find the total values get in a variable in php

Question

i am doing a shopping cart.i had a variable $total.when i add 1st product of price 200 i get $total=200 , when i add 2nd product of price 100 i get $total=200100 when i add 3rd product of price 400 i get $total=200100400 . i want to get $total=700. how to fix this problem?

 session_start();
           $ar20=Db::getInstance()->ExecuteS('SELECT * FROM '._DB_PREFIX_.'price WHERE id="'.$_SESSION['y'].'"  AND prodid="'.$row['id_product'].'" ORDER BY size DESC');

            for($i=0;$i < count($ar20);$i++)
            {
                $test=$test+$ar20[$i]['price'];
            }




             $row['price'] += $test ;
 $row['total'] = $row['price'] * intval($row['quantity']);

if 3 products then echo the $row['total'] will get this 100200300

In php + does an implicit conversion to a numeric value. This might sound like a stupid question, but are you sure you're adding the numbers together in PHP an not in a Javascript code block embedded in a PHP doc? — leepowers
– leepowers, Commented Apr 2, 2010 at 5:21
session_start(); $ar20=Db::getInstance()->ExecuteS('SELECT * FROM '._DB_PREFIX_.'price WHERE id="'.$_SESSION['y'].'" AND prodid="'.$row['id_product'].'" ORDER BY size DESC'); for($i=0;$i < count($ar20);$i++) { $test=$test+$ar20[$i]['price']; } echo $test; $row['price'] += $test ; $row['total'] = $row['price'] * intval($row['quantity']); if 3 products then echo the $row['total'] will get this 100200300 — Hmwd
– Hmwd, Commented Apr 2, 2010 at 5:36
Please edit your question instead of posting code in comments, it's messy. — deceze
– deceze ♦, Commented Apr 2, 2010 at 5:38

Your Common Sense · Accepted Answer · 2010-04-02 05:06:25Z

1

Use + instead of . operator

answered Apr 2, 2010 at 5:06

Your Common Sense

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1 Comment

Your Common Sense Over a year ago

@Hmwd So, you're doing your math in javascript, not in PHP

OM The Eternity · Accepted Answer · 2010-04-02 05:09:08Z

1

Just get the value of price each time and add it in previous on like wise, the below logic may help u say $newprice is a variable where u get the price..

$total = 0;
$total +=$newprice;

answered Apr 2, 2010 at 5:09

OM The Eternity

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Hmwd Over a year ago

i tried like as you shown but no changes, i gt the same like $total=200100400

Christian Mann · Accepted Answer · 2010-04-02 05:22:42Z

0

Are you doing this?

$total = '100'; $total += '300'; $total += '300'; echo $total;

PHP is forgiving to a point, especially to the JavaScript fans: It allows you to use the + operator as concatenation if the operands are both strings. Make sure that you don't surround them in quotes, single or double, like so:

$total = 100; $total += 300; $total += 300; echo $total;

answered Apr 2, 2010 at 5:22

Christian Mann

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1 Comment

Hmwd Over a year ago

session_start(); $ar20=Db::getInstance()->ExecuteS('SELECT * FROM '._DB_PREFIX_.'price WHERE id="'.$_SESSION['y'].'" AND prodid="'.$row['id_product'].'" ORDER BY size DESC'); for($i=0;$i < count($ar20);$i++) { $test=$test+$ar20[$i]['price']; } echo $test; $row['price'] += $test ; $row['total'] = $row['price'] * intval($row['quantity']); if 3 products then echo the $row['total'] will get this 100200300

Amy B · Accepted Answer · 2010-04-02 05:41:54Z

0

Your code isn't clear because it has no context, but this should work:

$test = (integer) $test;

for($i=0; $i < count($ar20); $i++)
{
    $test = $test + (integer) $ar20[$i]['price'];
}

$row['price'] = (integer) $row['price'] + (integer) $test;

$row['total'] = $row['price'] * intval($row['quantity']);

answered Apr 2, 2010 at 5:41

Amy B

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2 Comments

Hmwd Over a year ago

i tried but the result is same. i used the code inside a foreach loop

Amy B Over a year ago

I'm finding that kind of hard to believe, honestly.

Brant Messenger · Accepted Answer · 2010-04-02 14:32:20Z

0

session_start();

$records=Db::getInstance()->ExecuteS('SELECT * FROM '._DB_PREFIX_.'price WHERE id="'.$_SESSION['y'].'"  AND prodid="'.$row['id_product'].'" ORDER BY size DESC');

$row = array();
$test = '0';

if (count($records) < 1) {
    $row['price'] = '0';
    $row['total'] = '0';
} else {
    foreach($records as $record)
    {
        $test = bcadd($test, $record['price']);
    }

    $row['price'] = bcadd($row['price'], $test);
    $row['total'] = bcmul($row['price'], $row['quantity']);
}

PHP: bcadd

edited Apr 2, 2010 at 14:32

answered Apr 2, 2010 at 5:19

Brant Messenger

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1 Comment

Brant Messenger Over a year ago

Check the bcadd, bcmul specification for scale.

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