Error in count query!

PHP
1

Can anyone tell me why the following query is returning an error?
I can’t see anything wrong with it. I’ve echoed the query and it doesn’t seem to be taking any notice of the if statements.

$search_type = $_GET['search_type'];
$top_make = $_GET['top_make'];
$top_model = $_GET['top_model'];
$thisYear = date("Y")-1;
$today = date("Y-m-d");

if($search_type = "used"){ $MyTable = 'Used_Stock';}
if($search_type = "vans"){ $MyTable = 'Used_Stock';}
if($search_type = "bikes"){ $MyTable = 'Used_Stock';}
if($search_type = "nearlynew"){ $MyTable = 'Used_Stock';}
if($search_type = "new"){ $MyTable = 'New_Offers';}
if($search_type = "motability"){ $MyTable = 'Motability_Offers';}

	//build query
$howmanyquery = "SELECT COUNT(*) as NUM FROM $MyTable";
if($search_type = "used")
	$howmanyquery .= " WHERE Type='2'";
if($search_type = "vans")
	$howmanyquery .= " WHERE Type='4'";

if($search_type = "bikes")
	$howmanyquery .= " WHERE Type='6'";

if($search_type = "nearlynew")
	$howmanyquery .= " WHERE Type='2' AND Year = '$thisYear'";

if($search_type = "new")
	$howmanyquery .= " WHERE Make !='Harley-Davidson' AND Valid >='$today'";

if($search_type = "motability")
	$howmanyquery .= " WHERE Expiry >='$today'";

if($top_make != "")
	$howmanyquery .= " AND Make = '$top_make'";

if($top_model != "")
	$howmanyquery .= " AND Model LIKE '%$model%'";

	//Execute query
$total_results = mysql_result(mysql_query("$howmanyquery"),0);

echo "" .$total_results. "";//Shows found number of results in db
echo "Query:" .$howmanyquery. "";//Shows found number of results in db
2

There should be a song about this somewhere.
You’ve given every one of those IF statements a definition, not a comparison. (= vs ==) !

3

Thanks for that, daft mistake to make though after another straight 14hrs coding I am starting to miss things!! Appreciate the help once agian.

4

A switch statement is ideal for what your doing, not stacking multiple if conditions. To that end I took the liberty of running clean-up including adding proper escaping and input checking to make the script more robust. Never assume user input exists, that is a bad assumption to make. Always check make sure the input exists before assigning it to a variable. Also always use mysql_real_escape_string() when using the standard library to protect against SQL injection.

(untested)


$search_type = isset($_GET['search_type'])?$_GET['search_type']:null;
$top_make = isset($_GET['top_make'])?$_GET['top_make']:null;
$top_model = isset($_GET['top_make'])?$_GET['top_model']:null;

$thisYear = date("Y")-1;
$today = date("Y-m-d");

$querytpl = 'SELECT COUNT(*) as NUM FROM %s %s';

$MyTable = 'Used_Stock';
$myFilter = '';

switch((string) $search_type) {

	case 'used':
		$myFilter = "WHERE Type='2'";
		break;
	
	case 'vans':
		$myFilter = "WHERE Type='4'";
		break;
	
	case 'bikes':
		$myFilter = "WHERE Type='6'";
		break;
	
	case 'nearlynew':
		$myFilter = "WHERE Type='2' AND Year = '$thisYear'";
		break;
	
	case 'new':
		$MyTable = 'New_Offers';
		$myFilter = "WHERE Make !='Harley-Davidson' AND Valid >='$today'";
		break;
	
	case 'motability':
		$MyTable = 'Motability_Offers';
		$myFilter = "WHERE Expiry >='$today'";
		break;
	
	default:

}

if($top_make !== null) {
    $myFilter.= ($myFilter == ''?'WHERE ':' AND ')."Make = '".mysql_real_escape_string($top_make)."'";
}

if($top_model !== null)
    $myFilter.= ($myFilter == ''?'WHERE ':' AND '),"Model LIKE '%".mysql_real_escape_string($top_model)."%'";
}


$howmanyquery = sprintf($querytpl,$MyTable,$myFilter);

//Execute query
$total_results = mysql_result(mysql_query($howmanyquery),0);

echo "" .$total_results. "";//Shows found number of results in db
echo "Query:" .$howmanyquery. "";//Shows found number of results in db