Question

I have a variable ($date) holding the start date of an event which is in this format:

 2012-11-20 (YYYY-M-D)

And I have another variable ($duration) holding a number e.g ‘3’ which is number of days or duration of the event. How can i increment the day in the date variable by the duration variable and then display the event date in this format

 20th - 23th November

Thanks!

Answer 1

Try this:

echo date('F d Y');

See php date() for more details

If you have 2012-11-20 in a variable, you can do this:

$date = '2012-11-20';
echo date('F d Y', strtotime($date));
//=> November 20 2012

Two things to consider:

1. If you don’t have control over the date, you might want to take extra precaution using strtotime(). If you give strtotime() an invalid date string, it will return false
2. Every call to a date/time function will generate a E_NOTICE if the time zone is not valid, and/or a E_STRICT or E_WARNING message if using the system settings or the TZ environment variable. See also date_default_timezone_set()

A more robust solution:

// date input
$date = '2012-11-20';

// set timezone
date_default_timezone_set('America/Chicago');

// validate date
if ( ($time=strtotime($date)) === false ) {
    throw new Exception("invalid date given: {$date}");
}

// output
echo date('F d Y', $time);

---

EDIT

You updated your question asking how to get a date range. For this, I’d recommend the following:

// input
$date = '2012-11-20';
$duration = 3;

// set timezone
date_default_timezone_set('America/Chicago');

// parse date
try {

    // init range
    $start = new DateTime($date);
    $end = clone $start;

    // modify end date based on $duration
    $end->modify(sprintf("+%d days", $duration));
    
    // output range
    echo $start->format('jS') . ' - ' . $end->format('jS F Y');

    //=> 20th - 23rd November 2012

}
catch (Exception $e) {
    die($e->getMessage());
}