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Need help with PHP/MySQL drop down menu

Posted on 16th Feb 2014 07:03 pm by admin

I need help on how I can implement a drop down menu which queries mysql database and output the available data based on price range. This feature has been used here http://www.vebra.com - I will appreciate your help.

$searchSQL = "SELECT * FROM simple_search WHERE";


// grab the search types.
$types = array();
$types[] = isset($_GET['price'])?"`price` LIKE '%{$searchTermDB}%'":'';
$types[] = isset($_GET['location'])?"`location` LIKE '%{$searchTermDB}%'":'';

$types = array_filter($types, "removeEmpty"); // removes any item that was empty (not checked)

if (count($types) < 1)
$types[] = "`id` LIKE '%{$searchTermDB}%'"; // use the estate as a default search if none are checked


$andOr = isset($_GET['matchall'])?'AND':'OR';
$searchSQL .= implode(" {$andOr} ", $types) . " ORDER BY `price`"; // order by price.

$searchResult = mysql_query($searchSQL) or die("There was an error.<br/>" . mysql_error() . "<br />SQL Was: {$searchSQL}");

if (mysql_num_rows($searchResult) < 1) {
$error[] = "The search term provided <i>{$searchTerms}</i> yielded no results.";
}else {

$results = array(); // the result array
$i = 1;
while ($row = mysql_fetch_assoc($searchResult)) {
$results[] = "{$row['location']} <br/>{$row['image']}<br/>{$row['price']}";

$i++;

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