What regular expression will match valid international phone numbers? - regex
I need to determine whether a phone number is valid before attempting to dial it. The phone call can go anywhere in the world.
What regular expression will match valid international phone numbers?
\+(9[976]\d|8[987530]\d|6[987]\d|5[90]\d|42\d|3[875]\d|
2[98654321]\d|9[8543210]|8[6421]|6[6543210]|5[87654321]|
4[987654310]|3[9643210]|2[70]|7|1)\d{1,14}$
Is the correct format for matching a generic international phone number. I replaced the US land line centric international access code 011 with the standard international access code identifier of '+', making it mandatory. I also changed the minimum for the national number to at least one digit.
Note that if you enter numbers in this format into your mobile phone address book, you may successfully call any number in your address book no matter where you travel. For land lines, replace the plus with the international access code for the country you are dialing from.
Note that this DOES NOT take into account national number plan rules - specifically, it allows zeros and ones in locations that national number plans may not allow and also allows number lengths greater than the national number plan for some countries (e.g., the US).
All country codes are defined by the ITU. The following regex is based on ITU-T E.164 and Annex to ITU Operational Bulletin No. 930 – 15.IV.2009. It contains all current country codes and codes reserved for future use. While it could be shortened a bit, I decided to include each code independently.
This is for calls originating from the USA. For other countries, replace the international access code (the 011 at the beginning of the regex) with whatever is appropriate for that country's dialing plan.
Also, note that ITU E.164 defines the maximum length of a full international telephone number to 15 digits. This means a three digit country code results in up to 12 additional digits, and a 1 digit country code could contain up to 14 additional digits. Hence the
[0-9]{0,14}$
a the end of the regex.
Most importantly, this regex does not mean the number is valid - each country defines its own internal numbering plan. This only ensures that the country code is valid.
^011(999|998|997|996|995|994|993|992|991|
990|979|978|977|976|975|974|973|972|971|970|
969|968|967|966|965|964|963|962|961|960|899|
898|897|896|895|894|893|892|891|890|889|888|
887|886|885|884|883|882|881|880|879|878|877|
876|875|874|873|872|871|870|859|858|857|856|
855|854|853|852|851|850|839|838|837|836|835|
834|833|832|831|830|809|808|807|806|805|804|
803|802|801|800|699|698|697|696|695|694|693|
692|691|690|689|688|687|686|685|684|683|682|
681|680|679|678|677|676|675|674|673|672|671|
670|599|598|597|596|595|594|593|592|591|590|
509|508|507|506|505|504|503|502|501|500|429|
428|427|426|425|424|423|422|421|420|389|388|
387|386|385|384|383|382|381|380|379|378|377|
376|375|374|373|372|371|370|359|358|357|356|
355|354|353|352|351|350|299|298|297|296|295|
294|293|292|291|290|289|288|287|286|285|284|
283|282|281|280|269|268|267|266|265|264|263|
262|261|260|259|258|257|256|255|254|253|252|
251|250|249|248|247|246|245|244|243|242|241|
240|239|238|237|236|235|234|233|232|231|230|
229|228|227|226|225|224|223|222|221|220|219|
218|217|216|215|214|213|212|211|210|98|95|94|
93|92|91|90|86|84|82|81|66|65|64|63|62|61|60|
58|57|56|55|54|53|52|51|49|48|47|46|45|44|43|
41|40|39|36|34|33|32|31|30|27|20|7|1)[0-9]{0,
14}$
This is a further optimisation.
\+(9[976]\d|8[987530]\d|6[987]\d|5[90]\d|42\d|3[875]\d|
2[98654321]\d|9[8543210]|8[6421]|6[6543210]|5[87654321]|
4[987654310]|3[9643210]|2[70]|7|1)
\W*\d\W*\d\W*\d\W*\d\W*\d\W*\d\W*\d\W*\d\W*(\d{1,2})$
(i) allows for valid international prefixes
(ii) followed by 9 or 10 digits, with any type or placing of delimeters (except between the last two digits)
This will match:
+1-234-567-8901
+61-234-567-89-01
+46-234 5678901
+1 (234) 56 89 901
+1 (234) 56-89 901
+46.234.567.8901
+1/234/567/8901
You can use the library libphonenumber from Google.
PhoneNumberUtil phoneNumberUtil = PhoneNumberUtil.getInstance();
String decodedNumber = null;
PhoneNumber number;
try {
number = phoneNumberUtil.parse(encodedHeader, null);
decodedNumber = phoneNumberUtil.format(number, PhoneNumberFormat.E164);
} catch (NumberParseException e) {
e.printStackTrace();
}
Modified #Eric's regular expression - added a list of all country codes (got them from xxxdepy # Github.
