How can I replace any year from 1990-2050 with a space ?
I can replace any 4 digit number as follows
select regexp_replace('sdfg 2000', '(\y(\d{4})\y)', '', 'g');
But how additionally I can check the range?
Any help are welcome
I have discovered an alternative way to solve your problem. Take a look.
You wish to replace year from 1990-2050. Let's break that range into
1990-1999
2000-2049
2050
All three ranges can be matched by following regex.
Regex: [1][9][9][0-9]|[2][0][0-4][0-9]|2050
Explanation:
[1][9][9][0-9] will match years from 1990 to 1999.
[2][0][0-4][0-9] will match years from 2000 to 2049.
2050 will match 2050 literally
| means alteration. It will check either of these three patterns.
Regex101 Demo
You can use a CASE expression to extract and test for the year and only replace if the year falls into the range you want:
with test_data (col1) as (
values ('sdfg 2000'), ('foo 1983'), ('bar 2010'), ('bla 1940')
)
select col1,
case
when nullif(regexp_replace(col1, '[^0-9]+',''),'')::int between 1990 and 2050
then regexp_replace(col1, '\d{4}', '', 'g')
else col1
end as replaced
from test_data;
Results in:
col1 | replaced
----------+---------
sdfg 2000 | sdfg
foo 1983 | foo 1983
bar 2010 | bar
bla 1940 | bla 1940
The nullif(..) is necessary for values that do not contain any numbers. If you don't have values like that, you can leave it out.
You can't, from Wikipedia (emphasis mine):
Each character in a regular expression (that is, each character in the
string describing its pattern) is understood to be: a metacharacter
(with its special meaning), or a regular character (with its literal
meaning).
In your case, the letters do not have literal meaning, their meaning depends on characters around it.
Related
I have the following raw data:
1.1.2.2.4.4.4.5.5.9.11.15.16.16.19 ...
I'm using this regex to remove duplicates:
([^.]+)(.[ ]*\1)+
which results in the following:
1.2.4.5.9.115.16.19 ...
The problem is how the regex handles 1.1 in the substring .11.15. What should be 9.11.15.16 becomes 9.115.16. How do I fix this?
The raw values are sorted in numeric order to accommodate the regex used for processing the duplicate values.
The regex is being used within Oracle's REGEXP_REPLACE
The decimal is a delimiter. I've tried commas and pipes but that doesn't fix the problem.
Oracle's REGEX does not work the way you intended. You could split the string and find distinct rows using the general method Splitting string into multiple rows in Oracle. Another option is to use XMLTABLE , which works for numbers and also strings with proper quoting.
SELECT LISTAGG(n, '.') WITHIN
GROUP (
ORDER BY n
) AS n
FROM (
SELECT DISTINCT TO_NUMBER(column_value) AS n
FROM XMLTABLE(replace('1.1.2.2.4.4.4.5.5.9.11.15.16.16.19', '.', ','))
);
Demo
Unfortunately Oracle doesn't provide a token to match a word boundary position. Neither familiar \b token nor ancient [[:<:]] or [[:>:]].
But on this specific set you can use:
(\d+\.)(\1)+
Note: You forgot to escape dot.
Your regex caught:
a 1 - the second digit in 11,
then a dot,
and finally 1 - the first digit in 15.
So your regex failed to catch the whole sequence of digits.
The most natural way to write a regex catching the whole sequence
of digits would be to use:
a loobehind for either the start of the string or a dot,
then catch a sequence of digits,
and finally a lookahead for a dot.
But as I am not sure whether Oracle supports lookarounds, I wrote
the regex another way:
(^|\.)(\d+)(\.(\2))+
Details:
(^|\.) - Either start of the string or a dot (group 1), instead of
the loobehind.
(\d+) - A sequence of digits (group 2).
( - Start of group 3, containing:
\.(\2) - A dot and the same sequence of digits which caught group 2.
)+ - End of group 3, it may occur multiple times.
