I know how to make a regex case-insensitive. This is not about case but about character width. I'm looking for something that is width-insensitive. In Japanese, you have half-width and full-width characters (consider ０１２３ vs 0123 or ABCD vs ＡＢＣＤ). You can make SQL Server databases width-insensitive with _WI (or width-sensitive with _WS). I was hoping there would be something similar for a regex.
I'm trying to find birth dates where the numbers can be half and full width. Here's an illustration of the problem
For a more specific date matching problem, here's another illustration:
So while \d{4} finds instances of 4 digits, it will not find 4 full-width digits, etc. The workaround I've found is to do something like [０１２３４５６７８９|\d]{4} like so:
But that feels really dirty. Is there a better way to do this?
To match ASCII or full-width digits you can use
[０-９\d]
Or, [\uFF10-\uFF19\d] if you need to abstract from using Unicode text in your source.
Note that in ECMAScript 2018 and later, you can use \p{N} or \p{Nd} to match all Unicode digits.
The current \p{N} range (encompassing Number, Decimal Digit (Nd), Number, Letter (Nl) and Number, Other (No) categories) regex matching 1,791 code points is
(?:[0-9\xB2\xB3\xB9\xBC-\xBE\u0660-\u0669\u06F0-\u06F9\u07C0-\u07C9\u0966-\u096F\u09E6-\u09EF\u09F4-\u09F9\u0A66-\u0A6F\u0AE6-\u0AEF\u0B66-\u0B6F\u0B72-\u0B77\u0BE6-\u0BF2\u0C66-\u0C6F\u0C78-\u0C7E\u0CE6-\u0CEF\u0D58-\u0D5E\u0D66-\u0D78\u0DE6-\u0DEF\u0E50-\u0E59\u0ED0-\u0ED9\u0F20-\u0F33\u1040-\u1049\u1090-\u1099\u1369-\u137C\u16EE-\u16F0\u17E0-\u17E9\u17F0-\u17F9\u1810-\u1819\u1946-\u194F\u19D0-\u19DA\u1A80-\u1A89\u1A90-\u1A99\u1B50-\u1B59\u1BB0-\u1BB9\u1C40-\u1C49\u1C50-\u1C59\u2070\u2074-\u2079\u2080-\u2089\u2150-\u2182\u2185-\u2189\u2460-\u249B\u24EA-\u24FF\u2776-\u2793\u2CFD\u3007\u3021-\u3029\u3038-\u303A\u3192-\u3195\u3220-\u3229\u3248-\u324F\u3251-\u325F\u3280-\u3289\u32B1-\u32BF\uA620-\uA629\uA6E6-\uA6EF\uA830-\uA835\uA8D0-\uA8D9\uA900-\uA909\uA9D0-\uA9D9\uA9F0-\uA9F9\uAA50-\uAA59\uABF0-\uABF9\uFF10-\uFF19]|\uD800[\uDD07-\uDD33\uDD40-\uDD78\uDD8A\uDD8B\uDEE1-\uDEFB\uDF20-\uDF23\uDF41\uDF4A\uDFD1-\uDFD5]|\uD801[\uDCA0-\uDCA9]|\uD802[\uDC58-\uDC5F\uDC79-\uDC7F\uDCA7-\uDCAF\uDCFB-\uDCFF\uDD16-\uDD1B\uDDBC\uDDBD\uDDC0-\uDDCF\uDDD2-\uDDFF\uDE40-\uDE48\uDE7D\uDE7E\uDE9D-\uDE9F\uDEEB-\uDEEF\uDF58-\uDF5F\uDF78-\uDF7F\uDFA9-\uDFAF]|\uD803[\uDCFA-\uDCFF\uDD30-\uDD39\uDE60-\uDE7E\uDF1D-\uDF26\uDF51-\uDF54\uDFC5-\uDFCB]|\uD804[\uDC52-\uDC6F\uDCF0-\uDCF9\uDD36-\uDD3F\uDDD0-\uDDD9\uDDE1-\uDDF4\uDEF0-\uDEF9]|\uD805[\uDC50-\uDC59\uDCD0-\uDCD9\uDE50-\uDE59\uDEC0-\uDEC9\uDF30-\uDF3B]|\uD806[\uDCE0-\uDCF2\uDD50-\uDD59]|\uD807[\uDC50-\uDC6C\uDD50-\uDD59\uDDA0-\uDDA9\uDFC0-\uDFD4]|\uD809[\uDC00-\uDC6E]|\uD81A[\uDE60-\uDE69\uDEC0-\uDEC9\uDF50-\uDF59\uDF5B-\uDF61]|\uD81B[\uDE80-\uDE96]|\uD834[\uDEE0-\uDEF3\uDF60-\uDF78]|\uD835[\uDFCE-\uDFFF]|\uD838[\uDD40-\uDD49\uDEF0-\uDEF9]|\uD83A[\uDCC7-\uDCCF\uDD50-\uDD59]|\uD83B[\uDC71-\uDCAB\uDCAD-\uDCAF\uDCB1-\uDCB4\uDD01-\uDD2D\uDD2F-\uDD3D]|\uD83C[\uDD00-\uDD0C]|\uD83E[\uDFF0-\uDFF9])
and the \p{Nd} (with 660 code points) converts to
(?:[0-9\u0660-\u0669\u06F0-\u06F9\u07C0-\u07C9\u0966-\u096F\u09E6-\u09EF\u0A66-\u0A6F\u0AE6-\u0AEF\u0B66-\u0B6F\u0BE6-\u0BEF\u0C66-\u0C6F\u0CE6-\u0CEF\u0D66-\u0D6F\u0DE6-\u0DEF\u0E50-\u0E59\u0ED0-\u0ED9\u0F20-\u0F29\u1040-\u1049\u1090-\u1099\u17E0-\u17E9\u1810-\u1819\u1946-\u194F\u19D0-\u19D9\u1A80-\u1A89\u1A90-\u1A99\u1B50-\u1B59\u1BB0-\u1BB9\u1C40-\u1C49\u1C50-\u1C59\uA620-\uA629\uA8D0-\uA8D9\uA900-\uA909\uA9D0-\uA9D9\uA9F0-\uA9F9\uAA50-\uAA59\uABF0-\uABF9\uFF10-\uFF19]|\uD801[\uDCA0-\uDCA9]|\uD803[\uDD30-\uDD39]|\uD804[\uDC66-\uDC6F\uDCF0-\uDCF9\uDD36-\uDD3F\uDDD0-\uDDD9\uDEF0-\uDEF9]|\uD805[\uDC50-\uDC59\uDCD0-\uDCD9\uDE50-\uDE59\uDEC0-\uDEC9\uDF30-\uDF39]|\uD806[\uDCE0-\uDCE9\uDD50-\uDD59]|\uD807[\uDC50-\uDC59\uDD50-\uDD59\uDDA0-\uDDA9]|\uD81A[\uDE60-\uDE69\uDEC0-\uDEC9\uDF50-\uDF59]|\uD835[\uDFCE-\uDFFF]|\uD838[\uDD40-\uDD49\uDEF0-\uDEF9]|\uD83A[\uDD50-\uDD59]|\uD83E[\uDFF0-\uDFF9])
Related
I'm not sure if using regex is the correct way to go about this here, but I wanted to try solving this with regex first (if it's possible)
I have an edifact file, where the data (in bold) in certain fields in some segments need to be substituted (with different dates, same format)
UNA:+,? '
UNB+UNOC:3+000000000+000000000+20190801:1115+00001+DDMP190001'
UNH+00001+BRKE:01+00+0'
INV+ED Format 1+Brustkrebs+19880117+E000000001+**20080702**+++1+0'
FAL+087897044+0000000++name+000000000+0+**20080702**++1+++J+N+N+N+N+N+++0'
INL+181095200+385762115+++0'
BEE+20080702++++0'
BAA+++J+J++++++J+++++++J++0'
BBA++++++++J++++++J+J++++++J+++++J+++J+J++++++++J+0'
BHP+J+++++J+++++J+++++0'
BLA+++J+++++++++0'
BFA++++++++++++J++0'
BSA++J+++J+J+++0'
BAT+20190801+0'
DAT+**20080702**++++0'
UNT+000014+00001'
UNZ+00001+00001'
at first I was able to match those fields using a positive lookahead and a lookbehind (I had different expressions for matching each date).
