如何使用 Dynamic SQL 或 mySQL Prepared 语句从下拉列表中过滤搜索结果？

Question

I only have about 2 months worth of PHP under my belt, but I am having trouble trying to find an answer to a problem. I want to at least have four dropdown menus that will have options of selecting: car make, model, manufacturer, etc, and then take the selected option from each dropdown box and create a custom query to pull from a car database that I have created. I believe the answer lies somewhere with Prepared Statements, but I am having trouble finding an answer. Any reference/help would be most appreciated.

Answer 1

Can you please provide code what you have so far. I don't think this is related to Prepared Statements. If your dropdown's are not related in the way that second dropdown will be populated based on selection from first, then just do simple POST request, get data from dropdown's and make something like this

$first = $_POST['ddown1'];
$sedond = $_POST['ddown2'];
$third = $_POST['ddown3'];
$fourth = $_POST['ddown4'];
 $sql = "SELECT * FROM cars WHERE model = '$first' AND manufacturer = '$sedond' AND color = '$third'";

If you don't know how many data will be sent then just use concatenating assignment operator (.=)

$sql .= "WHERE ";