PHP calculation from html form

Question

I have created a reservations page for a restaurant website, with a form present to book a dinner reservation. It contains a dropdown box which allows the user to select the amount of people who will be dining in their party. It also contains a radio button to allow the user to select if they wish to be seated in the VIP area (Yes/ No).

Each person in the party should cost an extra £5 towards their booking fee and if they wish to be seated in the VIP area, they will be charged an additional £5. Once the form validation is successful, the user will be sent to a confirmation page, where the reservation details they have entered, will be displayed to them.

Currently I have a calculation present on the confirmation page of my website, which correctly calculates the party size costs, but my issue is that even if VIP area selection is "No", it is still adding on the £5 fee as if the user has selected "Yes".

Here is the relevant html code on my reservations page:

<?php
 session_start();
 if (isset($_POST['submit'])) { 
    $_SESSION['party'] = $_POST['party'];
}

 if ( !empty($_POST['vip'])) 
    $_SESSION['vip'] = $_POST['vip'];
?>

...

 <strong>Select Party Size* :</strong>
 <br>
 <select name="party" id="party" value="<?php echo $party;?>">
 <option value="">Please Select</option>
 <option <?php if (isset($party) && $party=="1") echo "selected";?> value="1">1 Person (+£5)</option>
 <option <?php if (isset($party) && $party=="2") echo "selected";?> value="2">2 People (+£10)</option>
 <option <?php if (isset($party) && $party=="3") echo "selected";?> value="3">3 People (+£15)</option>
 <option <?php if (isset($party) && $party=="4") echo "selected";?> value="4">4 People (+£20)</option>
 <option <?php if (isset($party) && $party=="5") echo "selected";?> value="5">5 People (+£25)</option>
 <option <?php if (isset($party) && $party=="6") echo "selected";?> value="6">6 People (+£30)</option>
 <option <?php if (isset($party) && $party=="7") echo "selected";?> value="7">7 People (+£35)</option>
 <option <?php if (isset($party) && $party=="8") echo "selected";?> value="8">8 People (+£40)</option>
 <option <?php if (isset($party) && $party=="9") echo "selected";?> value="9">9 People (+£45)</option>
 <option <?php if (isset($party) && $party=="10") echo "selected";?> value="10">10+ People (+£50)</option>
 </select>

 <strong> VIP area* : </strong> <br><br>
 Yes (+£5) <input type="radio" name="vip" <?php if (isset($vip) && $vip=="Yes") echo "checked";?> value="Yes">
 <br><span id="vip" class="error"><?php echo $vipErr;?></span><br>
 No <input type="radio" name="vip" <?php if (isset($vip) && $vip=="No") echo "checked";?> value="No">

Here is the relevant calculation on the confirmation page:

<b>Total Reservation Costs: </b><br><br> £
<?php
     if (isset($_SESSION['party']) && is_numeric($_SESSION['party'])) {
         $party = (int)$_SESSION['party'];
         $vip = isset($_SESSION['vip']) ? 5 : 0;
         echo (($party * 5) + $vip);
     }
?>

Does anyone know how I can enable the calculation to only add the additional £5 for VIP area seating if the user has selected "Yes"? I am new to web languages so any help will be appreciated. Thank you!

You could improve your code by adding a loop "for" from 1 to 10 to make your <option>s ;) — Veve
– Veve, Commented Nov 22, 2014 at 14:42

Nathan · Accepted Answer · 2014-11-22 15:45:21Z

1

Your ternary statement is currently only checking if $_SESSION['vip'] is set. The problem is that it is set no matter what the user chooses ("yes" or "no") because in the part where you set the session variable, the conditional statement sets the session variable if $_POST['vip'] isn't empty.

Try checking if it's equal to "Yes" like this:

<b>Total Reservation Costs: </b><br><br> £
<?php
     if (isset($_SESSION['party']) && is_numeric($_SESSION['party'])) {
         $party = (int)$_SESSION['party'];
         $vip = (isset($_SESSION['vip']) && $_SESSION['vip'] === 'Yes') ? 5 : 0;
         echo (($party * 5) + $vip);
     }
?>

answered Nov 22, 2014 at 15:45

Nathan

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Comments

DarkBee · Accepted Answer · 2014-11-22 14:44:45Z

1

There is no need to complicate your code with sessions. Just use the $_POST :

<?php

$options =  array(
    1 => '1 Person (+£5)',
    2 => '2 Person (+£10)',
    3 => '3 Person (+£15)',
    ...
    10 => '1 Person (+£50)',
);
?>
...