Here, for example is the expression I intially used to match the date in the "FAL" segment: (?<=\+[\d]{1}\+)\d{8}(?=\+\+), but then i saw that this date is sometimes preceeded by 9 digits, and sometimes by 1 (based on version) and followed by a either ++ or a + and a date so I added a logiacl OR like this: (?<=\+[\d]{9}\+|\+[\d]{1}\+)\d{8}(?=\+[\d]{8}\+|\+\+)and quickly realized it's not sustainable because I saw that these edifact files vary (far beyond only either 9 and 1 digits)
(I have 6 versions for each type, and i have 6 types total)
Because I have a scheme/map indicating what each version should be built like and I know on what position (based on the + separator) the date is written in each version, I thought about maybe matching the date based on the +, so after the 7th occurence (say in the FAL segment) of plus in a certain line, match the next 8 digits.
is this possible to achieve with regex? and if yes, could someone please tell me how?
I suggest using a pattern like
^((?:[^+\n]*\+){7})\d{8}(?=\+(?:\d{8})?\+)
where {7} can be adjusted to the value you need for each type of segments, and replace with the backreference to Group 1. In Python, it is \g<1>20200101 (where 20200101 is your new date), in PHP/.NET, it is ${1}20200101. In JS, it will be just $1.
To run on a multiline text, use m flag. In Python regex, you may embed it like (?m)^((?:[^+\n]*\+){7})\d{8}(?=\+(?:\d{8})?\+).
See the Python regex demo
Details
^ - start of string/line
((?:[^+\n]*\+){7}) - Group 1: 7 repetitions of any chars other than + and newline, and then a +
\d{8} - 8 digits
(?=\+(?:\d{8})?\+) - that are followed with +, and optional chunk of 8 digits and a +.
I need to validate uk numbers
Below are sample type of number
01457 341235
0229 111111
+1213 3133143
Optional Plus should be allowed at first postion only
Using this regex but not working
^(?:\W*\d){11}\W*$
An actual UK phone number will start with 0 or +44 (the latter being the UK country code), or possibly just 44, followed by nine or ten digits. A regex to capture that would look something like:
^(?:0|\+?44)(?:\d\s?){9,10}$
In this regex, I have allowed the digits to be separated by spaces in any way, because there isn't a single standardized way of breaking down the numbers. You could further narrow this down to certain allowed groupings, if you like, but it would greatly increase the complexity of the regex.
Your question implies you might want something broader or different. As some of your examples aren't valid UK numbers (+1213 3133143, 12345 123456).
You could use something like this to simply match between 10 and 12 digits, with arbitrary spacing, possibly preceded by a +:
^\+?(?:\d\s?){10,12}$
How can you create a regular expression that checks if a user input matches characters formally found in a currency syntax? (number, period/decimal place, comma, or dollar sign?).
The following can find all characters listed above except for the dollar sign, any idea how to properly structure this?
/([0-9.,])/g
The regex I use for currency validation is as follows:
^(\$)?([1-9]{1}[0-9]{0,2})(\,\d{3})*(\.\d{2})?$|^(\$)?([1-9]{1}[0-9]{0,2})(\d{3})*(\.\d{2})?$|^(0)?(\.\d{2})?$|^(\$0)?(\.\d{2})?$|^$
RegExr is a great website for testing and reviewing these strings (perhaps you could make a regex string that's less of a beast!)
Are you just trying to test the characters? In that case
[0-9,.$]+
will suffice. Or are you testing for the format $1,123,123.12 with the correct placements of commas and everything?
In that case you would need something more like
(\$?\d{1,3}(?:,\d{3})*(?:.\d{2})?)
should do.
You need to define what you want your regex to match, more formally than "matches characters formally found in a currency syntax". We don't know which currencies you're interested in. We don't know how strict you need it to be.
Maybe you'll come up with something like:
These elements must come in this order:
A currency symbol ('£', '€' or '$') (your requirement might specify more currencies)
1 or more numeric digits
A period or a comma
Exactly two numeric digits
Once you have a specification like that, it's easy to translate into a regular expression:
[£€$] // one of these chars.
\d+ // '+' means 'one or more'
[.,] // '[]' means 'any one of these'.
\d\d // Two digits. Could also be written as '\d{2}'
Or concatenated together:
[£€$]\d+[.,]\d\d
If you've learned about escaping special characters like $ and ., you may be surprised not to see it done here. Within [], they lose their special meaning.