Group the repeating pattern and remove it
As revo has indicated, a big source of your difficulties came with not escaping the period. In addition, the resulting string having a 115 included can be explained as follows (Valdi_Bo made a similar observation earlier):
([^.]+)(.[ ]*\1)+ will match 11.15 as follow:
SCOTT#DB>SELECT
2 '11.15' val,
3 regexp_replace('11.15','([^.]+)(\.[ ]*\1)+','\1') deduplicated
4 FROM
5 dual;
VAL DEDUPLICATED
11.15 115
Here is a similar approach to address those problems:
matching pattern composition
-Look for a non-period matching list of length 0 to N (subexpression is referenced by \1).
'19' which matches ([^.]*)
-Look for the repeats which form our second matching list associated with subexression 2, referenced by \2.
'19.19.19' which matches ([^.]*)([.]\1)+
-Look for either a period or end of string. This is matching list referenced by \3. This fixes the match of '11.15' by '115'.
([.]|$)
replacement string
I replace the match pattern with a replacement string composed of the first instance of the non-period matching list.
\1\3
Solution
regexp_replace(val,'([^.]*)([.]\1)+([.]|$)','\1\3')
Here is an example using some permutations of your examples:
SCOTT#db>WITH tst AS (
2 SELECT
3 '1.1.2.2.4.4.4.5.5.9.11.15.16.16.19' val
4 FROM
5 dual
6 UNION ALL
7 SELECT
8 '1.1.1.1.2.2.4.4.4.4.4.5.5.9.11.11.11.15.16.16.19' val
9 FROM
10 dual
11 UNION ALL
12 SELECT
13 '1.1.2.2.4.4.4.5.5.9.11.15.16.16.19.19.19' val
14 FROM
15 dual
16 ) SELECT
17 val,
18 regexp_replace(val,'([^.]*)([.]\1)+([.]|$)','\1\3') deduplicate
19 FROM
20 tst;
VAL DEDUPLICATE
------------------------------------------------------------------------
1.1.2.2.4.4.4.5.5.9.11.15.16.16.19 1.2.4.5.9.11.15.16.19
1.1.1.1.2.2.4.4.4.4.4.5.5.9.11.11.11.15.16.16.19 1.2.4.5.9.11.15.16.19
1.1.2.2.4.4.4.5.5.9.11.15.16.16.19.19.19 1.2.4.5.9.11.15.16.19
My approach does not address possible spaces in the string. One could just remove them separately (e.g. through a separate replace statement).
I have following SQL result entries.
Result
---------
TW - 5657980 Due Date updated : to <strong>2017-08-13 10:21:00</strong> by <strong>System</strong>
TW - 5657980 Priority updated from <strong> Medium</strong> to <strong>Low</strong> by <strong>System</strong>
TW - 5657980 Material added: <strong>1000 : Cash in Bank - Operating (Old)/ QTY:2</strong> by <strong>System</strong>#9243
TW - 5657980 Labor added <strong>Kelsey Franks / 14:00 hours </strong> by <strong>System</strong>#65197
Now I am trying to extract a short description from this result and trying to migrate it to the another column in the same table.
Expected result
--------------
Due Date Updated
Priority Updated
Material Added
Labor Added
Ignore first 13 characters. For most of the cases it ends with 'updated'. Few ends with 'added'. It should be case insensitive.
Is there any way to get the expected result.
Solution with substring() using a regular expression. It skips the first 13 characters, then takes the string up to the first ' updated' or ' added', case-insensitive, with leading blank. Else NULL:
SELECT substring(result, '(?i)^.{13}(.*? (?:updated|added))')
FROM tbl;
The regexp explained:
(?i) .. meta-syntax to switch to case-insensitive matching
^ .. start of string
.{13} .. skip the first 13 characters
() .. capturing parenthesis (captures payload)
.*? .. any number of characters (non-greedy)
(?:) .. non-capturing parenthesis
(?:updated|added) .. 2 branches (string ends in 'updated' or 'added')
If we cannot rely on 13 leading characters like you later commented, we need some other reliable definition instead. Your difficulty seems with hazy requirements more than with the actual implementation.
Say, we are dealing with 1 or more non-digits, followed by 1 or more digits, a space and then the payload as defined above:
SELECT substring(result, '(?i)^\D+\d+ (.*? (?:updated|added))') ...
\d .. class shorthand for digits
\D .. non-digits, the opposite of \d
I'm using some Regex to find date strings of the form Jan 12, 2015 or Feb 3, 1999.
