SEEMS AS IF MY ARTICLE HAS BEEN SUBMITTED TWICE. I DO APOLOGISE FOR THAT
Hi guys,
I'm working on a small website and I'm currently stuck on a small issue.
Ive got a set of dropdown boxes created and populated in HTML, for example:
<select name="heatingType" id="heatingType" required>
<option value="" disabled selected>Select Your Option</option>
<option value = "Gas">Gas</option>
<option value = "Electricity">Electricity</option>
<option value = "Other">Other</option>
</select>
I'm able to store the values in a variable once the form has been posted/submitted, These are stored in my Controller Class eg:
$newCalc = new ConCalc();
// instantiate drawing tool
$draw = new DrawTool();
// parse (render) appliance view
$renderedView = $draw->render('../View/calculator.php', array('calcvalues' => $newCalc->getValues()));
if(isset($_POST['btn-calcCon'])){
$heatType = $_POST['heatingType'];
$meterType = $_POST['meterType'];
$bedrooms = $_POST['noBedrooms'];
$house = $_POST['houseType'];
$age = $_POST['houseAge'];
echo $heatType;
echo $meterType;
echo $bedrooms;
echo $house;
echo $age;
}
echo $renderedView;
If i echo out $heatType then it will display the value that was selected.
My table structure is as follows:
HeatingType MeterType Bedrooms HouseType HouseAge Consumption
Gas Standard 1 or 2 Flat Less than 11 years 5430
Gas Standard 1 or 2 Flat More than 11 years 7270
Now the problem I'm facing is how to use these posted values in a select statement,
for example
SELECT Consumption fron ConTable WHERE HeatingType = heatingTypeDropdownValue AND MeterType = MeterTypeDropDownValue etc etc.
I have tired it but kept getting quite a few errors.
Any help will be appreciated
Thanks!