right to list the properties I want.. Can anybody give me a hand ?

my old code;

if ($bedroom)
{
$bedroomsearch = " and bedrooms > '$bedroom' ";
}

if ($seaview)
{
$seaviewsearch = " and seaview like '%on%' ";
}

if ($security)
{
$securitysearch = " and security like '%on%' ";
}

if ($furniture)
{
$furnituresearch = " and furniture like '%on%' ";
}

if ($state)
{
$statesearch = " and state like '%$state%' ";
}

if ($type1)
{
$typesearch = " and type1 like '%$type1%' ";
}

if ($type2)
{
$typesearch = " and (type1 like '%$type1%' and type2 like '%$type2%') ";
}

if ($type3)
{
$typesearch = "  and (type1 like '%$type1%' and type2 like '%$type2%' and type3 like '%$type3%') ";
}
if ($type4)
{
$typesearch = "  and (type1 like '%$type1%' and type2 like '%$type2%' and type3 like '%$type3%' and type4 like '%$type4%') ";
}


//----------------------------------------------------------------------
if (!isset($oteleme) || $oteleme < 0) $oteleme=0;
if (!isset($dizeindex) || $dizeindex < 0) $dizeindex=0;
//----------------------------------------------------------------------

if (!$pricemin)
        {
        $check = mysql_query("select * from $table where location like '%$location%' and state like '%$state%' and status='4' $seaviewsearch $securitysearch $furnituresearch $statesearch $bedroomsearch $typesearch $typesearch2 $typesearch3 $typesearch4 order by ID desc LIMIT $oteleme,$KayitLimiti");
        }

elseif ($pricemin)
        {
        $check = mysql_query("select * from $table where location like '%$location%' and state like '%$state%' and price >= '$pricemin' and price <= '$pricemax' and status='4' $seaviewsearch $securitysearch $furnituresearch $statesearch $bedroomsearch $typesearch $typesearch2 $typesearch3 $typesearch4 order by ID desc LIMIT $oteleme,$KayitLimiti");
        }

        $numf = mysql_num_rows($check);

$toplam_satiri_al = $check;
$toplam_satir     = mysql_num_rows($toplam_satiri_al);

if ($numf == 0)
        {
        echo "<center><br><br><br><b><font face=verdana size=5 color=red>Nothing found..!</b></font>";
        exit;
        }

PHP:

What should my new code be ???

 

there are 2 solutions :

1. you simply put ' ' around your vars

$check = mysql_query("select * from $table where location like '%$location%' and state like '%$state%' and price >= '$pricemin' and price <= '$pricemax' and status='4' '$seaviewsearch' '$securitysearch' etc.

Code (markup):

2. You do a better solution:

lets say that $check is your query var, declare it first then you simply add to it as things turns out true like this

$check = "SELECT blabla..";

if ($seaview == "1")
{
$check .= " and seaview like '%on%' ";
}

//Finally to wrap it up
$check .= "order by ID desc LIMIT $oteleme,$KayitLimiti";

Code (markup):

I prefer the last method as it is more versitile..

 

Thanks for the tip;

//----------------------------------------------------------------------
if (!isset($oteleme) || $oteleme < 0) $oteleme=0;
if (!isset($dizeindex) || $dizeindex < 0) $dizeindex=0;
//----------------------------------------------------------------------

$check = mysql_query("select * from $table where location like '%$location%' and state like '%$state%' and status='4'");

if ($type1 != "")
{
$check .= " and type1 like '%$type1%' ";
}
if ($type2 != "")
{
$check .= " and type2 like '%$type2%' ";
}
if ($type3 != "")
{
$check .= " and type3 like '%$type3%' ";
}
if ($pricemin != "")
{
$check .= " and price >= '$pricemin' and price <= '$pricemax' ";
}
else 
{ 
$check .= "order by ID desc LIMIT $oteleme,$KayitLimiti";

$numf = mysql_num_rows($check);
}
$toplam_satiri_al = $check;
$toplam_satir     = mysql_num_rows($toplam_satiri_al);

if ($numf == 0)
        {
        echo "<center><br><br><br><b><font face=verdana size=5 color=red>Nothing found..!</b></font>";
        exit;
        }

PHP:

but this code returns ;

Warning: mysql_num_rows(): supplied argument is not a valid MySQL result resource in /home/trealest/public_html/test/properties.php on line 80


What did I do wrong ?
line 80 ; $numf = mysql_num_rows($check);