PHP Passing variables to a function and losing values

Question

I am passing some variables to a function to round and add a dollars sign but when the values get into the function they have lost their values.

formatDollars($cost, $taxedCost, $shipping, $total)

function formatDollars($cost, $taxedCost, $shipping, $total) {
    $taxedCost = '$'.round($taxedCost, 2);
    $shipping = '$'.round($shipping, 2);
    $total = '$'.round($total, 2);
    $cost = '$'.round($cost, 2);
    return array($cost, $taxedCost, $shipping, $total);
}

list($cost, $taxedCost, $shipping, $total) = formatDollars();

When I output I get the dollar signs but all my numbers have become zero.

Please make an example how you use it, what output you get and what you would expect! formatDollars(); ?! Where are the arguments ? — Rizier123
– Rizier123, Commented Mar 26, 2015 at 13:57

kainaw · Accepted Answer · 2015-03-26 14:10:10Z

1

What you are doing is a very round-about way of handling this. You want to change the values of the parameters, so make them pass-by-reference.

formatDollars($cost, $taxedCost, $shipping, $total);
function formatDollars(&$cost, &$taxedCost, &$shipping, &$total)
{
    $taxedCost = '$'.round($taxedCost, 2);
    $shipping = '$'.round($shipping, 2);
    $total = '$'.round($total, 2);
    $cost = '$'.round($cost, 2);
}

Now, the variables are passed in and any changes made to them inside the function actually change them. You don't need to return anything.

By the way - your function failed because the second call (with the list command) did not pass any parameters into the function.

Also - I would read up on number_format. If you round(3,2), you get 3, not 3.00.

edited Mar 26, 2015 at 14:10

answered Mar 26, 2015 at 14:00

kainaw

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6 Comments

BlueBarren Over a year ago

I tried what you suggested but it hates my line formatDollars($cost, $taxedCost, $shipping, $total);

tmt Over a year ago

You have a syntax error on the first line (missing ;).

BlueBarren Over a year ago

@cascaval I have the semi colon on the actual code and it still doesn't like that line

kainaw Over a year ago

Computers do not have the capacity to "hate". If it throws an error, you must tell us what the error is. Otherwise, you just look foolish.

BlueBarren Over a year ago

The error I get when I try your code , "Fatal error: Call to undefined function formatDollars() in C:\inetpub\wwwroot\BTassign4.php on line 52" and line 52 is the line I mentioned early

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Sumesh mani · Accepted Answer · 2015-03-26 14:14:14Z

1

When followed by $ sign it may considered as variable... You may follow below code

formatDollars($cost, $taxedCost, $shipping, $total)

 function formatDollars($cost, $taxedCost, $shipping, $total) {
 setlocale(LC_MONETARY, 'en_US');
 $taxedCost = round($taxedCost, 2);
 $taxedCost =money_format('%i', $taxedCost) 
 return array( $taxedCost);
  }

edited Mar 26, 2015 at 14:14

answered Mar 26, 2015 at 14:04

Sumesh mani

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Comments

BlueBarren · Accepted Answer · 2015-03-26 17:34:28Z

0

I ended up finding the answer to my own quesion

In the line that I send variables to the function it did not like not having a variable to equal to, so when I ended up saying my variables equalled sending to the function.

function formatDollars($numberFormatted)
    {
        $numberFormat = '$'.round($numberFormatted, 2);
        return $numberFormat;
    }
    $cost = formatDollars($cost);
    $taxedCost = formatDollars($taxedCost);
    $shipping = formatDollars($shipping);
    $total = formatDollars($total);

answered Mar 26, 2015 at 17:34

BlueBarren

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