I have date in my .txt file which comes like either of the below:
mmddyyyy
OR
mm/dd/yyyy
Below is the regex which works fine for mm/dd/yyyy.
^02\/(?:[01]\d|2\d)\/(?:19|20)(?:0[048]|[13579][26]|[2468][048])|(?:0[13578]|10|12)\/(?:[0-2]\d|3[01])\/(?:19|20)\d{2}|(?:0[469]|11)\/(?:[0-2]\d|30)\/(?:19|20)\d{2}|02\/(?:[0-1]\d|2[0-8])\/(?:19|20)\d{2}$
However, unable to build the regex for mmddyyyy. I just want to understand is there any generic regex that would work for both cases?
Why use regex for this? Seems like a case of "Now you have two problems"
It would be more effective (and easier to understand) to use a DateTimeFormatter (assuming you are on the JVM and not using scala-js)
The format patterns support using [] to surround optional sections, such as the /, and the formatters inherently perform input validation so if you plug in a month or day that can't exist, it'll throw an exception.
import java.time.format.DateTimeFormatter
import java.time.LocalDate
val mdy = DateTimeFormatter.ofPattern("MM[/]dd[/]yyyy")
def parse(rawDate: String) = LocalDate.parse(rawDate, mdy)
scala> parse("12252022")
res7: java.time.LocalDate = 2022-12-25
scala> parse("12/25/2022")
res8: java.time.LocalDate = 2022-12-25
scala> parse("25/12/2022")
java.time.format.DateTimeParseException: Text '25/12/2022' could not be parsed: Invalid value for MonthOfYear (valid values 1 - 12): 25
scala> parse("abc123")
java.time.format.DateTimeParseException: Text 'abc123' could not be parsed at index 0
If you want to match all those variations with either 2 forward slashes or only digits, you can use a positive lookahead to assert either only digits or 2 forward slashes surrounded by digits.
Then in the pattern itself you can make matching the / optional.
Note that you don't have to escape the \/
^(?=\d+(?:/\d+/\d+)?$)(?:02/?(?:[01]\d|2\d)/?(?:19|20)(?:0[048]|[13579][26]|[2468][048])|(?:0[13578]|10|12)/?(?:[0-2]\d|3[01])/?(?:19|20)\d{2}|(?:0[469]|11)/?(?:[0-2]\d|30)/?(?:19|20)\d{2}|02/?(?:[0-1]\d|2[0-8])\?(?:19|20)\d{2})$
Regex demo
Another option is to write an alternation | matching the same pattern without the / in it.
First of all, there is a tiny shortcoming in your regex: the ^ anchor only applies to the first part of your regex, not to the other alternatives that are separated by |. Similarly the final $ applies only to the final alternative. You should put all alternatives in a non-capturing group, like ^(?: | | | )$
Then for the question itself, you could make the forward slash that follows the month optional and put it in a capture group. Then what comes between the day and the year could be a backreference to that capture group. So (\/?) and \1.
^(?:02(\/?)(?:[01]\d|2\d)\1(?:19|20)(?:0[048]|[13579][26]|[2468][048])|(?:0[13578]|10|12)(\/?)(?:[0-2]\d|3[01])\2(?:19|20)\d{2}|(?:0[469]|11)(\/?)(?:[0-2]\d|30)\3(?:19|20)\d{2}|02(\/?)(?:[0-1]\d|2[0-8])\4(?:19|20)\d{2})$
Related
My input is of this format: (xxx)yyyy(zz)(eee)fff where {x,y,z,e,f} are all numbers. But fff is optional though.
Input: x = (123)4567(89)(660)
Expected output: Only the eeepart i.e. the number inside 3rd "()" i.e. 660 in my example.
I am able to achieve this so far:
re.search("\((\d*)\)", x).group()
Output: (123)
Expected: (660)
I am surely missing something fundamental. Please advise.
Edit 1: Just added fff to the input data format.
You could find all those matches that have round braces (), and print the third match with findall
import re
n = "(123)4567(89)(660)999"
r = re.findall("\(\d*\)", n)
print(r[2])
Output:
(660)
The (eee) part is identical to the (xxx) part in your regex. If you don't provide an anchor, or some sequencing requirement, then an unanchored search will match the first thing it finds, which is (xxx) in your case.
If you know the (eee) always appears at the end of the string, you could append an "at-end" anchor ($) to force the match at the end. Or perhaps you could append a following character, like a space or comma or something.
