Format a date string in PHP

Question

If I have a string which represents a date, like "2011/07/01" (which is 1st July 2011) , how would I output that in more readable forms, like:

1 July 2011
1 Jul 2011  (month as three letters)

And also, how could I make it intelligently show date ranges like "2011/07/01" to "2011/07/11" as

1 - 11 July 2001

(without repeating the 'July' and '2011' in this case)

Answer 1

You can convert your date to a timestamp using strtotime() and then use date() on that timestamp. On your example:

$date = date("j F Y", strtotime("2011/07/01")); // 1 July 2011
$date = date("j M Y", strtotime("2011/07/01")); // 1 Jul 2011

Answer 2

As for the second one:

$time1 = time();
$time2 = $time1 + 345600; // 4 days
if( date("j",$time1) != date("j",$time2) && date("FY",$time1) == date("FY",$time2) ){
   echo date("j",$time1)." - ".date("j F Y",$time2);
}

Can be seen in action here

Just make up more conditions

Answer 3

As NullUserException mentioned, you can use strtotime to convert the date strings to timestamps. You can output 'intelligent' ranges by using a different date format for the first date, determined by comparing the years, months and days:

$date1 = "2011/07/01";
$date2 = "2011/07/11";

$t1 = strtotime($date1);
$t2 = strtotime($date2);

// get date and time information from timestamps
$d1 = getdate($t1);
$d2 = getdate($t2);

// three possible formats for the first date
$long = "j F Y";
$medium = "j F";
$short = "j";

// decide which format to use
if ($d1["year"] != $d2["year"]) {
    $first_format = $long;
} elseif ($d1["mon"] != $d2["mon"]) {
    $first_format = $medium;
} else {
    $first_format = $short;
}

printf("%s - %s\n", date($first_format, $t1), date($long, $t2));

Answer 4

Nice i have an tutorial for all type of php date formats see http://www.motyar.info/2010/08/format-php-date.html