How to use Variables in PHP Function Argument?

Question

I currently have this code.

function outputCalendarByDateRange($client, $startDate="2011-06-22", 
                               $endDate='2011-06-26')

I want $startDate and $endDate to reflect today's date and the date three days from now with it automatically updating. I've tried using

$startDate=date("Y-m-D")
$endDate=strtotime(date("Y-m-d", strtotime($todayDate)) . " +3 days");

and

$date1=date("Y-m-D")
$date2=strtotime(date("Y-m-d", strtotime($todayDate)) . " +3 days");
    function outputCalendarByDateRange($client, $startDate=$date1, 
                               $endDate=$date2)

none of these work. How do I make it work?

Thanks!

Your first $endDate is highly overdone. A simple $endDate = date('Y-m-D', strtotime('+3 days')) will do. — Marc B
– Marc B, Commented Jun 23, 2011 at 3:34
@Marc B: Actually, you'd run into problems right around midnight each day doing it that way - better to base it off of $startDate: $endDate = date('Y-m-d', strtotime($startDate . '+3 days')) to handle date changes during execution of the two statements. It will still be off a day by the time the script gets to the third line of code, but at least the two dates will be off uniformly... — ETWW-Dave
– ETWW-Dave, Commented Jun 23, 2011 at 3:40
@Et: in that case, it'd be best to store the raw time value for today, rather than roundtripping everything through strtotime extra times. It's a magical function, but it's NOT efficient. — Marc B
– Marc B, Commented Jun 23, 2011 at 4:01

zzzzBov · Accepted Answer · 2011-06-23 03:22:05Z

4

you can't use a statement in a function declaration, but you can set the value to null and check it at runtime:

function foo( $bar = null )
{
  if (is_null($bar))
  {
    $bar = 'baz';
  }
  ...code...
}

answered Jun 23, 2011 at 3:22

zzzzBov

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1 Comment

zzzzBov Over a year ago

also, empty could be used for the check in some cases where array(), "", null, false, etc should trigger the default.

mario · Accepted Answer · 2011-06-23 03:31:46Z

3

You cannot have expressions in the function declaration. But constants could be a workaround for what you want to do.

define("FUNC_CAL_DATE1", date("Y-m-D"));
define("FUNC_CAL_DATE2", strtotime(date("Y-m-d",strtotime($to...

function outputCalendarByDateRange($client,
          $startDate=FUNC_CAL_DATE1, $endDate=FUNC_CAL_DATE2) {

They are actually expressions too, but are specially handled in this context and work where the =$date1 wouldn't.

answered Jun 23, 2011 at 3:31

mario

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1 Comment

zzzzBov Over a year ago

I'm curious as to whether there could be issues due to caching.

ETWW-Dave · Accepted Answer · 2011-06-23 03:37:08Z

You can't pass variables as default values. See below for a possible solution to what you're trying to achieve:

<?php
    error_reporting(E_ALL);

    $defaultStartDate = date("Y-m-d");
    $defaultEndDate   = date("Y-m-d", strtotime($defaultStartDate . " + 3 days"));

    function outputCalendarByDateRange($client, $startDate="", $endDate="") {
        global $defaultStartDate, $defaultEndDate;

        if ($startDate === "") {
            $startDate = $defaultStartDate;
        }
        if ($endDate === "") {
            $endDate = $defaultEndDate;
        }
        echo "Client: " . $client . "<br />";
        echo "Start Date: " . $startDate . "<br />";
        echo "End Date: " . $endDate . "<br />";
    }

    outputCalendarByDateRange("Test Client");

    echo "<br />";

    outputCalendarByDateRange("Test Client #2", date("Y-m-d", strtotime("2011-06-01")), date("Y-m-d", strtotime("2011-07-01")));

?>

Output:

Client: Test Client
Start Date: 2011-06-23
End Date: 2011-06-26

Client: Test Client #2
Start Date: 2011-06-01
End Date: 2011-07-01

deceze · Accepted Answer · 2011-06-23 03:23:45Z

0

You can't have variable default argument values, you'll have to solve this in code:

function outputCalendarByDateRange($client, $startDate = null, $endDate = null) {
    $startDate = $startDate ? $startDate : date('Y-m-d');
    $endDate = $endDate ? $endDate : date('Y-m-d', strtotime('+3 days'));

    ...
}

Calling this function without the second and third argument will use the current date/current date +3, calling it with arguments you can specify your own values.

answered Jun 23, 2011 at 3:23

deceze♦

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Comments

Steve Robbins · Accepted Answer · 2011-06-23 03:43:29Z

-1

I'm gonna take a guess and say that you're trying to set $date1 equal to the variable in the function. That's not necessary, just list them in order. function outputCalendarByDateRange($client, $date1, $date2)

answered Jun 23, 2011 at 3:43

Steve Robbins

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