I hope you will find it helpful:
/(\+|00)(297|93|244|1264|358|355|376|971|54|374|1684|1268|61|43|994|257|32|229|226|880|359|973|1242|387|590|375|501|1441|591|55|1246|673|975|267|236|1|61|41|56|86|225|237|243|242|682|57|269|238|506|53|5999|61|1345|357|420|49|253|1767|45|1809|1829|1849|213|593|20|291|212|34|372|251|358|679|500|33|298|691|241|44|995|44|233|350|224|590|220|245|240|30|1473|299|502|594|1671|592|852|504|385|509|36|62|44|91|246|353|98|964|354|972|39|1876|44|962|81|76|77|254|996|855|686|1869|82|383|965|856|961|231|218|1758|423|94|266|370|352|371|853|590|212|377|373|261|960|52|692|389|223|356|95|382|976|1670|258|222|1664|596|230|265|60|262|264|687|227|672|234|505|683|31|47|977|674|64|968|92|507|64|51|63|680|675|48|1787|1939|850|351|595|970|689|974|262|40|7|250|966|249|221|65|500|4779|677|232|503|378|252|508|381|211|239|597|421|386|46|268|1721|248|963|1649|235|228|66|992|690|993|670|676|1868|216|90|688|886|255|256|380|598|1|998|3906698|379|1784|58|1284|1340|84|678|681|685|967|27|260|263)(9[976]\d|8[987530]\d|6[987]\d|5[90]\d|42\d|3[875]\d|2[98654321]\d|9[8543210]|8[6421]|6[6543210]|5[87654321]|4[987654310]|3[9643210]|2[70]|7|1)\d{4,20}$/
No criticism regarding those great answers I just want to present the simple solution I use for our admin content creators:
^(\+|00)[1-9][0-9 \-\(\)\.]{7,32}$
Force start with a plus or two zeros and use at least a little bit of numbers. White space, brackets, minus and point are optional, no other characters allowed.
You can safely remove all non-numbers and use this in a tel: input. Numbers will have a common form of representation and I do not have to worry about being to restrictive.
I use this one:
/([0-9\s\-]{7,})(?:\s*(?:#|x\.?|ext\.?|extension)\s*(\d+))?$/
Advantages: recognizes + or 011 beginnings, lets it be as long as needed, and handles many extension conventions. (#,x,ext,extension)
This will work for international numbers;
C#:
#"^((\+\d{1,3}(-| )?\(?\d\)?(-| )?\d{1,5})|(\(?\d{2,6}\)?))(-| )?(\d{3,4})(-| )?(\d{4})(( x| ext)\d{1,5}){0,1}$"
JS:
/^((\+\d{1,3}(-| )?\(?\d\)?(-| )?\d{1,5})|(\(?\d{2,6}\)?))(-| )?(\d{3,4})(-| )?(\d{4})(( x| ext)\d{1,5}){0,1}$/
Here's an "optimized" version of your regex:
^011(9[976]\d|8[987530]\d|6[987]\d|5[90]\d|42\d|3[875]\d|
2[98654321]\d|9[8543210]|8[6421]|6[6543210]|5[87654321]|
4[987654310]|3[9643210]|2[70]|7|1)\d{0,14}$
You can replace the \ds with [0-9] if your regex syntax doesn't support \d.
For iOS SWIFT I found this helpful,
let phoneRegEx = "^((\\+)|(00)|(\\*)|())[0-9]{3,14}((\\#)|())$"
Here is a regex for the following most common phone number scenarios. Although this is tailored from a US perspective for area codes it works for international scenarios.
The actual number should be 10 digits only.
For US numbers area code may be surrounded with parentheses ().
The country code can be 1 to 3 digits long. Optionally may be preceded by a + sign.
There may be dashes, spaces, dots or no spaces between country code, area code and the rest of the number.
A valid phone number cannot be all zeros.
^(?!\b(0)\1+\b)(\+?\d{1,3}[. -]?)?\(?\d{3}\)?([. -]?)\d{3}\3\d{4}$
Explanation:
^ - start of expression
(?!\b(0)\1+\b) - (?!)Negative Look ahead. \b - word boundary around a '0' character. \1 backtrack to previous capturing group (zero). Basically don't match all zeros.
(\+?\d{1,3}[. -]?)? - '\+?' plus sign before country code is optional.\d{1,3} - country code can be 1 to 3 digits long. '[. -]?' - spaces,dots and dashes are optional. The last question mark is to make country code optional.
\(?\d{3}\)? - '\)?' is to make parentheses optional. \d{3} - match 3 digit area code.
([. -]?) - optional space, dash or dot
$ - end of expression
More examples and explanation - https://regex101.com/r/hTH8Ct/2/
I have used this below:
^(\+|00)[0-9]{1,3}[0-9]{4,14}(?:x.+)?$
The format +CCC.NNNNNNNNNNxEEEE or 00CCC.NNNNNNNNNNxEEEE
Phone number must start with '+' or '00' for an international call.
where C is the 1–3 digit country code,
N is up to 14 digits,
and E is the (optional) extension.
The leading plus sign and the dot following the country code are required. The literal “x” character is required only if an extension is provided.
