I'm trying to develop a regex where it will take in format i.e:
date: 1:10 #7 (correct)
date: 1:10 (correct)
1:10 #7 (correct)
13.01.06 (incorrect)
Here is my regex developed on pythex:
(date)? ?\D? ?(1|4) ?(:|-|\.) ?[-+]?[0-9]+( ?(#) ?[a-zA-Z0-9]?)?
I'm working in python projects that uses OCR so sometimes the ":" between 1 and 10 is not translated correctly. Do you guys have a better way to tackle to regex problem?
Do you specifically need regex? If not, you could solve your problem by parsing the string to datetime.
from datetime import datetime
def test_date(date_string):
try:
datetime.strptime(date_string, '%d.%m.%y')
return # or do something else to skip the further processing
except ValueError:
pass
# Process valid date string
print('Valid date: {}'.format(date_string))
test_date('13.01.06') # Does not print anything
test_date('1:10 #7') # Works!
You need to make the non-digit pattern obligatory by removing ? after \D and wrap the whole part before the (1|4) pattern with an optional non-capturing group (to match date, : and space optionally), and in the end add a word boundary before the (1|4) pattern so that it could only be matched as a whole word, when the digit is not preceded with a digit, letter or _.
(?:(date)? ?\D ?)?\b([14]) ?([-:.]) ?[-+]?([0-9]+)( ?# ?([a-zA-Z0-9]*))?
^^^ ^ ^^^
See the regex demo.
Related
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})$
I am trying to extract just the emails from text column in openrefine. some cells have just the email, but others have the name and email in john doe <john#doe.com> format. I have been using the following GREL/regex but it does not return the entire email address. For the above exaple I'm getting ["n#doe.com"]
value.match(
/.*([a-zA-Z0-9_\-\+]+#[\._a-zA-Z0-9-]+).*/
)
Any help is much appreciated.
The n is captured because you are using .* before the capturing group, and since it can match any 0+ chars other than line break chars greedily the only char that can land in Group 1 during backtracking is the char right before #.
If you can get partial matches git rid of the .* and use
/[^<\s]+#[^\s>]+/
See the regex demo
Details
[^<\s]+ - 1 or more chars other than < and whitespace
# - a # char
[^\s>]+ - 1 or more chars other than whitespace and >.
Python/Jython implementation:
import re
res = ''
m = re.search(r'[^<\s]+#[^\s>]+', value)
if m:
res = m.group(0)
return res
There are other ways to match these strings. In case you need a full string match .*<([^<]+#[^>]+)>.* where .* will not gobble the name since it will stop before an obligatory <.
If some cells contain just the email, it's probably better to use the #wiktor-stribiżew's partial match. In the development version of Open Refine, there is now a value.find() function that can do this, but it will only be officially implemented in the next version (2.9). In the meantime, you can reproduce it using Python/Jython instead of GREL:
import re
return re.findall(r"[^<\s]+#[^\s>]+", value)[0]
Result :
How can I match any characters between a period and a semi-colon in Regex, including whitespace, carriage returns, and other periods and special characters including brackets?
For example, given:
SomeMethod().AnotherMethod("argument")
.AnotherMethod(1, "arg");
I want to be able to match:
.AnotherMethod("argument")
.AnotherMethod(1, "arg");
\.([^;]*); seems to fit the bill.
Note that the question you asked is probably simpler than what you really need: presumably you want to get a similar result from SomeMethod().AnotherMethod("foo; bar; and baz").AnotherMethod(1);, but the regex you've asked for will stop at the first ;.
Either [.](.*?); or [.]([^;]*);.
[.] is just another way to escape the . char, traditionally escaped using \.
\.([^;]*); works per your original spec, but you'll quickly run into problems...
In practice you'll need named groups and a recursive regex to parse this properly.