Otherwise, you might do well to match the other parts of the pattern and not capture them:
pattern = r'[0-9()]{13}\((\d{3})\)'
If you want to get the third group of numbers in brackets, you need to skip the first two groups which you can do with a repeating non-capturing group which looks for a set of digits enclosed in () followed by some number of non ( characters:
x = '(123)4567(89)(660)'
print(re.search("(?:\(\d+\)[^(]*){2}(\(\d+\))", x).group(1))
Output:
(660)
Demo on rextester
I'd like to duplicate a multiple matches in a line, substituting part of the match, but keeping the runs of matches together (that seems to be the tricky part).
e.g.:
Regex:
(x(\d)(,)?)
Replacement:
X$2,O$2$3
Input:
x1,x2,Z3,x4,Z5,x6
Output: (repeated groups broken apart)
X1,O1,X2,O2,Z3,X4,O4,Z5,X6,O6
Desired output (repeated groups, "X1,X2" kept together):
X1,X2,O1,O2,Z3,X4,O4,Z5,X6,O6
Demo: https://regex101.com/r/gH9tL9/1
Is this possible with regex or do I need to use something else?
Update: Wills answer is what I expected. It occurs to me that it might be possible with multiple passes of regex.
You would have to capture the repeating patterns as one match and write out replacements for the whole repeating pattern at once. your current pattern cannot tell that your first and second matches, x1, and x2, respectively, are adjacent.
Im going to say no, this is not possible with one pure regex.
This is because of two important facts about capture groups and replacing.
Repeated capture groups will return the last capture:
Regex's are able to capture patterns which repeat an arbitrary amount of time by using the form <PATTERN>{1,},<PATTERN>+ or <PATTERN>*. However any capture group within <PATTERN> would only return the captures from the last iteration of the pattern. This would prevent your desired ability to capture matches that arbitrarily repeat.
"Hold on", you might say, "I only want to capture patterns that repeat one or two times, I could use (x(\d)(,)?)(x(\d)(,)?)?", which brings us to point 2.
There is no conditional replacement
Using the above pattern we could get your desired output for the repeated match, but not without mangling the solo match replacement.
See: https://regex101.com/r/gH9tL9/2 Without the ability to turn off sections of the replacement based on the existence of capture groups, we cannot achieve the desired output.
But "No, you can't do that" is a challenge to a hacker, I hope I am shown up by a true regex ninja.
Solution with 2 regexes and some code
There's definitely ways to achieve this goal with some code.
Here's a quick and dirty python hack using two regexes http://pythonfiddle.com/wip-soln-for-so-q/
This makes use of python's re.sub(), which can pass matches to one regex to a function ordered_repl which returns the replacement string. By using your original regex within the ordered_repl we can extract the information we want and get the right order by buffering our lists of Xs and Os.
import re
input_string="x1,x2,Z3,x4,Z5,x6"
re1 = re.compile("(?:x\d,?)+") # captures the general thing you want to match using a repeating non-capturing group
re2 = re.compile("(x(\d)(,)?)") # your actual matcher
def ordered_repl(m): # m is a matchobj
buf1 = []
buf2 = []
cap_iter = re.finditer(re2,m.group(0)) # returns an iterator of MatchObjects for all non-overlapping matches
for cap_group in cap_iter:
capture = cap_group.group(2) # capture the digit
buf1.append("X%s" % capture) # buffer X's of this submatch group
buf2.append("O%s" % capture) # buffer O's of this submatch group
return "%s,%s," % (",".join(buf1),",".join(buf2)) # concatenate the buffers and return
print re.sub(re1,ordered_repl,input_string).rstrip(',') # searches string for matches to re1 and passes them to the ordered_repl function
In my specific case I'm using powershell, so I was able to come up with the following:
(linebreaks added for readability)
("x1,x2,z3,x4,z5,x6"
-split '((?<=x\d),(?!x)|(?<!x\d),(?=x))'
| Foreach-Object {
if ($_ -match 'x') {
$_ + ',' + ($_ -replace 'x','y')
} else {$_}
}
) -join ''
Outputs:
x1,x2,y1,y2,z3,x4,y4,z5,x6,y6
Where:
-split '((?<=x\d),(?!x)|(?<!x\d),(?=x))'
breaks apart the string into these groups:
x1,x2
,
z3
,
x4
,
z5
,
x6
using positive and negative lookahead and lookbehind:
comma with x\d before and without x after:
(?<=x\d),(?!x)
comma without x\d before and with x after:
(?<!x\d),(?=x)
For a string such as:
abzyxcabkmqfcmkcde
Notice that there are string patterns between ab and c in bold. To capture the first string pattern:
ab([a-z]{3,5})c
Is it possible to match both of the groups from the sample string? Actually, there should be 1 to 5 groups.
