Adding two variables - php

Question

I am in the process of designing a php validated form for a restaurant website which directs the user to a thank you page, displaying their reservation details, once they have pressed submit. Currently I have 2 variables, one named 'party', the other named 'vip'. These variables both display a string value when echoed. However, I wish to assign them an int value and then add them together to display a total price to the user.

Currently I have this code which does not seem to function:

<b>Total Reservation Costs: </b> £
 <?php
 if ( $party == "1 Person (+£5)" )
   {
        $party = 5;
   }
    elseif if ( $party == "2 People (+£10)" )
   {
         $party = 10;
   }
   elseif if ( $party == "3 People (+£15)" )
   {
         $party = 15;
   }
   elseif if ( $party == "4 People (+£20)" )
   {
         $party = 20;
   }
    elseif if ( $party == "5 People (+£25)" )
   {
        $party = 25;
   }
   elseif if ( $party == "6 People (+£30)" )
   {
        $party = 30;
   }
    elseif if ( $party == "7 People (+£35)" )
   {
        $party = 35;
   }
   elseif if ( $party == "8 People (+£40)" )
   {
        $party = 40;
   }
   elseif if ( $party == "9 People (+£45)" )
   {
        $party = 45;
   }
   elseif if ( $party == "10+ People (+£50)" )
   {
        $party = 50;
   }

 if ( $vip == "Yes" )
   {
        $vip = 5;
   }
    elseif if ( $vip == "No" )
   {
        $vip = 0;
   }

 echo $party + $vip;
?>

If anyone knows how I can get this to work I would be very grateful, I am new to web languages so I apologise in advance if my answer isn't very clear. Thank you.

ummm elseif if I guess it works but it's not what you mean to write. — robbmj
– robbmj, Commented Nov 20, 2014 at 22:32

robbmj · Accepted Answer · 2014-11-21 14:21:19Z

2

Instead of using a string to determine the number of people why not just use an int

Here is a complete program

<form action="" method="POST">
    <select name="party">
        <option value="1">1 People (+£5)</option> 
        <option value="2">2 People (+£10)</option>
        <!-- ... -->
        <option value="10">10+ People (+£50)</option>
    </select>
    <input type="checkbox" name="vip" />
    <input type="submit" value="Confirm Reservation" /> 
</form>

<?php
if (isset($_POST['party']) && is_numeric($_POST['party'])) {
    $party = (int)$_POST['party'];
    $vip = isset($_POST['vip']) ? 5 : 0;
    echo "Total is: " . (($party * 5) + $vip);
}

and elseif if should be just else if

edited Nov 21, 2014 at 14:21

answered Nov 20, 2014 at 22:35

robbmj

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7 Comments

Oscar Over a year ago

Because as the user is displayed their reservation details I wish for them to be told on page "Party size: 4 people (+£20)". I then wish to take this 20 from their selection and add it to the 5 if they have selected VIP area seating. Although, if you have a simplified method to make this work, please let me know

robbmj Over a year ago

@JamesPatterson please see the updated answer for a nice way of implementing your form and form substitution script.

Oscar Over a year ago

I applied your suggestions however I still am not presented with a total cost :( Although, no errors are thrown up either. Why might this be happening?

robbmj Over a year ago

@JamesPatterson Yeah, is_int should have been is_numeric. Give it a go now.

Oscar Over a year ago

Still the same result, also I notice that after the "Total Reservation Costs: " line, the page is cut off? :(

|

T90 · Accepted Answer · 2014-11-20 22:44:40Z

0

It's not elseif if, its elseif

Total Reservation Costs:  £
<?php
    if ( $party == "1 Person (+£5)" ){
    $party = 5;
    }
    elseif( $party == "2 People (+£10)" ){
    $party = 10;
    }
    elseif( $party == "3 People (+£15)" ){
    $party = 15;
    }
    elseif( $party == "4 People (+£20)" ){
    $party = 20;
    }
    elseif( $party == "5 People (+£25)" ){
    $party = 25;
    }
    elseif ( $party == "6 People (+£30)" ){
    $party = 30;
    }
    elseif( $party == "7 People (+£35)" ){
    $party = 35;
    }
    elseif( $party == "8 People (+£40)" ){
    $party = 40;
    }
    elseif( $party == "9 People (+£45)" ){
    $party = 45;
    }
    elseif( $party == "10+ People (+£50)" ){
    $party = 50;
    }
    else{    $party = 0;
    }
    if ( $vip == "Yes" ){
    $vip = 5;
    }
    elseif ( $vip == "No" ){
    $vip = 0;
    }
    else{
    $vip = 0;
    }
    echo $party + $vip;
?>

`

I'd suggest adding an else statement too, as i have added in my answer, to handle any other cases that might occur.

answered Nov 20, 2014 at 22:44

T90

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4 Comments

Oscar Over a year ago

Hi friend, I tried your suggestion however I just got no such variable errors on screen once the thank you page loads. Why might this be happening? :(

Oscar Over a year ago

I get an "Notice: Undefined variable: party in" error on each line the $party is mentioned in your code and also an "Notice: Undefined variable: vip" error where $vip is mentioned :(

T90 Over a year ago

did you initialize thos variables as per your code $party = $_POST['party']; $vip = isset($_POST['vip']) ? 5 : 0; ?

Oscar Over a year ago

Sorry I am really new to weblanguages, can you explain what you mean by this? For help, here is my reservations page in its entirety: pastebin.com/UNDkAEYT - here is also my thank you page in its entirety: pastebin.com/31pdBnD5

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