Here's a starting point (in ruby):
re = %r{
(?<sstr> '(?:[^']|(?<=\\)')*?') {0} # single quoted
(?<dstr> "(?:[^"]|(?<=\\)")*?") {0} # double quoted
(?<nstr> [^;]*?) {0} # non-string
(?<arg> (?:\g<sstr>|\g<dstr>|\g<nstr>|)) {0} # optional arg
(?<func> \s*[a-zA-Z0-9_]+\s*\(\s*\g<arg>\s*\)\s*) {0} # foo(arg)
(?<chain> \g<func>(?:\.\g<func>)*) {0} # chained foo(arg)
^\g<chain>; # match until ;
}x
pp re.match 'SomeMethod("With (a; Funky\', arg\"ument");'
Note that the above is only a partial solution. For instance, it wouldn't cover things like:
foo(cond()? 'even' : "more", "; funky")
But it'll hopefully get you started in the right direction, if you fail to find a built-in function/argument parser in the language that you're coding into.
I don't write many regular expressions so I'm going to need some help on the one.
I need a regular expression that can validate that a string is an alphanumeric comma delimited string.
Examples:
123, 4A67, GGG, 767 would be valid.
12333, 78787&*, GH778 would be invalid
fghkjhfdg8797< would be invalid
This is what I have so far, but isn't quite right: ^(?=.*[a-zA-Z0-9][,]).*$
Any suggestions?
Sounds like you need an expression like this:
^[0-9a-zA-Z]+(,[0-9a-zA-Z]+)*$
Posix allows for the more self-descriptive version:
^[[:alnum:]]+(,[[:alnum:]]+)*$
^[[:alnum:]]+([[:space:]]*,[[:space:]]*[[:alnum:]]+)*$ // allow whitespace
If you're willing to admit underscores, too, search for entire words (\w+):
^\w+(,\w+)*$
^\w+(\s*,\s*\w+)*$ // allow whitespaces around the comma
Try this pattern: ^([a-zA-Z0-9]+,?\s*)+$
I tested it with your cases, as well as just a single number "123". I don't know if you will always have a comma or not.
The [a-zA-Z0-9]+ means match 1 or more of these symbols
The ,? means match 0 or 1 commas (basically, the comma is optional)
The \s* handles 1 or more spaces after the comma
and finally the outer + says match 1 or more of the pattern.
This will also match
123 123 abc (no commas) which might be a problem
This will also match 123, (ends with a comma) which might be a problem.
Try the following expression:
/^([a-z0-9\s]+,)*([a-z0-9\s]+){1}$/i
This will work for:
test
test, test
test123,Test 123,test
I would strongly suggest trimming the whitespaces at the beginning and end of each item in the comma-separated list.
You seem to be lacking repetition. How about:
^(?:[a-zA-Z0-9 ]+,)*[a-zA-Z0-9 ]+$
I'm not sure how you'd express that in VB.Net, but in Python:
>>> import re
>>> x [ "123, $a67, GGG, 767", "12333, 78787&*, GH778" ]
>>> r = '^(?:[a-zA-Z0-9 ]+,)*[a-zA-Z0-9 ]+$'
>>> for s in x:
... print re.match( r, s )
...
<_sre.SRE_Match object at 0xb75c8218>
None
>>>>
You can use shortcuts instead of listing the [a-zA-Z0-9 ] part, but this is probably easier to understand.
Analyzing the highlights:
[a-zA-Z0-9 ]+ : capture one or more (but not zero) of the listed ranges, and space.
(?:[...]+,)* : In non-capturing parenthesis, match one or more of the characters, plus a comma at the end. Match such sequences zero or more times. Capturing zero times allows for no comma.
[...]+ : capture at least one of these. This does not include a comma. This is to ensure that it does not accept a trailing comma. If a trailing comma is acceptable, then the expression is easier: ^[a-zA-Z0-9 ,]+
Yes, when you want to catch comma separated things where a comma at the end is not legal, and the things match to $LONGSTUFF, you have to repeat $LONGSTUFF:
$LONGSTUFF(,$LONGSTUFF)*
If $LONGSTUFF is really long and contains comma repeated items itself etc., it might be a good idea to not build the regexp by hand and instead rely on a computer for doing that for you, even if it's just through string concatenation. For example, I just wanted to build a regular expression to validate the CPUID parameter of a XEN configuration file, of the ['1:a=b,c=d','2:e=f,g=h'] type. I... believe this mostly fits the bill: (whitespace notwithstanding!)