The regex I'm using is \w+\s\d{1,2},\s\d{4} and it's working correctly, but the thing is that on the file are also some strings with the form:
Weg 58, 4047 or Strasse 1, 4482 and I also match them.
How can I avoid those non-date matches? My approach is:
The first string (the one of the month, Jan, Feb, etc.) has to have always length 3.
The year has to start with 1 or 2.
The thing is that I dont know how can I add these two options to my regex. Any help please?
You can make the test right here: https://regex101.com/r/bN2pO0/1
Thanks in advance.
Since the months won't change (ie: consistent values between January - Decemeber, we can put the 3 starting characters).
We can then use a OR | operator to select years starting with 1 or 2
/((Jan|Feb|Mar|Apr|May|Jun|Jul|Aug|Sep|Oct|Nov|Dec)\s\d{1,2},\s(1|2)\d{3})/ig
https://regex101.com/r/bN2pO0/3
Just as you used \d{1,2} to match a digit 1 or 2 times and \d{4} to match a digit 4 times, you can use \w{3} to match a word character 3 times.
For the year, you can use the pipe "or" operator |.
\w{3}\s\d{1,2},\s(?:1|2)\d{3}
Although, this will also match non-dates of form Abc xy, 1xyz
If you want, you can go with brute force approach or just get rid of regex and use code to capture the dates.
Brute force:
(?:Jan|Feb|Mar|Apr|May|Jun|Jul|Aug|Sep|Oct|Nov|Dec)\s[0-2]?[0-9],\s[12]\d{3}
Given a value I want to validate it to check if it is a valid year. My criteria is simple where the value should be an integer with 4 characters. I know this is not the best solution as it will not allow years before 1000 and will allow years such as 5000. This criteria is adequate for my current scenario.
What I came up with is
\d{4}$
While this works it also allows negative values.
How do I ensure that only positive integers are allowed?
Years from 1000 to 2999
^[12][0-9]{3}$
For 1900-2099
^(19|20)\d{2}$
You need to add a start anchor ^ as:
^\d{4}$
Your regex \d{4}$ will match strings that end with 4 digits. So input like -1234 will be accepted.
By adding the start anchor you match only those strings that begin and end with 4 digits, which effectively means they must contain only 4 digits.
The "accepted" answer to this question is both incorrect and myopic.
It is incorrect in that it will match strings like 0001, which is not a valid year.
It is myopic in that it will not match any values above 9999. Have we already forgotten the lessons of Y2K? Instead, use the regular expression:
^[1-9]\d{3,}$
If you need to match years in the past, in addition to years in the future, you could use this regular expression to match any positive integer:
^[1-9]\d*$
Even if you don't expect dates from the past, you may want to use this regular expression anyway, just in case someone invents a time machine and wants to take your software back with them.
Note: This regular expression will match all years, including those before the year 1, since they are typically represented with a BC designation instead of a negative integer. Of course, this convention could change over the next few millennia, so your best option is to match any integer—positive or negative—with the following regular expression:
^-?[1-9]\d*$
This works for 1900 to 2099:
/(?:(?:19|20)[0-9]{2})/
Building on #r92 answer, for years 1970-2019:
(19[789]\d|20[01]\d)
To test a year in a string which contains other words along with the year you can use the following regex: \b\d{4}\b
In theory the 4 digit option is right. But in practice it might be better to have 1900-2099 range.
Additionally it need to be non-capturing group. Many comments and answers propose capturing grouping which is not proper IMHO. Because for matching it might work, but for extracting matches using regex it will extract 4 digit numbers and two digit (19 and 20) numbers also because of paranthesis.
This will work for exact matching using non-capturing groups:
(?:19|20)\d{2}
Use;
^(19|[2-9][0-9])\d{2}$
for years 1900 - 9999.
No need to worry for 9999 and onwards - A.I. will be doing all programming by then !!! Hehehehe
You can test your regex at https://regex101.com/
Also more info about non-capturing groups ( mentioned in one the comments above ) here http://www.manifold.net/doc/radian/why_do_non-capture_groups_exist_.htm
you can go with sth like [^-]\d{4}$: you prevent the minus sign - to be before your 4 digits.
you can also use ^\d{4}$ with ^ to catch the beginning of the string. It depends on your scenario actually...