Note: python style regex.
You can verify that a given string conforms to the 1-5 repetitions of ab([a-z]{3,5})c using this regex
(?:ab([a-z]{3,5})c){1,5}
or this one if there are characters expected between the groups
(?:ab([a-z]{3,5})c.*?){1,5}
You will only be able to extract the last matching group from that string however, not any of the previous ones. to get a previous one you need to use hsz's approach
Just match all results - i.e. with g flag:
/ab([a-z]{3,5})c/g
or some method like in Python:
re.findall(pattern, string, flags=0)
I am trying to capture the bold part of strings like this:
'capture a year range at the end of a string 1995-2010'
'if there's no year range just capture the single year 2005'
'capture a year/year range followed by a parenthesis, including the parenthesis 2007-2012 (58 months)'
This regex works for 1 and 2, but I can't get it to work for 3:
/(\d+([-–— ]\d+( \(\d+ months\))?)?$)/
What am I doing wrong?
Try this regex:
/\d{4}(?:[-–— ]\d{4})?(?:\s*\([^)]+\))?$/gm
This one captures everything in the brackets.
If you need a regex specific to the text "(number) months" in the brackets, then you can use this: \d{4}(?:[-–— ]\d{4})?(?:\s+\(\d+\smonths\))?$
Link to test: RegexPal or RegExr
Sample text:
capture a year range at the end of a string 1995-2010
if there's no year range just capture the single year 2005
capture a year/year range followed by a parenthesis, including the
parenthesis 2007-2012 (58 months)
trying out another example 1990 (23 weeks)
trying out another example 1995-2002 (x days)
trying out another example 2050 (blah blah)
trying out another example 2050—3000
trying out another example 2050-3000
trying out another example 2050–3000
And the JavaScript code:
var regex = /\d{4}(?:[-–— ]\d{4})?(?:\s*\([^)]+\))?$/gm; //multiline enabled
var input = "your input string";
if(regex.test(input)) {
var matches = input.match(regex);
for(var match in matches) {
alert(matches[match]);
}
} else {
alert("No matches found!");
}
This Regex works nicely. :)
/(?:(?:\d{4}[-–— ])?\d{4})(?: \(\d+ months\))?$/
The main difference between my Regex and Jonah's is that mine contains ?: which means not to capture the sub-groups. When you group in a Regex it automatically returns what is in that group unless you tell it not to, and I've found that sometimes when those groups get captured when using methods such as replace or split, that it can be a little buggy which may be your problem as well.
The following regex works for me in a sample Perl script. It should be workable in JavaScript:
/(\d{4}([-–— ]\d{4})?( \(\d+ months\))?)$/
We first match a 4-digit year: \d{4}
Then we match an optional separator followed by another 4-digit year: ([-–— ]\d{4})?
Finally, we match the optional months portion: ( \(\d+ months\))?
You may need to insert whitespace matches (\s*) where needed, if your data doesn't always follow this strict template.
It actually works fine here, if I understand your needs correctly: Gskinner RegExr
Just alternate which sentence is the last, as $ will not count for newlines, just the end of the string.
I'm trying to learn something about regular expressions.
Here is what I'm going to match:
/parent/child
/parent/child?
/parent/child?firstparam=abc123
/parent/child?secondparam=def456
/parent/child?firstparam=abc123&secondparam=def456
/parent/child?secondparam=def456&firstparam=abc123
/parent/child?thirdparam=ghi789&secondparam=def456&firstparam=abc123
/parent/child?secondparam=def456&firstparam=abc123&thirdparam=ghi789
/parent/child?thirdparam=ghi789
/parent/child/
/parent/child/?
/parent/child/?firstparam=abc123
/parent/child/?secondparam=def456
/parent/child/?firstparam=abc123&secondparam=def456
/parent/child/?secondparam=def456&firstparam=abc123
/parent/child/?thirdparam=ghi789&secondparam=def456&firstparam=abc123
/parent/child/?secondparam=def456&firstparam=abc123&thirdparam=ghi789
/parent/child/?thirdparam=ghi789
My expression should "grabs" abc123 and def456.
