Python regex to match dates

I find the below RE working fine for Date in the following format;

1. 14-11-2017
2. 14.11.2017
3. 14|11|2017

It can accept year from 2000-2099

Please do not forget to add $ at the end,if not it accept 14-11-201 or 20177

date="13-11-2017"

x=re.search("^([1-9] |1[0-9]| 2[0-9]|3[0-1])(.|-)([1-9] |1[0-2])(.|-|)20[0-9][0-9]$",date)

x.group()

output = ’13-11-2017′

Using this regular expression you can validate different kinds of Date/Time samples, just a little change is needed.

^dddd/(0?[1-9]|1[0-2])/(0?[1-9]|[12][0-9]|3[01]) (00|[0-9]|1[0-9]|2[0-3]):([0-9]|[0-5][0-9]):([0-9]|[0-5][0-9])$ –>validate this: 2018/7/12 13:00:00

for your format you cad change it to:

^(0?[1-9]|[12][0-9]|3[01])/(0?[1-9]|1[0-2])/dd$ –> validates this: 11/12/98

Well, from my understanding, simply for matching this format in a given string, I prefer this regular expression:

pattern='[0-9|/]+'

to match the format in a more strict way, the following works:

pattern='(?:[0-9]{2}/){2}[0-9]{2}'

Personally, I cannot agree with unutbu’s answer since sometimes we use regular expression for “finding” and “extract”, not only “validating”.

Sometimes we need to get the date from a string.
One example with grouping:

record = '1518-09-06 00:57 some-alphanumeric-charecter'
pattern_date_time = ([0-9]{4}-[0-9]{2}-[0-9]{2} [0-9]{2}:[0-9]{2}) .+
match = re.match(pattern_date_time, record)
if match is not None:
  group = match.group()
  date = group[0]
  print(date) // outputs 1518-09-06 00:57

I built my solution on top of @aditya Prakash appraoch:

 print(re.search("^([1-9]|0[1-9]|1[0-9]|2[0-9]|3[0-1])(.|-|/)([1-9]|0[1-9]|1[0-2])(.|-|/)([0-9][0-9]|19[0-9][0-9]|20[0-9][0-9])$|^([0-9][0-9]|19[0-9][0-9]|20[0-9][0-9])(.|-|/)([1-9]|0[1-9]|1[0-2])(.|-|/)([1-9]|0[1-9]|1[0-9]|2[0-9]|3[0-1])$",'01/01/2018'))

The first part (^([1-9]|0[1-9]|1[0-9]|2[0-9]|3[0-1])(.|-|/)([1-9]|0[1-9]|1[0-2])(.|-|/)([0-9][0-9]|19[0-9][0-9]|20[0-9][0-9])$) can handle the following formats:

• 01.10.2019
• 1.1.2019
• 1.1.19
• 12/03/2020
• 01.05.1950

The second part (^([0-9][0-9]|19[0-9][0-9]|20[0-9][0-9])(.|-|/)([1-9]|0[1-9]|1[0-2])(.|-|/)([1-9]|0[1-9]|1[0-9]|2[0-9]|3[0-1])$) can basically do the same, but in inverse order, where the year comes first, followed by month, and then day.

• 2020/02/12

As delimiters it allows ., /, -. As years it allows everything from 1900-2099, also giving only two numbers is fine.

If you have suggestions for improvement please let me know in the comments, so I can update the answer.

As the question title asks for a regex that finds many dates, I would like to propose a new solution, although there are many solutions already.

In order to find all dates of a string that are in this millennium (2000 – 2999), for me it worked the following:

dates = re.findall('([1-9]|1[0-9]|2[0-9]|3[0-1]|0[0-9])(.|-|/)([1-9]|1[0-2]|0[0-9])(.|-|/)(20[0-9][0-9])',dates_ele)

dates = [''.join(dates[i]) for i in range(len(dates))]

This regex is able to find multiple dates in the same string, like bla Bla 8.05/2020 n BLAH bla15/05-2020 blaa. As one could observe, instead of / the date can have . or –, not necessary at the same time.

Some explaining

More specifically it can find dates of format day , moth year. Day is an one digit integer or a zero followed by one digit integer or 1 or 2 followed by an one digit integer or a 3 followed by 0 or 1. Month is an one digit integer or a zero followed by one digit integer or 1 followed by 0, 1, or 2. Year is the number 20 followed by any number between 00 and 99.

Useful notes

One can add more date splitting symbols by adding | symbol at the end of both (.|-|/). For example for adding — one would do (.|-|/|--)

To have years outside of this millennium one has to modify (20[0-9][0-9]) to ([0-9][0-9][0-9][0-9])

I use something like this

>>> import datetime
>>> regex = datetime.datetime.strptime
>>>
>>> # TEST
>>> assert regex('2020-08-03', '%Y-%m-%d')
>>>

>>> assert regex('2020-08', '%Y-%m-%d')
ValueError: time data '2020-08' does not match format '%Y-%m-%d'

>>> assert regex('08/03/20', '%m/%d/%y')
>>>

>>> assert regex('08-03-2020', '%m/%d/%y')
ValueError: time data '08-03-2020' does not match format '%m/%d/%y'

I use something like this :

string="text 24/02/2021 ... 24-02-2021 ... 24_02_2021 ... 24|02|2021 text"
new_string = re.sub(r"[0-9]{1,4}[_|-|/||][0-9]{1,2}[_|-|/||][0-9]{1,4}", ' ', string)
print(new_string)

out : text … … … text

If you don’t want to raise ValueError exception like in methods with datetime, you can use re. Maybe you should also check that day of month lower than 31 and month number is lower than 12, inclusive:

from re import search as re_search


date_input = '31.12.1998'
re_search(r'^(3[01]|[12][0-9]|0[1-9]).(1[0-2]|0[1-9]).[0-9]{4}$', date_input)

With datetime good answer gave @unutbu earlier.

This regular expression for matching dates in this format "22/10/2021" works for me :

import re
date = "WHATEVER 22/10/2029 WHATEVER"
match = re.search("([0-9]|1[0-9]|2[0-9]|3[0-5])/([0-9]|1[0-9]|2[0-9]|3[0-5])/([0-9][0-9][0-9][0-9])", date)
print(match)

OUTPUT = <re.Match object; span=(9, 19), match=’22/10/2029′>

You can see in the fourth line that there is this string ([0-9]|1[0-9]|2[0-9]|3[0-5])/([0-9]|1[0-9]|2[0-9]|3[0-5])/([0-9][0-9][0-9][0-9]), this is the regular expression that I made based in this page.

In case anyone wants to match this type of date "24 November 2008"
you can use

import re
date = "24 November 2008"
regex = re.compile("d+sw+sd+")
matchDate = regex.findall(date)
print(matchDate)

Or

import re
date = "24 November 2008"
matchDate = re.findall("d+sw+sd+", date)
print(matchDate)