/^\d{4}$/
This will check if a string consists of only 4 numbers. In this scenario, to input a year 989, you can give 0989 instead.
You could convert your integer into a string. As the minus sign will not match the digits, you will have no negative years.
I use this regex in Java ^(0[1-9]|1[012])[/](0[1-9]|[12][0-9]|3[01])[/](19|[2-9][0-9])[0-9]{2}$
Works from 1900 to 9999
If you need to match YYYY or YYYYMMDD you can use:
^((?:(?:(?:(?:(?:[1-9]\d)(?:0[48]|[2468][048]|[13579][26])|(?:(?:[2468][048]|[13579][26])00))(?:0?2(?:29)))|(?:(?:[1-9]\d{3})(?:(?:(?:0?[13578]|1[02])(?:31))|(?:(?:0?[13-9]|1[0-2])(?:29|30))|(?:(?:0?[1-9])|(?:1[0-2]))(?:0?[1-9]|1\d|2[0-8])))))|(?:19|20)\d{2})$
You can also use this one.
([0-2][0-9]|3[0-1])\/([0-1][0-2])\/(19[789]\d|20[01]\d)
In my case I wanted to match a string which ends with a year (4 digits) like this for example:
Oct 2020
Nov 2020
Dec 2020
Jan 2021
It'll return true with this one:
var sheetName = 'Jan 2021';
var yearRegex = new RegExp("\b\d{4}$");
var isMonthSheet = yearRegex.test(sheetName);
Logger.log('isMonthSheet = ' + isMonthSheet);
The code above is used in Apps Script.
Here's the link to test the Regex above: https://regex101.com/r/SzYQLN/1
You can try the following to capture valid year from a string:
.*(19\d{2}|20\d{2}).*
Works from 1950 to 2099 and value is an integer with 4 characters
^(?=.*?(19[56789]|20\d{2}).*)\d{4}$
I'm trying to write a regular expression that validates a date. The regex needs to match the following
M/D/YYYY
MM/DD/YYYY
Single digit months can start with a leading zero (eg: 03/12/2008)
Single digit days can start with a leading zero (eg: 3/02/2008)
CANNOT include February 30 or February 31 (eg: 2/31/2008)
So far I have
^(([1-9]|1[012])[-/.]([1-9]|[12][0-9]|3[01])[-/.](19|20)\d\d)|((1[012]|0[1-9])(3[01]|2\d|1\d|0[1-9])(19|20)\d\d)|((1[012]|0[1-9])[-/.](3[01]|2\d|1\d|0[1-9])[-/.](19|20)\d\d)$
This matches properly EXCEPT it still includes 2/30/2008 & 2/31/2008.
Does anyone have a better suggestion?
Edit: I found the answer on RegExLib
^((((0[13578])|([13578])|(1[02]))[\/](([1-9])|([0-2][0-9])|(3[01])))|(((0[469])|([469])|(11))[\/](([1-9])|([0-2][0-9])|(30)))|((2|02)[\/](([1-9])|([0-2][0-9]))))[\/]\d{4}$|^\d{4}$
It matches all valid months that follow the MM/DD/YYYY format.
Thanks everyone for the help.
This is not an appropriate use of regular expressions. You'd be better off using
[0-9]{2}/[0-9]{2}/[0-9]{4}
and then checking ranges in a higher-level language.
Here is the Reg ex that matches all valid dates including leap years. Formats accepted mm/dd/yyyy or mm-dd-yyyy or mm.dd.yyyy format
^(?:(?:(?:0?[13578]|1[02])(\/|-|\.)31)\1|(?:(?:0?[1,3-9]|1[0-2])(\/|-|\.)(?:29|30)\2))(?:(?:1[6-9]|[2-9]\d)?\d{2})$|^(?:0?2(\/|-|\.)29\3(?:(?:(?:1[6-9]|[2-9]\d)?(?:0[48]|[2468][048]|[13579][26])|(?:(?:16|[2468][048]|[3579][26])00))))$|^(?:(?:0?[1-9])|(?:1[0-2]))(\/|-|\.)(?:0?[1-9]|1\d|2[0-8])\4(?:(?:1[6-9]|[2-9]\d)?\d{2})$
courtesy Asiq Ahamed
I landed here because the title of this question is broad and I was looking for a regex that I could use to match on a specific date format (like the OP). But I then discovered, as many of the answers and comments have comprehensively highlighted, there are many pitfalls that make constructing an effective pattern very tricky when extracting dates that are mixed-in with poor quality or non-structured source data.