And now just an example about what I'm not going to match ("question mark" is missing):
/parent/child/firstparam=abc123&secondparam=def456
Well, I built the following expression:
^(?:/parent/child){1}(?:^(?:/\?|\?)+(?:firstparam=([^&]*)|secondparam=([^&]*)|[^&]*)?)?
But that doesn't work.
Could you help me to understand what I'm doing wrong?
Thanks in advance.
UPDATE 1
Ok, I made other tests.
I'm trying to fix the previous version with something like this:
/parent/child(?:(?:\?|/\?)+(?:firstparam=([^&]*)|secondparam=([^&]*)|[^&]*)?)?$
Let me explain my idea:
Must start with /parent/child:
/parent/child
Following group is optional
(?: ... )?
The previous optional group must starts with ? or /?
(?:\?|/\?)+
Optional parameters (I grab values if specified parameters are part of querystring)
(?:firstparam=([^&]*)|secondparam=([^&]*)|[^&]*)?
End of line
$
Any advice?
UPDATE 2
My solution must be based just on regular expressions.
Just for example, I previously wrote the following one:
/parent/child(?:[?&/]*(?:firstparam=([^&]*)|secondparam=([^&]*)|[^&]*))*$
And that works pretty nice.
But it matches the following input too:
/parent/child/firstparam=abc123&secondparam=def456
How could I modify the expression in order to not match the previous string?
You didn't specify a language so I'll just usre Perl. So basically instead of matching everything, I just matched exactly what I thought you needed. Correct me if I am wrong please.
while ($subject =~ m/(?<==)\w+?(?=&|\W|$)/g) {
# matched text = $&
}
(?<= # Assert that the regex below can be matched, with the match ending at this position (positive lookbehind)
= # Match the character “=” literally
)
\\w # Match a single character that is a “word character” (letters, digits, and underscores)
+? # Between one and unlimited times, as few times as possible, expanding as needed (lazy)
(?= # Assert that the regex below can be matched, starting at this position (positive lookahead)
# Match either the regular expression below (attempting the next alternative only if this one fails)
& # Match the character “&” literally
| # Or match regular expression number 2 below (attempting the next alternative only if this one fails)
\\W # Match a single character that is a “non-word character”
| # Or match regular expression number 3 below (the entire group fails if this one fails to match)
\$ # Assert position at the end of the string (or before the line break at the end of the string, if any)
)
Output:
This regex will work as long as you know what your parameter names are going to be and you're sure that they won't change.
\/parent\/child\/?\?(?:(?:firstparam|secondparam|thirdparam)\=([\w]+)&?)(?:(?:firstparam|secondparam|thirdparam)\=([\w]+)&?)?(?:(?:firstparam|secondparam|thirdparam)\=([\w]+)&?)?
Whilst regex is not the best solution for this (the above code examples will be far more efficient, as string functions are way faster than regexes) this will work if you need a regex solution with up to 3 parameters. Out of interest, why must the solution use only regex?
In any case, this regex will match the following strings:
/parent/child?firstparam=abc123
/parent/child?secondparam=def456
/parent/child?firstparam=abc123&secondparam=def456
/parent/child?secondparam=def456&firstparam=abc123
/parent/child?thirdparam=ghi789&secondparam=def456&firstparam=abc123
/parent/child?secondparam=def456&firstparam=abc123&thirdparam=ghi789
/parent/child?thirdparam=ghi789
/parent/child/?firstparam=abc123
/parent/child/?secondparam=def456
/parent/child/?firstparam=abc123&secondparam=def456
/parent/child/?secondparam=def456&firstparam=abc123
/parent/child/?thirdparam=ghi789&secondparam=def456&firstparam=abc123
/parent/child/?secondparam=def456&firstparam=abc123&thirdparam=ghi789
/parent/child/?thirdparam=ghi789
It will now only match those containing query string parameters, and put them into capture groups for you.
What language are you using to process your matches?
If you are using preg_match with PHP, you can get the whole match as well as capture groups in an array with
preg_match($regex, $string, $matches);
Then you can access the whole match with $matches[0] and the rest with $matches[1], $matches[2], etc.
If you want to add additional parameters you'll also need to add them in the regex too, and add additional parts to get your data. For example, if you had
/parent/child/?secondparam=def456&firstparam=abc123&fourthparam=jkl01112&thirdparam=ghi789
The regex will become
\/parent\/child\/?\?(?:(?:firstparam|secondparam|thirdparam|fourthparam)\=([\w]+)&?)(?:(?:firstparam|secondparam|thirdparam|fourthparam)\=([\w]+)&?)?(?:(?:firstparam|secondparam|thirdparam|fourthparam)\=([\w]+)&?)?(?:(?:firstparam|secondparam|thirdparam|fourthparam)\=([\w]+)&?)?
