What is the regex pattern for datetime (2008-09-01 12:35:45 )?

Question

What is the RegEx pattern for DateTime (2008-09-01 12:35:45 ) ?

I get this error:

No ending delimiter '^' found

Using:

preg_match('(?n:^(?=\d)((?<day>31(?!(.0?[2469]|11))|30(?!.0?2)|29(?(.0?2)(?=.{3,4}(1[6-9]|[2-9]\d)(0[48]|[2468][048]|[13579][26])|(16|[2468][048]|[3579][26])00))|0?[1-9]|1\d|2[0-8])(?<sep>[/.-])(?<month>0?[1-9]|1[012])\2(?<year>(1[6-9]|[2-9]\d)\d{2})(?:(?=\x20\d)\x20|$))?(?<time>((0?[1-9]|1[012])(:[0-5]\d){0,2}(?i:\ [AP]M))|([01]\d|2[0-3])(:[0-5]\d){1,2})?$)', '2008-09-01 12:35:45');

Gives this error:

Warning: preg_match() [function.preg-match]: Compilation failed: nothing to repeat at offset 0 in E:\www\index.php on line 19

Answer 1

@Espo: I just have to say that regex is incredible. I'd hate to have to write the code that did something useful with the matches, such as if you wanted to actually find out what date and time the user typed.

It seems like Tom's solution would be more tenable, as it is about a zillion times simpler and with the addition of some parentheses you can easily get at the values the user typed:

(\d{4})-(\d{2})-(\d{2}) (\d{2}):(\d{2}):(\d{2})

If you're using perl, then you can get the values out with something like this:

$year = $1;
$month = $2;
$day = $3;
$hour = $4;
$minute = $5;
$second = $6;

Other languages will have a similar capability. Note that you will need to make some minor mods to the regex if you want to accept values such as single-digit months.

Answer 2

A simple version that will work for the format mentioned, but not all the others as per @Espos:

(\d{4}-\d{2}-\d{2} \d{2}:\d{2}:\d{2}) 

Answer 3

http://regexlib.com/REDetails.aspx?regexp_id=610

^(?=\d)(?:(?:31(?!.(?:0?[2469]|11))|(?:30|29)(?!.0?2)|29(?=.0?2.(?:(?:(?:1[6-9]|[2-9]\d)?(?:0[48]|[2468][048]|[13579][26])|(?:(?:16|[2468][048]|[3579][26])00)))(?:\x20|$))|(?:2[0-8]|1\d|0?[1-9]))([-./])(?:1[012]|0?[1-9])\1(?:1[6-9]|[2-9]\d)?\d\d(?:(?=\x20\d)\x20|$))?(((0?[1-9]|1[012])(:[0-5]\d){0,2}(\x20[AP]M))|([01]\d|2[0-3])(:[0-5]\d){1,2})?$

---

This RE validates both dates and/or times patterns. Days in Feb. are also validated for Leap years. Dates: in dd/mm/yyyy or d/m/yy format between 1/1/1600 - 31/12/9999. Leading zeroes are optional. Date separators can be either matching dashes(-), slashes(/) or periods(.) Times: in the hh:MM:ss AM/PM 12 hour format (12:00 AM - 11:59:59 PM) or hh:MM:ss military time format (00:00:00 - 23:59:59). The 12 hour time format: 1) may have a leading zero for the hour. 2) Minutes and seconds are optional for the 12 hour format 3) AM or PM is required and case sensitive. Military time 1) must have a leading zero for all hours less than 10. 2) Minutes are manditory. 3) seconds are optional. Datetimes: combination of the above formats. A date first then a time separated by a space. ex) dd/mm/yyyy hh:MM:ss

---

Edit: Make sure you copy the RegEx from the regexlib.com website as StackOverflow sometimes removes/destroys special chars.

Answer 4

^([2][0]\d{2}\/([0]\d|[1][0-2])\/([0-2]\d|[3][0-1]))$|^([2][0]\d{2}\/([0]\d|[1][0-2])\/([0-2]\d|[3][0-1])\s([0-1]\d|[2][0-3])\:[0-5]\d\:[0-5]\d)$

Answer 5

$date = "2014-04-01 12:00:00";

preg_match('/(\d{4})-(\d{2})-(\d{2}) (\d{2}):(\d{2}):(\d{2})/',$date, $matches);

print_r($matches);

$matches will be:

Array ( 
   [0] => 2014-04-01 12:00:00 
   [1] => 2014 
   [2] => 04 
   [3] => 01 
   [4] => 12 
   [5] => 00 
   [6] => 00
)

An easy way to break up a datetime formated string.

