I'm having troubles trying to match a pattern of dates. Any of the following dates are legal:
- 121212
- 4 9 12
- 5-3-2000
- 62502
- 3/3/11
- 09-08-2001
- 8 6 07
- 12 10 2004
- 4-16-08
- 3/7/2005
What makes this date matching really challenging is that the year doesn't have to be 4 digits (a 2 digit year is assumed to be in the 21st century i.e. 02 = 2002), the month/date can either be written with a beginning 0 if it is a one digit month, and the dates may or may not be separated by spaces, dashes, or slashes.
This is what I currently have:
/((((0[13578])|([13578])|(1[02]))[\/-]?\s*(([1-9])|(0[1-9])|([12][0-9])|(3[01])))|(((0[469])|([469])|(11))[\/-]?\s*(([1-9])|(0[1-9])|([12][0-9])|(30)))|((2|02)[\/](([1-9])|(0[1-9])|([12][0-9])))[\/-]?\s*(20[0-9]{2})|([0-9]{2}))/g
This almost works, except right now I'm not exactly sure if I'm assuming the length of the dates and months. For example, in the case 121212, I might be assuming the month is 1 instead of 12. Also, for some reason when I'm printing out $1 and $2, it is the same value. In the case of 121212, $1 is 1212, $2 is 1212 and $3 is 12. However, I just want $1 to be 121212.
Your task is ambiguous, since you may not be able to tell mmd from mdd or mdccyy from mmddyy.
You left off the option for spaces or dashes in one place where you match /.
You aren't checking for leap years.
This is doable, but it's awfully easy to make a mistake; how about not trying to do it with a regex.
The CPAN modules Time::ParseDate and DateTime are probably what you're looking for, except the 62502 pattern:
use DateTime;
use Time::ParseDate;
foreach my $str (<DATA>) {
chomp $str;
$str =~ tr{ }{/};
my $epoch = parsedate($str, GMT => 1);
next unless $epoch; # skip 62502
my $dt = DateTime->from_epoch ( epoch => $epoch );
print $dt->ymd, "\n";
}
__DATA__
121212
4 9 12
5-3-2000
62502
3/3/11
09-08-2001
8 6 07
12 10 2004
4-16-08
3/7/2005
Once you have the DateTime object, you can extract year, month, and day information easily.
This solution handles all of the cases that you provided. But the solution isn't foolproof because the problem has ambiguities. E.g. how do we interpret the date 12502? Is it 1/25/02 or 12/5/02?
use 5.010;
while (my $line = <DATA>) {
chomp $line;
my #date = $line =~ /
\A
([01]?\d) # month is 1-2 digits, but the first digit may only be 0 or 1
[ \-\/]? # may or may not have a separator
([0123]?\d) # day is 1-2 digits
[ \-\/]?
(\d{2,4}) # year is 2-4 digits
\z
/x;
say join '_', #date;
}
__DATA__
121212
4 9 12
5-3-2000
12502
3/3/11
09-08-2001
8 6 07
12 10 2004
4-16-08
3/7/2005
This is the best I could come up with based on what info you've given. It matches all possibilities, and has error checking for month/day ranges and also the year (from 1900 to 2099)
/(1[012]|0?\d)([-\/ ]?)([12]\d|3[01]|0?\d)\2((19|20)?\d\d)/
Related
I have a line in a text doc that im trying to pull data from.
Example
Line im searching for in Text file:
Valid from: Sun May 17 19:00:00 CDT 1998
I want to find the key words "Valid from:" and then get only Sun May 17 and 1998
So end result should look like this:
Sun May 17 1998
I think im close to getting it right. This is what I have. It finds the keyword Valid From: but it returns more than I need
Sun May 17 19:00:00 CDT 1998
(?<=Valid from:)\s+\w+\s+\w+\s+\d+\s+\d+:\d+:\d+\s+\w+\s+\d+
Thank you in advance for any assistance.
I would use two capture groups (…) instead of lookaround construct:
$sampleText = 'Valid from: Sun May 17 19:00:00 CDT 1998'
$regEx = 'Valid from:\s+(\w+\s+\w+\s+\d+)\s+\d+:\d+:\d+\s+\w+\s+(\d+)'
if( $sampleText -match $regEx ) {
# Combine the matched values of both capture groups into a single string
$matches[1,2] -join ' '
}
Output:
Sun May 17 1998
If the -match operator successfully matches the pattern on the right-hand-side with the input text on the left-hand-side, the automatic variable $matches is set.
$matches contains the full match at index 0 and the matched values of any capture groups at subsequent indices, in this case 1 and 2.
Using the -join operator we combine the matched values of the capture groups into a single string.
