Javascript date regex DD MM YYYY

Question

I know there are a lot of regex threads out there by I need a specific pattern I couldn t fin anywhere   This regex validates in a YYYY-MM-DD format      d 4     -  0  1-9  1 012      -  0  1-9   12  0-9  3 01       I need the pattern to be DD MM YYYY  day first since it s in spanish and only       -  should not be allowed   I searched several regex libraries and I think this one should work    but since I m not familiar with regex I m not sure it validates like that   0 1-9   12  0-9  3 01      -  0 1-9  1 012      -  19 20   d d   I also don t know ho to escape the slashes  I try to  see  the logic in the string but it s like trying  see  the Matrix code for me  I m placing the regex string in a options  js             date                          regex    0 1-9   12  0-9  3 01      -  0 1-9  1 012      -  19 20   d d                       alertText    Alert text AAAA-MM-DD                      other type            So  if the regex is ok  how would I escape it   if it s not  what s the correct regex and how do I escape it   P  Thanks a lot

User · Accepted Answer

You could take the regex that validates YYYY/MM/DD and flip it around to get what you need for DD/MM/YYYY:

/^(0?[1-9]|[12][0-9]|3[01])[\/\-](0?[1-9]|1[012])[\/\-]\d{4}$/

BTW - this regex validates for either DD/MM/YYYY or DD-MM-YYYY

P.S. This will allow dates such as 31/02/4899

User · Answer

Do the following change to the jquery validationengine-en js file and update the dd mm yyyy inline validation by including leap year    date            Check if date is valid by leap year      func   function  field          var pattern   new RegExp     d 4      -    0  1-9  1 012      -    0  1-9   12  0-9  3 01           var pattern   new RegExp    0  1-9   12  0-9  3 01      -    0  1-9  1 012      -     d 4           var match   pattern exec field val         if  match    null      return false         var year   match 1         var month   match 2  1        var day   match 3  1      var year   match 3       var month   match 2  1      var day   match 1  1      var date   new Date year  month - 1  day      because months starts from 0       return  date getFullYear      year  amp  amp  date getMonth       month - 1   amp  amp  date getDate      day       alertText      Invalid date  must be in DD-MM-YYYY format

User · Answer

I build this regular to check month 30 31 and let february to 29   new RegExp     0 1-9   12  0-9  3 01       0 13578  1 02      0 1-9   12  0-9       02     0 1-9   12  0-9  3 0       0 469  11       d 4       I think  it s more simple and more flexible and enough full   Perhaps first part can be contract but I Don t find properly

User · Answer

Take a look from here https   www regextester com  fam 114662  Use this following Regular Expression Details  This will support leap year also   var reg        0 1-9   12  d 3 01     0 13578  1 02      19  2-9  d  d 2      0 1-9   12  d 30    0 13456789  1 012      19  2-9  d  d 2      0 1-9  1 d 2 0-8    02    19  2-9  d  d 2     29  02    1 6-9   2-9  d  0 48   2468  048   13579  26      1  26   2468  048   3579  26  00      g    Example

User · Answer

d 4   d 4      a-zA-Z  3   a-zA-Z  3   d 2             -      0-9  2   0-9  2      0-9  1 2   0-9  1 2   a-zA-Z  3             -      0-9  4   0-9  4      0-9  2   0-9  2   a-zA-Z  3     Regex Compile on it  2012 22 Jan 2012 22 12  2012 22 12 2012 22 12 2012 22 12 2012 22 12 2012 22 12 2012-Dec-22 2012-12-22 23 12 2012 23 12 2012 Dec-22-2012 12-2-2012 23-12-2012 23-12-2012

User · Answer

This validates date like dd-mm-yyyy   0-2  0-9   3  0-1    -    0  0-9     1  0-2     -   0-9  0-9  0-9  0-9    This can use with javascript like angular reactive forms

User · Answer

Try using this     0-9  2     0-9  2     0-9  4     this should work with this pattern DD DD DDDD where D is any digit  0-9

User · Answer

If you are in Javascript already   couldn t you just use Date Parse   to validate a date instead of using regEx   RegEx for date is actually unwieldy and hard to get right especially with leap years and all