Answer 6

regarding to Imran answer from Sep 1th 2008 at 12:33 there is a missing : in the pattern the correct patterns are

preg_match('/\d{4}-\d{2}-\d{2} \d{2}:\d{2}:\d{2}/', '2008-09-01 12:35:45', $m1);
print_r( $m1 );
preg_match('/\d{4}-\d{2}-\d{2} \d{1,2}:\d{2}:\d{2}/', '2008-09-01 12:35:45', $m2);
print_r( $m2 );
preg_match('/^\d{4}-\d{2}-\d{2} \d{2}:\d{2}:\d{2}$/', '2008-09-01 12:35:45', $m3);
print_r( $m3 );

this returns

Array ( [0] => 2008-09-01 12:35:45 )
Array ( [0] => 2008-09-01 12:35:45 )
Array ( [0] => 2008-09-01 12:35:45 ) 

Answer 7

Here is my solution:

/^(2[0-9]{3})-(0[1-9]|1[012])-(0[1-9]|[12][0-9]|3[01]) (0[0-9]|1[0-9]|2[0123])\:([012345][0-9])\:([012345][0-9])$/u

Answer 8

I have modified the regex pattern from http://regexlib.com/REDetails.aspx?regexp_id=610. The following pattern should match your case.

^(?=\d)(?:(?:1[6-9]|[2-9]\d)?\d\d([-.\/])(?:1[012]|0?[1-9])\1(?:31(?<!.(?:0[2469]|11))|(?:30|29)(?<!.02)|29(?=.0?2.(?:(?:(?:1[6-9]|[2-9]\d)?(?:0[48]|[2468][048]|[13579][26])|(?:(?:16|[2468][048]|[3579][26])00)))(?:\x20|$))|(?:2[0-8]|1\d|0?[1-9]))(?:(?=\x20\d)\x20|$))?(((0?[1-9]|1[012])(:[0-5]\d){0,2}(\x20[AP]M))|([01]\d|2[0-3])(:[0-5]\d){1,2})?$

YYYY-MM-DD HH:MM:SS

Answer 9

PHP preg functions needs your regex to be wrapped with a delimiter character, which can be any character. You can't use this delimiter character without escaping inside the regex. This should work (here the delimiter character is /):

preg_match('/\d{4}-\d{2}-\d{2} \d{2}:\d{2}:\d{2}/', '2008-09-01 12:35:45');

// or this, to allow matching 0:00:00 time too.
preg_match('/\d{4}-\d{2}-\d{2} \d{1,2}:\d{2}:\d{2}/', '2008-09-01 12:35:45');

If you need to match lines that contain only datetime, add ^ and $ at the beginning and end of the regex.

preg_match('/^\d{4}-\d{2}-\d{2} \d{2}:\d{2}:\d{2}$/', '2008-09-01 12:35:45');

Link to PHP Manual's preg_match()

Answer 10

Here is a simplified version (originated from Espo's answer). It checks the correctness of date (even leap year), and hh:mm:ss is optional
Examples that work:
- 31/12/2003 11:59:59
- 29-2-2004

^(?=\d)(?:(?:31(?!.(?:0?[2469]|11))|(?:30|29)(?!.0?2)|29(?=.0?2.(?:(?:(?:1[6-9]|[2-9]\d)?(?:0[48]|[2468][048]|[13579][26])|(?:(?:16|[2468][048]|[3579][26])00)))(?:\x20|$))|(?:2[0-8]|1\d|0?[1-9]))([-./])(?:1[012]|0?[1-9])\1(?:1[6-9]|[2-9]\d)?\d\d(?:(?=\x20\d)\x20|$))(|([01]\d|2[0-3])(:[0-5]\d){1,2})?$

Answer 11

Here is my solution:

[12]\d{3}-(0[1-9]|1[0-2])-(0[1-9]|[12]\d|3[01]) ([01][0-9]|2[0-3]):[0-5]\d

Debuggex Demo

https://regex101.com/r/lbthaT/4