Demo at regex101.
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.
I have a file like this:
3107 0.9 0.0 0.0 chr1 29312346 29312694 (219937927) C L1HS LINE/L1 (4) 6151 5803 54360
8095 0.5 0.0 0.0 chr1 31040661 31041597 (218209024) + L1HS LINE/L1 5203 6139 (16) 57249
...
When the 9th column is C, I need to subtract column 14 from 13, and when the 9th column is +, I need to subtract column 12 from 13.
I understand I can create arrays, but how can I use a regex, such as ($line =~/(\w+)\s+(\w+)/), to solve this instead?
You can split at white spaces into #F array(first value being $F[0]), subtract columns, and output values separated by space.
perl -lane'
$F[12] -= $F[13] if $F[8] eq "C";
$F[12] -= $F[11] if $F[8] eq "+";
print "#F";
' file
Since you wanted to use a regex, here is another solution. It is perhaps a bit unsharp, because you did not define your lines cleanly but with only two example lines, and for those, it works. I commented the regex so that you can see, which part of the expression is matching a certain group and which of them are captured.
#!/usr/bin/perl
use strict;
use warnings;
use v5.10;
while( <DATA> )
{
if( $_ =~ /[0-9]+ # 1
\s+
[0-9.]+ # 2
\s+
[0-9.]+ # 3
\s+
[0-9.]+ # 4
\s+
[a-z0-9]+ # 5
\s+
[0-9]+ # 6
\s+
[0-9]+ # 7
\s+
\([a-z0-9]+\) # 8
\s+
([c+]) # 9 -> capture group 1
\s+
[a-z0-9]+ # 10
\s+
[a-z0-9\/]+ # 11
\s+
\(?([0-9]+)\)? # 12 -> capture group 2
\s+
([0-9]+) # 13 -> capture group 3
\s+
\(?([0-9]+)\)? # 14 -> capture group 4
\s+
[0-9]+? # 15
/ix )
{
say "Matched: $_";
say "Operation: $1";
if( $1 eq "+" )
{
say "$2 - $3 = ".( $2 - $3 );
}
elsif( $1 eq "C" )
{
say "$4 - $3 = ".( $4 - $3 );
}
else
{
say "Nothing do to here...";
}
}
}
exit;
#1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
__DATA__
3107 0.9 0.0 0.0 chr1 29312346 29312694 (219937927) C L1HS LINE/L1 (4) 6151 5803 54360
8095 0.5 0.0 0.0 chr1 31040661 31041597 (218209024) + L1HS LINE/L1 5203 6139 (16) 57249
Update:
As you can see in the perl documentation, I used the x flag to have comments in my regex. The i flag makes it case insensitive.
Furthermore, I didn't just try to devide all the single columns by whitespaces but also by their types, which is an advantage of using a regular expression. While \s+ expressions are seperators for columns here, allowing arbitary amounts of whitespace all the single groups are kind of specified. That allows to find non-conforming lines. For example, by defining caputre group $1 as ([c+]) I was able to reduce the possible characters, that trigger an operation to C and + ( and c because of case-inesensitivity).
Binding a group to a variable (capturing it) is done by using parenthises.
This way, I was able to only pick the columns I really need (see the comments).
Do not use a regex for a problem like this.
If you're just working with columns separated by whitespace, the proper tool is split.
my #cols = split ' ', $line;
Is it possible to increment numbers using regex substitution? Not using evaluated/function-based substitution, of course.
This question was inspired by another one, where the asker wanted to increment numbers in a text editor. There are probably more text editors that support regex substitution than ones that support full-on scripting, so a regex might be convenient to float around, if one exists.
Also, often I've learned neat things from clever solutions to practically useless problems, so I'm curious.
Assume we're only talking about non-negative decimal integers, i.e. \d+.
Is it possible in a single substitution? Or, a finite number of substitutions?
If not, is it at least possible given an upper bound, e.g. numbers up to 9999?
Of course it's doable given a while-loop (substituting while matched), but we're going for a loopless solution here.
This question's topic amused me for one particular implementation I did earlier. My solution happens to be two substitutions so I'll post it.