In my exploration of the issues, I have come up with a system that enables you to build a regular expression by arranging together four simpler sub-expressions that match on the delimiter, and valid ranges for the year, month and day fields in the order you require.
These are :-
Delimeters
[^\w\d\r\n:]
This will match anything that is not a word character, digit character, carriage return, new line or colon. The colon has to be there to prevent matching on times that look like dates (see my test Data)
You can optimise this part of the pattern to speed up matching, but this is a good foundation that detects most valid delimiters.
Note however; It will match a string with mixed delimiters like this 2/12-73 that may not actually be a valid date.
Year Values
(\d{4}|\d{2})
This matches a group of two or 4 digits, in most cases this is acceptable, but if you're dealing with data from the years 0-999 or beyond 9999 you need to decide how to handle that because in most cases a 1, 3 or >4 digit year is garbage.
Month Values
(0?[1-9]|1[0-2])
Matches any number between 1 and 12 with or without a leading zero - note: 0 and 00 is not matched.
Date Values
(0?[1-9]|[12]\d|30|31)
Matches any number between 1 and 31 with or without a leading zero - note: 0 and 00 is not matched.
This expression matches Date, Month, Year formatted dates
(0?[1-9]|[12]\d|30|31)[^\w\d\r\n:](0?[1-9]|1[0-2])[^\w\d\r\n:](\d{4}|\d{2})
But it will also match some of the Year, Month Date ones. It should also be bookended with the boundary operators to ensure the whole date string is selected and prevent valid sub-dates being extracted from data that is not well-formed i.e. without boundary tags 20/12/194 matches as 20/12/19 and 101/12/1974 matches as 01/12/1974
Compare the results of the next expression to the one above with the test data in the nonsense section (below)
\b(0?[1-9]|[12]\d|30|31)[^\w\d\r\n:](0?[1-9]|1[0-2])[^\w\d\r\n:](\d{4}|\d{2})\b
There's no validation in this regex so a well-formed but invalid date such as 31/02/2001 would be matched. That is a data quality issue, and as others have said, your regex shouldn't need to validate the data.
Because you (as a developer) can't guarantee the quality of the source data you do need to perform and handle additional validation in your code, if you try to match and validate the data in the RegEx it gets very messy and becomes difficult to support without very concise documentation.
Garbage in, garbage out.
Having said that, if you do have mixed formats where the date values vary, and you have to extract as much as you can; You can combine a couple of expressions together like so;
This (disastrous) expression matches DMY and YMD dates
(\b(0?[1-9]|[12]\d|30|31)[^\w\d\r\n:](0?[1-9]|1[0-2])[^\w\d\r\n:](\d{4}|\d{2})\b)|(\b(0?[1-9]|1[0-2])[^\w\d\r\n:](0?[1-9]|[12]\d|30|31)[^\w\d\r\n:](\d{4}|\d{2})\b)
BUT you won't be able to tell if dates like 6/9/1973 are the 6th of September or the 9th of June. I'm struggling to think of a scenario where that is not going to cause a problem somewhere down the line, it's bad practice and you shouldn't have to deal with it like that - find the data owner and hit them with the governance hammer.
Finally, if you want to match a YYYYMMDD string with no delimiters you can take some of the uncertainty out and the expression looks like this
\b(\d{4})(0[1-9]|1[0-2])(0[1-9]|[12]\d|30|31)\b
But note again, it will match on well-formed but invalid values like 20010231 (31th Feb!) :)
Test data
In experimenting with the solutions in this thread I ended up with a test data set that includes a variety of valid and non-valid dates and some tricky situations where you may or may not want to match i.e. Times that could match as dates and dates on multiple lines.
I hope this is useful to someone.
