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)
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)); 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 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); } Just make up more conditions
I would use strtotime AND strftime. Is a much simpler way of doing it.
By example, if a have a date string like "Oct 20 18:29:50 2001 GMT" and I want to get it in format day/month/year I could do:
$mystring = "Oct 20 18:29:50 2001 GMT"; printf("Original string: %s\n", $mystring); $newstring = strftime("%d/%m/%Y", strtotime($mystring)); printf("Data in format day/month/year is: %s\n", $newstring);