User · Answer

A regex is good for matching the general format but I think you should move parsing to the Date class  e g    function parseDate str      var m   str match     d 1 2      d 1 2      d 4         return  m    new Date m 3   m 2 -1  m 1     null      Now you can use this function to check for valid dates  however  if you need to actually validate without rolling  e g   31 2 2010  doesn t automatically roll to  3 3 2010   then you ve got another problem    Edit  If you also want to validate without rolling then you could add a check to compare against the original string to make sure it is the same date   function parseDate str      var m   str match     d 1 2      d 1 2      d 4            d    m    new Date m 3   m 2 -1  m 1     null       nonRolling    d amp  amp  str   d getDate   d getMonth   1 d getFullYear    join           return  nonRolling    d   null       Edit2  If you want to match against zero-padded dates  e g   08 08 2013   then you could do something like this   function parseDate str      function pad x  return       x  length  2          0     x      var m   str match     d 1 2      d 1 2      d 4            d    m    new Date m 3   m 2 -1  m 1     null       matchesPadded    d amp  amp  str   pad d getDate    pad d getMonth   1  d getFullYear    join              matchesNonPadded    d amp  amp  str   d getDate   d getMonth   1 d getFullYear    join           return  matchesPadded    matchesNonPadded    d   null      However  it will still fail for inconsistently padded dates  e g   8 08 2013

User · Answer

Scape slashes is simply use   before   and it will be escaped       gt      Otherwise you re regex DD MM YYYY could be next     0-9  2      1  0-9  2      1  0-9  4   g Explanation    0-9   Just Numbers  2  or  4   Length 2 or 4  You could do  2 4  as well to length between two numbers  2 and 4 in this case        Character   g   Global -- Or m  Multiline  Optional  see your requirements     Anchor to end of string   Optional  see your requirements     Start of string   Optional  see your requirements   An example of use   x000D   x000D  var regex      0-9  2      0-9  2      0-9  4   g   var dates     2009-10-09    2009 10 09    2009 10 09    200910-09    1990 10 09         2016 0 09    2017 10 09    2016 09 09    20 09 2016    21 09 2016    22 09 2016        23 09 2016    19 09 2016    18 09 2016    25 09 2016    21 09 2018       Iterate array dates forEach      function date           console log date     matches with regex           console log regex test date            x000D   x000D   x000D   Of course you can use as boolean  if regex test date           do something

User · Answer

For people who needs to validate years earlier than year 1900  following should do the trick  Actually this is same as the above answer given by   OammieR  1  BUT with years including 1800 - 1899        0 1-9   12  d 3 01     0 13578  1 02      19  2-9  d  d 2      0 1-9   12  d 3 01     0 13578  1 02      18  2-9  d  d 2      0 1-9   12  d 30    0 13456789  1 012      19  2-9  d  d 2      0 1-9   12  d 30    0 13456789  1 012      18  2-9  d  d 2      0 1-9  1 d 2 0-8    02    19  2-9  d  d 2     29  02    1 6-9   2-9  d  0 48   2468  048   13579  26     16  2468  048   3579  26  00         Hope this helps someone who needs to validate years earlier than 1900  such as 01 01 1855  etc   Thanks  OammieR for the initial idea

User · Answer

I use this function for dd mm yyyy format         new Date    fromString  3 9 2013     3 of september     new Date    fromString  3 9 2013   false    9 of march Date prototype fromString   function str  ddmmyyyy        var m   str match    d   -      d     -         d   s   d     d        d           d              if m 2                   if ddmmyyyy     false              return new Date  m 4    m 1  - 1   m 3   m 5     m 5    0  m 6     m 6    0  m 7     m 7    0  m 8     m 8    100   0           return new Date  m 4    m 3  - 1   m 1   m 5     m 5    0  m 6     m 6    0  m 7     m 7    0  m 8     m 8    100   0             return new Date  m 1    m 3  - 1   m 4   m 5     m 5    0  m 6     m 6    0  m 7     m 7    0  m 8     m 8    100   0

[javascript] Javascript date regex DD/MM/YYYY

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