Valid Dates in various formats
Day, month, year
2/11/73
02/11/1973
2/1/73
02/01/73
31/1/1973
02/1/1973
31.1.2011
31-1-2001
29/2/1973
29/02/1976
03/06/2010
12/6/90
month, day, year
02/24/1975
06/19/66
03.31.1991
2.29.2003
02-29-55
03-13-55
03-13-1955
12\24\1974
12\30\1974
1\31\1974
03/31/2001
01/21/2001
12/13/2001
Match both DMY and MDY
12/12/1978
6/6/78
06/6/1978
6/06/1978
using whitespace as a delimiter
13 11 2001
11 13 2001
11 13 01
13 11 01
1 1 01
1 1 2001
Year Month Day order
76/02/02
1976/02/29
1976/2/13
76/09/31
YYYYMMDD sortable format
19741213
19750101
Valid dates before Epoch
12/1/10
12/01/660
12/01/00
12/01/0000
Valid date after 2038
01/01/2039
01/01/39
Valid date beyond the year 9999
01/01/10000
Dates with leading or trailing characters
12/31/21/
31/12/1921AD
31/12/1921.10:55
12/10/2016 8:26:00.39
wfuwdf12/11/74iuhwf
fwefew13/11/1974
01/12/1974vdwdfwe
01/01/99werwer
12321301/01/99
Times that look like dates
12:13:56
13:12:01
1:12:01PM
1:12:01 AM
Dates that runs across two lines
1/12/19
74
01/12/19
74/13/1946
31/12/20
08:13
Invalid, corrupted or nonsense dates
0/1/2001
1/0/2001
00/01/2100
01/0/2001
0101/2001
01/131/2001
31/31/2001
101/12/1974
56/56/56
00/00/0000
0/0/1999
12/01/0
12/10/-100
74/2/29
12/32/45
20/12/194
2/12-73
Maintainable Perl 5.10 version
/
(?:
(?<month> (?&mon_29)) [\/] (?<day>(?&day_29))
| (?<month> (?&mon_30)) [\/] (?<day>(?&day_30))
| (?<month> (?&mon_31)) [\/] (?<day>(?&day_31))
)
[\/]
(?<year> [0-9]{4})
(?(DEFINE)
(?<mon_29> 0?2 )
(?<mon_30> 0?[469] | (11) )
(?<mon_31> 0?[13578] | 1[02] )
(?<day_29> 0?[1-9] | [1-2]?[0-9] )
(?<day_30> 0?[1-9] | [1-2]?[0-9] | 30 )
(?<day_31> 0?[1-9] | [1-2]?[0-9] | 3[01] )
)
/x
You can retrieve the elements by name in this version.
say "Month=$+{month} Day=$+{day} Year=$+{year}";
( No attempt has been made to restrict the values for the year. )
To control a date validity under the following format :
YYYY/MM/DD or YYYY-MM-DD
I would recommand you tu use the following regular expression :
(((19|20)([2468][048]|[13579][26]|0[48])|2000)[/-]02[/-]29|((19|20)[0-9]{2}[/-](0[4678]|1[02])[/-](0[1-9]|[12][0-9]|30)|(19|20)[0-9]{2}[/-](0[1359]|11)[/-](0[1-9]|[12][0-9]|3[01])|(19|20)[0-9]{2}[/-]02[/-](0[1-9]|1[0-9]|2[0-8])))
Matches
2016-02-29 | 2012-04-30 | 2019/09/31
Non-Matches
2016-02-30 | 2012-04-31 | 2019/09/35
You can customise it if you wants to allow only '/' or '-' separators.
This RegEx strictly controls the validity of the date and verify 28,30 and 31 days months, even leap years with 29/02 month.
Try it, it works very well and prevent your code from lot of bugs !
FYI : I made a variant for the SQL datetime. You'll find it there (look for my name) : Regular Expression to validate a timestamp
Feedback are welcomed :)
Sounds like you're overextending regex for this purpose. What I would do is use a regex to match a few date formats and then use a separate function to validate the values of the date fields so extracted.
Perl expanded version
Note use of /x modifier.