This will become a bit more tedious to maintain as you add more parameters, though.
You can optionally include ^ $ at the start and end if the multi-line flag is enabled. If you also need to match the whole lines without query strings, wrap this whole regex in a non-capture group (including ^ $) and add
|(?:^\/parent\/child\/?\??$)
to the end.
You're not escaping the /s in your regex for starters and using {1} for a single repetition of something is unnecessary; you only use those when you want more than one repetition or a range of repetitions.
And part of what you're trying to do is simply not a good use of a regex. I'll show you an easier way to deal with that: you want to use something like split and put the information into a hash that you can check the contents of later. Because you didn't specify a language, I'm just going to use Perl for my example, but every language I know with regexes also has easy access to hashes and something like split, so this should be easy enough to port:
# I picked an example to show how this works.
my $route = '/parent/child/?first=123&second=345&third=678';
my %params; # I'm going to put those URL parameters in this hash.
# Perl has a way to let me avoid escaping the /s, but I wanted an example that
# works in other languages too.
if ($route =~ m/\/parent\/child\/\?(.*)/) { # Use the regex for this part
print "Matched route.\n";
# But NOT for this part.
my $query = $1; # $1 is a Perl thing. It contains what (.*) matched above.
my #items = split '&', $query; # Each item is something like param=123
foreach my $item (#items) {
my ($param, $value) = split '=', $item;
$params{$param} = $value; # Put the parameters in a hash for easy access.
print "$param set to $value \n";
}
}
# Now you can check the parameter values and do whatever you need to with them.
# And you can add new parameters whenever you want, etc.
if ($params{'first'} eq '123') {
# Do whatever
}
My solution:
/(?:\w+/)*(?:(?:\w+)?\?(?:\w+=\w+(?:&\w+=\w+)*)?|\w+|)
Explain:
/(?:\w+/)* match /parent/child/ or /parent/
(?:\w+)?\?(?:\w+=\w+(?:&\w+=\w+)*)? match child?firstparam=abc123 or ?firstparam=abc123 or ?
\w+ match text like child
..|) match nothing(empty)
If you need only query string, pattern would reduce such as:
/(?:\w+/)*(?:\w+)?\?(\w+=\w+(?:&\w+=\w+)*)
If you want to get every parameter from query string, this is a Ruby sample:
re = /\/(?:\w+\/)*(?:\w+)?\?(\w+=\w+(?:&\w+=\w+)*)/
s = '/parent/child?secondparam=def456&firstparam=abc123&thirdparam=ghi789'
if m = s.match(re)
query_str = m[1] # now, you can 100% trust this string
query_str.scan(/(\w+)=(\w+)/) do |param,value| #grab parameter
printf("%s, %s\n", param, value)
end
end
output
secondparam, def456
firstparam, abc123
thirdparam, ghi789
This script will help you.
First, i check, is there any symbol like ?.
Then, i kill first part of line (left from ?).
Next, i split line by &, where each value splitted by =.
my $r = q"/parent/child
/parent/child?
/parent/child?firstparam=abc123
/parent/child?secondparam=def456
/parent/child?firstparam=abc123&secondparam=def456
/parent/child?secondparam=def456&firstparam=abc123
/parent/child?thirdparam=ghi789&secondparam=def456&firstparam=abc123
/parent/child?secondparam=def456&firstparam=abc123&thirdparam=ghi789
/parent/child?thirdparam=ghi789
/parent/child/
/parent/child/?
/parent/child/?firstparam=abc123
/parent/child/?secondparam=def456
/parent/child/?firstparam=abc123&secondparam=def456
/parent/child/?secondparam=def456&firstparam=abc123
/parent/child/?thirdparam=ghi789&secondparam=def456&firstparam=abc123
/parent/child/?secondparam=def456&firstparam=abc123&thirdparam=ghi789
/parent/child/?thirdparam=ghi789";
for my $string(split /\n/, $r){
if (index($string,'?')!=-1){
substr($string, 0, index($string,'?')+1,"");
#say "string = ".$string;
if (index($string,'=')!=-1){
my #params = map{$_ = [split /=/, $_];}split/\&/, $string;
$"="\n";
say "$_->[0] === $_->[1]" for (#params);
say "######next########";
}
else{
#print "there is no params!"
}
}
else{
#say "there is no params!";
}
}