xend_fudge_item_re = r"""
e[a-d]x= #register of the call return value to fudge
(
0x[0-9A-F]+ | #either hardcode the reply
[10xks]{32} #or edit the bitfield directly
)
"""
xend_string_item_re = r"""
(0x)?[0-9A-F]+: #leafnum (the contents of EAX before the call)
%s #one fudge
(,%s)* #repeated multiple times
""" % (xend_fudge_item_re, xend_fudge_item_re)
xend_syntax = re.compile(r"""
\[ #a list of
'%s' #string elements
(,'%s')* #repeated multiple times
\]
$ #and nothing else
""" % (xend_string_item_re, xend_string_item_re), re.VERBOSE | re.MULTILINE)
Try ^(?!,)((, *)?([a-zA-Z0-9])\b)*$
Step by step description:
Don't match a beginning comma (good for the upcoming "loop").
Match optional comma and spaces.
Match characters you like.
The match of a word boundary make sure that a comma is necessary if more arguments are stacked in string.
Please use - ^((([a-zA-Z0-9\s]){1,45},)+([a-zA-Z0-9\s]){1,45})$
Here, I have set max word size to 45, as longest word in english is 45 characters, can be changed as per requirement
How do I get the substring " It's big \"problem " using a regular expression?
s = ' function(){ return " It\'s big \"problem "; }';
/"(?:[^"\\]|\\.)*"/
Works in The Regex Coach and PCRE Workbench.
Example of test in JavaScript:
var s = ' function(){ return " Is big \\"problem\\", \\no? "; }';
var m = s.match(/"(?:[^"\\]|\\.)*"/);
if (m != null)
alert(m);
This one comes from nanorc.sample available in many linux distros. It is used for syntax highlighting of C style strings
\"(\\.|[^\"])*\"
As provided by ePharaoh, the answer is
/"([^"\\]*(\\.[^"\\]*)*)"/
To have the above apply to either single quoted or double quoted strings, use
/"([^"\\]*(\\.[^"\\]*)*)"|\'([^\'\\]*(\\.[^\'\\]*)*)\'/
Most of the solutions provided here use alternative repetition paths i.e. (A|B)*.
You may encounter stack overflows on large inputs since some pattern compiler implements this using recursion.
Java for instance: http://bugs.java.com/bugdatabase/view_bug.do?bug_id=6337993
Something like this:
"(?:[^"\\]*(?:\\.)?)*", or the one provided by Guy Bedford will reduce the amount of parsing steps avoiding most stack overflows.
/(["\']).*?(?<!\\)(\\\\)*\1/is
should work with any quoted string
"(?:\\"|.)*?"
Alternating the \" and the . passes over escaped quotes while the lazy quantifier *? ensures that you don't go past the end of the quoted string. Works with .NET Framework RE classes
/"(?:[^"\\]++|\\.)*+"/
Taken straight from man perlre on a Linux system with Perl 5.22.0 installed.
As an optimization, this regex uses the 'posessive' form of both + and * to prevent backtracking, for it is known beforehand that a string without a closing quote wouldn't match in any case.
This one works perfect on PCRE and does not fall with StackOverflow.
"(.*?[^\\])??((\\\\)+)?+"
Explanation:
Every quoted string starts with Char: " ;
It may contain any number of any characters: .*? {Lazy match}; ending with non escape character [^\\];
Statement (2) is Lazy(!) optional because string can be empty(""). So: (.*?[^\\])??
Finally, every quoted string ends with Char("), but it can be preceded with even number of escape sign pairs (\\\\)+; and it is Greedy(!) optional: ((\\\\)+)?+ {Greedy matching}, bacause string can be empty or without ending pairs!
An option that has not been touched on before is:
Reverse the string.
Perform the matching on the reversed string.
Re-reverse the matched strings.
This has the added bonus of being able to correctly match escaped open tags.
Lets say you had the following string; String \"this "should" NOT match\" and "this \"should\" match"
Here, \"this "should" NOT match\" should not be matched and "should" should be.
On top of that this \"should\" match should be matched and \"should\" should not.
First an example.