(There are dialects of regex -- check the documentation for whatever implementation you're using)
Your requirements may be different though. The example I've given doesn't match:
$ 12.00
$12
USD12
¥200.00
25¢
$0.00005
20 μBTC
44 dollars
£1/19/11¾d ("one pound, nineteen shillings and elevenpence three farthings")
Work out your requirement, then write your code to meet it.
you should set \ before special chars, also you should set star(0+) or plus(1+) for match full currency chars, for example:
/([0-9\.,]*)/g
or for real price how 200,00 where all time exist 2 symbols after comma:
/(([0-9]+)(\.|,)([0-9]){2})/g
I need a RegEx pattern that will return the first N words using a custom word boundary that is the normal RegEx white space (\s) plus punctuation like .,;:!?-*_
EDIT #1: Thanks for all your comments.
To be clear:
I'd like to set the characters that would be the word delimiters
Lets call this the "Delimiter Set", or strDelimiters
strDelimiters = ".,;:!?-*_"
nNumWordsToFind = 5
A word is defined as any contiguous text that does NOT contain any character in strDelimiters
The RegEx word boundary is any contiguous text that contains one or more of the characters in strDelimiters
I'd like to build the RegEx pattern to get/return the first nNumWordsToFind using the strDelimiters.
EDIT #2: Sat, Aug 8, 2015 at 12:49 AM US CT
#maraca definitely answered my question as originally stated.
But what I actually need is to return the number of words ≤ nNumWordsToFind.
So if the source text has only 3 words, but my RegEx asks for 4 words, I need it to return the 3 words. The answer provided by maraca fails if nNumWordsToFind > number of actual words in the source text.
For example:
one,two;three-four_five.six:seven eight nine! ten
It would see this as 10 words.
If I want the first 5 words, it would return:
one,two;three-four_five.
I have this pattern using the normal \s whitespace, which works, but NOT exactly what I need:
([\w]+\s+){<NumWordsOut>}
where <NumWordsOut> is the number of words to return.
I have also found this word boundary pattern, but I don't know how to use it:
a "real word boundary" that detects the edge between an ASCII letter
and a non-letter.
(?i)(?<=^|[^a-z])(?=[a-z])|(?<=[a-z])(?=$|[^a-z])
However, I would want my words to allow numbers as well.
IAC, I have not been able how to use the above custom word boundary pattern to return the first N words of my text.
BTW, I will be using this in a Keyboard Maestro macro.
Can anyone help?
TIA.
All you have to do is to adapt your pattern ([\w]+\s+){<NumWordsOut>} to, including some special cases:
^[\s.,;:!?*_-]*([^\s.,;:!?*_-]+([\s.,;:!?*_-]+|$)){<NumWordsOut>}
1. 2. 3. 4. 5.
Match any amount of delimiters before the first word
Match a word (= at least one non-delimiter)
The word has to be followed by at least one delimiter
Or it can be at the end of the string (in case no delimiter follows at the end)
Repeat 2. to 4. <NumWordsOut> times
Note how I changed the order of the -, it has to be at the start or end, otherwise it needs to be escaped: \-.
Thanks to #maraca for providing the complete answer to my question.
I just wanted to post the Keyboard Maestro macro that I have built using #maraca's RegEx pattern for anyone interested in the complete solution.
See KM Forum Macro: Get a Max of N Words in String Using RegEx
I'm looking for a custom RegEx expression (that works!) to will validate common phone number with area code entries (no country code) such as:
111-111-1111
(111) 111-1111
(111)111-1111
111 111 1111
111.111.1111
1111111111
And combinations of these / anything else I may have forgotton.
Also, is it possible to have the RegEx expression itself reformat the entry? So take the 1111111111 and put it in 111-111-1111 format. The regex will most likely be entered in a Joomla / some type of CMS module, so I can't really add code to it aside from the expression itself.
\(?(\d{3})\)?[ .-]?(\d{3})[ .-]?(\d{4})
will match all your examples; after a match, backreference 1 will contain the area code, backreference 2 and 3 will contain the phone number.
I hope you don't need to handle international phone numbers, too.
If the phone number is in a string by itself, you could also use
^\s*\(?(\d{3})\)?[ .-]?(\d{3})[ .-]?(\d{4})\s*$
allowing for leading/trailing whitespace and nothing else.
Why not just remove spaces, parenthesis, dashes, and periods, then check that it is a number of 10 digits?
Depending on the language in question, you might be better off using a replace-like statement to replace non-numeric characters: ()-/. with nothing, and then just check if what is left is a 10-digit number.