My implementation environment is solaris, full example:
echo "0 1 2 3 7 8 9 10 19 99 109 199 909 999 1099 1909" |
perl -pe 's/\b([0-9]+)\b/0$1~01234567890/g' |
perl -pe 's/\b0(?!9*~)|([0-9])(?=9*~[0-9]*?\1([0-9]))|~[0-9]*/$2/g'
1 2 3 4 8 9 10 11 20 100 110 200 910 1000 1100 1910
Pulling it apart for explanation:
s/\b([0-9]+)\b/0$1~01234567890/g
For each number (#) replace it with 0#~01234567890. The first 0 is in case rounding 9 to 10 is needed. The 01234567890 block is for incrementing. The example text for "9 10" is:
09~01234567890 010~01234567890
The individual pieces of the next regex can be described seperately, they are joined via pipes to reduce substitution count:
s/\b0(?!9*~)/$2/g
Select the "0" digit in front of all numbers that do not need rounding and discard it.
s/([0-9])(?=9*~[0-9]*?\1([0-9]))/$2/g
(?=) is positive lookahead, \1 is match group #1. So this means match all digits that are followed by 9s until the '~' mark then go to the lookup table and find the digit following this number. Replace with the next digit in the lookup table. Thus "09~" becomes "19~" then "10~" as the regex engine parses the number.
s/~[0-9]*/$2/g
This regex deletes the ~ lookup table.
Wow, turns out it is possible (albeit ugly)!
In case you do not have the time or cannot be bothered to read through the whole explanation, here is the code that does it:
$str = '0 1 2 3 4 5 6 7 8 9 10 11 12 13 19 20 29 99 100 139';
$str = preg_replace("/\d+/", "$0~", $str);
$str = preg_replace("/$/", "#123456789~0", $str);
do
{
$str = preg_replace(
"/(?|0~(.*#.*(1))|1~(.*#.*(2))|2~(.*#.*(3))|3~(.*#.*(4))|4~(.*#.*(5))|5~(.*#.*(6))|6~(.*#.*(7))|7~(.*#.*(8))|8~(.*#.*(9))|9~(.*#.*(~0))|~(.*#.*(1)))/s",
"$2$1",
$str, -1, $count);
} while($count);
$str = preg_replace("/#123456789~0$/", "", $str);
echo $str;
Now let's get started.
So first of all, as the others mentioned, it is not possible in a single replacement, even if you loop it (because how would you insert the corresponding increment to a single digit). But if you prepare the string first, there is a single replacement that can be looped. Here is my demo implementation using PHP.
I used this test string:
$str = '0 1 2 3 4 5 6 7 8 9 10 11 12 13 19 20 29 99 100 139';
First of all, let's mark all digits we want to increment by appending a marker character (I use ~, but you should probably use some crazy Unicode character or ASCII character sequence that definitely will not occur in your target string.
$str = preg_replace("/\d+/", "$0~", $str);
Since we will be replacing one digit per number at a time (from right to left), we will just add that marking character after every full number.
Now here comes the main hack. We add a little 'lookup' to the end of our string (also delimited with a unique character that does not occur in your string; for simplicity I used #).
$str = preg_replace("/$/", "#123456789~0", $str);
We will use this to replace digits by their corresponding successors.
Now comes the loop:
do
{
$str = preg_replace(
"/(?|0~(.*#.*(1))|1~(.*#.*(2))|2~(.*#.*(3))|3~(.*#.*(4))|4~(.*#.*(5))|5~(.*#.*(6))|6~(.*#.*(7))|7~(.*#.*(8))|8~(.*#.*(9))|9~(.*#.*(~0))|(?<!\d)~(.*#.*(1)))/s",
"$2$1",
$str, -1, $count);
} while($count);
Okay, what is going on? The matching pattern has one alternative for every possible digit. This maps digits to successors. Take the first alternative for example:
0~(.*#.*(1))
This will match any 0 followed by our increment marker ~, then it matches everything up to our cheat-delimiter and the corresponding successor (that is why we put every digit there). If you glance at the replacement, this will get replaced by $2$1 (which will then be 1 and then everything we matched after the ~ to put it back in place). Note that we drop the ~ in the process. Incrementing a digit from 0 to 1 is enough. The number was successfully incremented, there is no carry-over.
The next 8 alternatives are exactly the same for the digits 1to 8. Then we take care of two special cases.
9~(.*#.*(~0))
When we replace the 9, we do not drop the increment marker, but place it to the left of our the resulting 0 instead. This (combined with the surrounding loop) is enough to implement carry-over propagation. Now there is one special case left. For all numbers consisting solely of 9s we will end up with the ~ in front of the number. That is what the last alternative is for:
(?<!\d)~(.*#.*(1))
If we encounter a ~ that is not preceded by a digit (therefore the negative lookbehind), it must have been carried all the way through a number, and thus we simply replace it with a 1. I think we do not even need the negative lookbehind (because this is the last alternative that is checked), but it feels safer this way.
A short note on the (?|...) around the whole pattern. This makes sure that we always find the two matches of an alternative in the same references $1 and $2 (instead of ever larger numbers down the string).