// The input string.
const myString = 'String \\"this "should" NOT match\\" and "this \\"should\\" match"';
// The RegExp.
const regExp = new RegExp(
// Match close
'([\'"])(?!(?:[\\\\]{2})*[\\\\](?![\\\\]))' +
'((?:' +
// Match escaped close quote
'(?:\\1(?=(?:[\\\\]{2})*[\\\\](?![\\\\])))|' +
// Match everything thats not the close quote
'(?:(?!\\1).)' +
'){0,})' +
// Match open
'(\\1)(?!(?:[\\\\]{2})*[\\\\](?![\\\\]))',
'g'
);
// Reverse the matched strings.
matches = myString
// Reverse the string.
.split('').reverse().join('')
// '"hctam "\dluohs"\ siht" dna "\hctam TON "dluohs" siht"\ gnirtS'
// Match the quoted
.match(regExp)
// ['"hctam "\dluohs"\ siht"', '"dluohs"']
// Reverse the matches
.map(x => x.split('').reverse().join(''))
// ['"this \"should\" match"', '"should"']
// Re order the matches
.reverse();
// ['"should"', '"this \"should\" match"']
Okay, now to explain the RegExp.
This is the regexp can be easily broken into three pieces. As follows:
# Part 1
(['"]) # Match a closing quotation mark " or '
(?! # As long as it's not followed by
(?:[\\]{2})* # A pair of escape characters
[\\] # and a single escape
(?![\\]) # As long as that's not followed by an escape
)
# Part 2
((?: # Match inside the quotes
(?: # Match option 1:
\1 # Match the closing quote
(?= # As long as it's followed by
(?:\\\\)* # A pair of escape characters
\\ #
(?![\\]) # As long as that's not followed by an escape
) # and a single escape
)| # OR
(?: # Match option 2:
(?!\1). # Any character that isn't the closing quote
)
)*) # Match the group 0 or more times
# Part 3
(\1) # Match an open quotation mark that is the same as the closing one
(?! # As long as it's not followed by
(?:[\\]{2})* # A pair of escape characters
[\\] # and a single escape
(?![\\]) # As long as that's not followed by an escape
)
This is probably a lot clearer in image form: generated using Jex's Regulex
Image on github (JavaScript Regular Expression Visualizer.)
Sorry, I don't have a high enough reputation to include images, so, it's just a link for now.
Here is a gist of an example function using this concept that's a little more advanced: https://gist.github.com/scagood/bd99371c072d49a4fee29d193252f5fc#file-matchquotes-js
here is one that work with both " and ' and you easily add others at the start.
("|')(?:\\\1|[^\1])*?\1
it uses the backreference (\1) match exactley what is in the first group (" or ').
http://www.regular-expressions.info/backref.html
One has to remember that regexps aren't a silver bullet for everything string-y. Some stuff are simpler to do with a cursor and linear, manual, seeking. A CFL would do the trick pretty trivially, but there aren't many CFL implementations (afaik).
A more extensive version of https://stackoverflow.com/a/10786066/1794894
/"([^"\\]{50,}(\\.[^"\\]*)*)"|\'[^\'\\]{50,}(\\.[^\'\\]*)*\'|“[^”\\]{50,}(\\.[^“\\]*)*”/
This version also contains
Minimum quote length of 50
Extra type of quotes (open “ and close ”)
If it is searched from the beginning, maybe this can work?
\"((\\\")|[^\\])*\"
I faced a similar problem trying to remove quoted strings that may interfere with parsing of some files.
I ended up with a two-step solution that beats any convoluted regex you can come up with:
line = line.replace("\\\"","\'"); // Replace escaped quotes with something easier to handle
line = line.replaceAll("\"([^\"]*)\"","\"x\""); // Simple is beautiful
Easier to read and probably more efficient.
If your IDE is IntelliJ Idea, you can forget all these headaches and store your regex into a String variable and as you copy-paste it inside the double-quote it will automatically change to a regex acceptable format.
example in Java:
String s = "\"en_usa\":[^\\,\\}]+";
now you can use this variable in your regexp or anywhere.
(?<="|')(?:[^"\\]|\\.)*(?="|')
" It\'s big \"problem "
match result:
It\'s big \"problem
("|')(?:[^"\\]|\\.)*("|')
" It\'s big \"problem "
match result:
" It\'s big \"problem "
Messed around at regexpal and ended up with this regex: (Don't ask me how it works, I barely understand even tho I wrote it lol)
"(([^"\\]?(\\\\)?)|(\\")+)+"