/^(
(
( # 31 day months
(0[13578])
| ([13578])
| (1[02])
)
[\/]
(
([1-9])
| ([0-2][0-9])
| (3[01])
)
)
| (
( # 30 day months
(0[469])
| ([469])
| (11)
)
[\/]
(
([1-9])
| ([0-2][0-9])
| (30)
)
)
| ( # 29 day month (Feb)
(2|02)
[\/]
(
([1-9])
| ([0-2][0-9])
)
)
)
[\/]
# year
\d{4}$
| ^\d{4}$ # year only
/x
Original
^((((0[13578])|([13578])|(1[02]))[\/](([1-9])|([0-2][0-9])|(3[01])))|(((0[469])|([469])|(11))[\/](([1-9])|([0-2][0-9])|(30)))|((2|02)[\/](([1-9])|([0-2][0-9]))))[\/]\d{4}$|^\d{4}$
if you didn't get those above suggestions working, I use this, as it gets any date I ran this expression through 50 links, and it got all the dates on each page.
^20\d\d-(Jan|Feb|Mar|Apr|May|Jun|Jul|Aug|Sep|Oct|Nov|Dec)-(0[1-9]|[1-2][0-9]|3[01])$
This regex validates dates between 01-01-2000 and 12-31-2099 with matching separators.
^(0[1-9]|1[012])([- /.])(0[1-9]|[12][0-9]|3[01])\2(19|20)\d\d$
var dtRegex = new RegExp(/[1-9\-]{4}[0-9\-]{2}[0-9\-]{2}/);
if(dtRegex.test(date) == true){
var evalDate = date.split('-');
if(evalDate[0] != '0000' && evalDate[1] != '00' && evalDate[2] != '00'){
return true;
}
}
Regex was not meant to validate number ranges(this number must be from 1 to 5 when the number preceding it happens to be a 2 and the number preceding that happens to be below 6).
Just look for the pattern of placement of numbers in regex. If you need to validate is qualities of a date, put it in a date object js/c#/vb, and interogate the numbers there.
I know this does not answer your question, but why don't you use a date handling routine to check if it's a valid date? Even if you modify the regexp with a negative lookahead assertion like (?!31/0?2) (ie, do not match 31/2 or 31/02) you'll still have the problem of accepting 29 02 on non leap years and about a single separator date format.
The problem is not easy if you want to really validate a date, check this forum thread.
For an example or a better way, in C#, check this link
If you are using another platform/language, let us know
Perl 6 version
rx{
^
$<month> = (\d ** 1..2)
{ $<month> <= 12 or fail }
'/'
$<day> = (\d ** 1..2)
{
given( +$<month> ){
when 1|3|5|7|8|10|12 {
$<day> <= 31 or fail
}
when 4|6|9|11 {
$<day> <= 30 or fail
}
when 2 {
$<day> <= 29 or fail
}
default { fail }
}
}
'/'
$<year> = (\d ** 4)
$
}
After you use this to check the input the values are available in $/ or individually as $<month>, $<day>, $<year>. ( those are just syntax for accessing values in $/ )
No attempt has been made to check the year, or that it doesn't match the 29th of Feburary on non leap years.
If you're going to insist on doing this with a regular expression, I'd recommend something like:
( (0?1|0?3| <...> |10|11|12) / (0?1| <...> |30|31) |
0?2 / (0?1| <...> |28|29) )
/ (19|20)[0-9]{2}
This might make it possible to read and understand.
/(([1-9]{1}|0[1-9]|1[0-2])\/(0[1-9]|[1-9]{1}|[12]\d|3[01])\/[12]\d{3})/
This would validate for following -
Single and 2 digit day with range from 1 to 31. Eg, 1, 01, 11, 31.
Single and 2 digit month with range from 1 to 12. Eg. 1, 01, 12.
4 digit year. Eg. 2021, 1980.
A slightly different approach that may or may not be useful for you.
I'm in php.
The project this relates to will never have a date prior to the 1st of January 2008. So, I take the 'date' inputed and use strtotime(). If the answer is >= 1199167200 then I have a date that is useful to me. If something that doesn't look like a date is entered -1 is returned. If null is entered it does return today's date number so you do need a check for a non-null entry first.
Works for my situation, perhaps yours too?