Lastly, we add the DOTALL modifier (s), to make this work with strings that contain line breaks (otherwise, only numbers in the last line will be incremented).
That makes for a fairly simple replacement string. We simply first write $2 (in which we captured the successor, and possibly the carry-over marker), and then we put everything else we matched back in place with $1.
That's it! We just need to remove our hack from the end of the string, and we're done:
$str = preg_replace("/#123456789~0$/", "", $str);
echo $str;
> 1 2 3 4 5 6 7 8 9 10 11 12 13 14 20 21 30 100 101 140
So we can do this entirely in regular expressions. And the only loop we have always uses the same regex. I believe this is as close as we can get without using preg_replace_callback().
Of course, this will do horrible things if we have numbers with decimal points in our string. But that could probably be taken care of by the very first preparation-replacement.
Update: I just realised, that this approach immediately extends to arbitrary increments (not just +1). Simply change the first replacement. The number of ~ you append equals the increment you apply to all numbers. So
$str = preg_replace("/\d+/", "$0~~~", $str);
would increment every integer in the string by 3.
I managed to get it working in 3 substitutions (no loops).
tl;dr
s/$/ ~0123456789/
s/(?=\d)(?:([0-8])(?=.*\1(\d)\d*$)|(?=.*(1)))(?:(9+)(?=.*(~))|)(?!\d)/$2$3$4$5/g
s/9(?=9*~)(?=.*(0))|~| ~0123456789$/$1/g
Explanation
Let ~ be a special character not expected to appear anywhere in the text.
If a character is nowhere to be found in the text, then there's no way to make it appear magically. So first we insert the characters we care about at the very end.
s/$/ ~0123456789/
For example,
0 1 2 3 7 8 9 10 19 99 109 199 909 999 1099 1909
becomes:
0 1 2 3 7 8 9 10 19 99 109 199 909 999 1099 1909 ~0123456789
Next, for each number, we (1) increment the last non-9 (or prepend a 1 if all are 9s), and (2) "mark" each trailing group of 9s.
s/(?=\d)(?:([0-8])(?=.*\1(\d)\d*$)|(?=.*(1)))(?:(9+)(?=.*(~))|)(?!\d)/$2$3$4$5/g
For example, our example becomes:
1 2 3 4 8 9 19~ 11 29~ 199~ 119~ 299~ 919~ 1999~ 1199~ 1919~ ~0123456789
Finally, we (1) replace each "marked" group of 9s with 0s, (2) remove the ~s, and (3) remove the character set at the end.
s/9(?=9*~)(?=.*(0))|~| ~0123456789$/$1/g
For example, our example becomes:
1 2 3 4 8 9 10 11 20 100 110 200 910 1000 1100 1910
PHP Example
$str = '0 1 2 3 7 8 9 10 19 99 109 199 909 999 1099 1909';
echo $str . '<br/>';
$str = preg_replace('/$/', ' ~0123456789', $str);
echo $str . '<br/>';
$str = preg_replace('/(?=\d)(?:([0-8])(?=.*\1(\d)\d*$)|(?=.*(1)))(?:(9+)(?=.*(~))|)(?!\d)/', '$2$3$4$5', $str);
echo $str . '<br/>';
$str = preg_replace('/9(?=9*~)(?=.*(0))|~| ~0123456789$/', '$1', $str);
echo $str . '<br/>';
Output:
0 1 2 3 7 8 9 10 19 99 109 199 909 999 1099 1909
0 1 2 3 7 8 9 10 19 99 109 199 909 999 1099 1909 ~0123456789
1 2 3 4 8 9 19~ 11 29~ 199~ 119~ 299~ 919~ 1999~ 1199~ 1919~ ~0123456789
1 2 3 4 8 9 10 11 20 100 110 200 910 1000 1100 1910
Is it possible in a single substitution?
No.
If not, is it at least possible in a single substitution given an upper bound, e.g. numbers up to 9999?
No.
You can't even replace the numbers between 0 and 8 with their respective successor. Once you have matched, and grouped this number:
/([0-8])/
you need to replace it. However, regex doesn't operate on numbers, but on strings. So you can replace the "number" (or better: digit) with twice this digit, but the regex engine does not know it is duplicating a string that holds a numerical value.
Even if you'd do something (silly) as this:
/(0)|(1)|(2)|(3)|(4)|(5)|(6)|(7)|(8)/
so that the regex engine "knows" that if group 1 is matched, the digit '0' is matched, it still cannot do a replacement. You can't instruct the regex engine to replace group 1 with the digit '1', group '2' with the digit '2', etc. Sure, some tools like PHP will let you define a couple of different patterns with corresponding replacement strings, but I get the impression that is not what you were thinking about.