<select name="party" id="party">
    <option value="-1">Please select</option>
    <?php
        foreach ($options as $persons => $text) {
            $selected = isset($_POST['party'] && $_POST['party'] == $persons) ? ' selected' : '';
            echo '<option value="'.$persons.'".'.$selected.'>'.$text.'</option>';
        }
    ?>
</select>
...
Yes <input type="radio" value="1" name="vip" id="vip_yes" <?php if (isset($_POST['vip']) && $_POST['vip'] == 1) { echo 'checked' } ?> />
No <input type="radio" value="0" name="vip" id="vip_no" <?php if (isset($_POST['vip']) && $_POST['vip'] == 0) { echo 'checked' } ?> />


...
<?php 
    if (isset($_POST['party']) && is_numeric($_POST['party']) ) {
        $price = $_POST['party'] * 5;
        if (isset($_POST['vip']) && $_POST['vip'] == 1) $price += 5;
    }else{
        //-- no post    
        $price = 0;
    }

    if ($price == 0) {
        echo 'Please select your options..';
    }else{
        echo 'Your price is £' . $price;
    }
?>

answered Nov 22, 2014 at 14:44

DarkBee

14.4k9 gold badges86 silver badges134 bronze badges

1 Comment

Oscar Over a year ago

Thanks for your feedback, is there any way to do this with sessions? I only ask because I dont have a lot of time to change my code and I have everything functioning apart from this one calculation :(

Glynne Turner · Accepted Answer · 2014-11-22 14:39:46Z

0

you need to redo the calculation and add a variable for the cost of being VIP. something like

if ($vip  =='Yes'){$vip_cost = +5)}else{$vip_cost=0}

then add the cost of each person and add the $vip_cost

answered Nov 22, 2014 at 14:39

Glynne Turner

1496 bronze badges

1 Comment

Oscar Over a year ago

Could you give me a better example? I am new to php and this is quite hard for me to understand sorry

Glynne Turner · Accepted Answer · 2014-11-22 15:19:55Z

<?php
$form = '<form action="'.$_SERVER['PHP_SELF'].'" method="post">
<select name="party" id="party" >
    <option value="">--Select People--</option>      
    <option value="1">1 People (+£5)</option>        
    <option value="2">2 People (+£10)</option>       
    <option value="3">3 People (+£15)</option>       
    <option value="4">4 People (+£20)</option>       
    <option value="5">5 People (+£25)</option>       
    <option value="6">6 People (+£30)</option>       
    <option value="7">7 People (+£35)</option>       
    <option value="8">8 People (+£40)</option>       
    <option value="9">9 People (+£45)</option> 
    <option value="10">10People (+£50)</option>
</select><br /><br />

<strong> VIP area* : </strong> <br><br>
Yes (+£5) <input type="radio" name="vip" value="Yes">
No <input type="radio" name="vip" value="No"><br /><br />

<input type="submit" value="Submit" name="submit" />';


if (isset($_POST['submit'])){
if (isset($_POST['party'])){
    $cost =  ($_POST['party'] *5); 
    if ($_POST['vip']  =='Yes'){$vip_cost = 5; $vip_text='with vip';}else{$vip_cost=0; $vip_text='';}
$total = ($cost+$vip_cost);
$display = 'Your reservation for '.$_POST['paty'].' '.$vip_text.' will cost '.$total; 
}
    else{
 $display = 'Please Select How Many People<br />'.$form;}

}else{

$display=$form;
}

?>
<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">
<html xmlns="http://www.w3.org/1999/xhtml">
<head>
<meta http-equiv="Content-Type" content="text/html; charset=utf-8" />
<title>Untitled Document</title>
</head>

<body>
<?php echo $display;?>
</body>
</html>

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