It is not possible by regular expression search and substitution alone.
You have to use use something else to help achieve that. You have to use the programming language at hand to increment the number.
Edit:
The regular expressions definition, as part of Single Unix Specification doesn't mention regular expressions supporting evaluation of aritmethic expressions or capabilities for performing aritmethic operations.
Nonetheless, I know some flavors ( TextPad, editor for Windows) allows you to use \i as a substitution term which is an incremental counter of how many times has the search string been found, but it doesn't evaluate or parse found strings into a number nor does it allow to add a number to it.
I have found a solution in two steps (Javascript) but it relies on indefinite lookaheads, which some regex engines reject:
const incrementAll = s =>
s.replaceAll(/(.+)/gm, "$1\n101234567890")
.replaceAll(/(?:([0-8]|(?<=\d)9)(?=9*[^\d])(?=.*\n\d*\1(\d)\d*$))|(?<!\d)9(?=9*[^\d])(?=(?:.|\n)*(10))|\n101234567890$/gm, "$2$3");
The key thing is to add a list of numbers in order at the end of the string in the first step, and in the second, to find the location relevant digit and capture the digit to its right via a lookahead. There are two other branches in the second step, one for dealing with initial nines, and the other for removing the number sequence.
Edit: I just tested it in safari and it throws an error, but it definately works in firefox.
I needed to increment indices of output files by one from a pipeline I can't modify. After some searches I got a hit on this page. While the readings are meaningful, they really don't give a readable solution to the problem. Yes it is possible to do it with only regex; no it is not as comprehensible.
Here I would like to give a readable solution using Python, so that others don't need to reinvent the wheels. I can imagine many of you may have ended up with a similar solution.
The idea is to partition file name into three groups, and format your match string so that the incremented index is the middle group. Then it is possible to only increment the middle group, after which we piece the three groups together again.
import re
import sys
import argparse
from os import listdir
from os.path import isfile, join
def main():
parser = argparse.ArgumentParser(description='index shift of input')
parser.add_argument('-r', '--regex', type=str,
help='regex match string for the index to be shift')
parser.add_argument('-i', '--indir', type=str,
help='input directory')
parser.add_argument('-o', '--outdir', type=str,
help='output directory')
args = parser.parse_args()
# parse input regex string
regex_str = args.regex
regex = re.compile(regex_str)
# target directories
indir = args.indir
outdir = args.outdir
try:
for input_fname in listdir(indir):
input_fpath = join(indir, input_fname)
if not isfile(input_fpath): # not a file
continue
matched = regex.match(input_fname)
if matched is None: # not our target file
continue
# middle group is the index and we increment it
index = int(matched.group(2)) + 1
# reconstruct output
output_fname = '{prev}{index}{after}'.format(**{
'prev' : matched.group(1),
'index' : str(index),
'after' : matched.group(3)
})
output_fpath = join(outdir, output_fname)
# write the command required to stdout
print('mv {i} {o}'.format(i=input_fpath, o=output_fpath))
except BrokenPipeError:
pass
if __name__ == '__main__': main()
I have this script named index_shift.py. To give an example of the usage, my files are named k0_run0.csv, for bootstrap runs of machine learning models using parameter k. The parameter k starts from zero, and the desired index map starts at one. First we prepare input and output directories to avoid overriding files
$ ls -1 test_in/ | head -n 5
k0_run0.csv
k0_run10.csv
k0_run11.csv
k0_run12.csv
k0_run13.csv
$ ls -1 test_out/
To see how the script works, just print its output:
$ python3 -u index_shift.py -r '(^k)(\d+?)(_run.+)' -i test_in -o test_out | head -n5
mv test_in/k6_run26.csv test_out/k7_run26.csv
mv test_in/k25_run11.csv test_out/k26_run11.csv
mv test_in/k7_run14.csv test_out/k8_run14.csv
mv test_in/k4_run25.csv test_out/k5_run25.csv
mv test_in/k1_run28.csv test_out/k2_run28.csv
It generates bash mv command to rename the files. Now we pipe the lines directly into bash.
$ python3 -u index_shift.py -r '(^k)(\d+?)(_run.+)' -i test_in -o test_out | bash
Checking the output, we have successfully shifted the index by one.
$ ls test_out/k0_run0.csv
ls: cannot access 'test_out/k0_run0.csv': No such file or directory
$ ls test_out/k1_run0.csv
test_out/k1_run0.csv
You can also use cp instead of mv. My files are kinda big, so I wanted to avoid duplicating them. You can also refactor how many you shift as input argument. I didn't bother, cause shift by one is most of my use cases.
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?