I made the regexp for european phone numbers, and it is specific against dial prefix vs length of number.
const PhoneEuropeRegExp = () => {
// eu phones map https://en.wikipedia.org/wiki/Telephone_numbers_in_Europe
const phonesMap = {
"43": [4, 13],
"32": [8, 10],
"359": [7, 9],
"385": [8, 9],
"357": 8,
"420": 9,
"45": 8,
"372": 7,
"358": [5, 12],
"33": 9,
"350": 8,
"49": [3, 12],
"30": 10,
"36": [8, 9],
"354": [7, 9],
"353": [7, 9],
"39": [6, 12],
"371": 8,
"423": [7, 12],
"370": 8,
"352": 8,
"356": 8,
"31": 9,
"47": [4, 12],
"48": 9,
"351": 9,
"40": 9,
"421": 9,
"386": 8,
"34": 9,
"46": [6, 9],
};
const regExpBuilt = Object.keys(phonesMap)
.reduce(function(prev, key) {
const val = phonesMap[key];
if (Array.isArray(val)) {
prev.push("(\\+" + key + `[0-9]\{${val[0]},${val[1]}\})`);
} else {
prev.push("(\\+" + key + `[0-9]\{${val}\})`);
}
return prev;
}, [])
.join("|");
return new RegExp(`^(${regExpBuilt})$`);
};
alert(PhoneEuropeRegExp().test("+420123456789"))
I only check for valid characters and allow up to 30 characters. Numbers that include an extension are also possible.
^[\+\(\s.\-\/\d\)]{5,30}$
Matches the following:
(0123) 123 456 1
555-555-5555
0049 1555 532-3455
123 456 7890
0761 12 34 56
+49 123 1-234-567-8901
+61-234-567-89-01
+46-234 5678901
+1 (234) 56 89 901
+1 (234) 56-89 901
+46.234.567.8901
+1/234/567/8901
It works pretty well with 00xx and +xx:
^(?:00|\+)(9[976]\d|8[987530]\d|6[987]\d|5[90]\d|42\d|3[875]\d|2[98654321]\d|9[8543210]|8[6421]|6[6543210]|5[87654321]|4[987654310]|3[9643210]|2[70]|7|1)\d{1,14}$
There's obviously a multitude of ways to do this, as evidenced by all of the different answers given thus far, but I'll throw my $0.02 worth in here and provide the regex below, which is a bit more terse than nearly all of the above, but more thorough than most as well. It also has the nice side-effect of leaving the country code in $1 and the local number in $2.
^\+(?=\d{5,15}$)(1|2[078]|3[0-469]|4[013-9]|5[1-8]|6[0-6]|7|8[1-469]|9[0-58]|[2-9]..)(\d+)$
A simple version for european numbers, that matches numbers like 0034617393211 but also long ones as 004401484172842.
^0{2}[0-9]{11,}
Hope it helps :·)
public static boolean validateInternationalPhoneNumberFormat(String phone) {
StringBuilder sb = new StringBuilder(200);
// Country code
sb.append("^(\\+{1}[\\d]{1,3})?");
// Area code, with or without parentheses
sb.append("([\\s])?(([\\(]{1}[\\d]{2,3}[\\)]{1}[\\s]?)|([\\d]{2,3}[\\s]?))?");
// Phone number separator can be "-", "." or " "
// Minimum of 5 digits (for fixed line phones in Solomon Islands)
sb.append("\\d[\\-\\.\\s]?\\d[\\-\\.\\s]?\\d[\\-\\.\\s]?\\d[\\-\\.\\s]?\\d[\\-\\.\\s]?");
// 4 more optional digits
sb.append("\\d?[\\-\\.\\s]?\\d?[\\-\\.\\s]?\\d?[\\-\\.\\s]?\\d?$");
return Pattern.compile(sb.toString()).matcher(phone).find();
}
The international numbering plan is based on the ITU E.164 numbering plan. I guess that's the starting point to your regular expression.
I'll update this if I get around to create a regular expression based on the ITU E.164 numbering.
This Regex Expression works for India, Canada, Europe, New Zealand, Australia, United States phone numbers, along with their country codes:
"^(\+(([0-9]){1,2})[-.])?((((([0-9]){2,3})[-.]){1,2}([0-9]{4,10}))|([0-9]{10}))$"
This works for me, without 00, 001, 0011 etc prefix though:
/^\+*(\d{3})*[0-9,\-]{8,}/
Try this, it works for me.
^(00|\+)[1-9]{1}([0-9][\s]*){9,16}$
^\+[1-9]\d{10,14}$
This will match "e164 phone numbers"
Related
Regex for UK phone number
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}$
Regular expression for address field validation
I am trying to write a regular expression that facilitates an address, example 21-big walk way or 21 St.Elizabeth's drive I came up with the following regular expression but I am not too keen to how to incorporate all the characters (alphanumeric, space dash, full stop, apostrophe) "regexp=^[A-Za-z-0-99999999'
See the answer to this question on address validating with regex:
regex street address match
The problem is, street addresses vary so much in formatting that it's hard to code against them. If you are trying to validate addresses, finding if one isn't valid based on its format is mighty hard to do.
This would return the following address (253 N. Cherry St. ), anything with its same format:
\d{1,5}\s\w.\s(\b\w*\b\s){1,2}\w*\.
This allows 1-5 digits for the house number, a space, a character followed by a period (for N. or S.), 1-2 words for the street name, finished with an abbreviation (like st. or rd.).
Because regex is used to see if things meet a standard or protocol (which you define), you probably wouldn't want to allow for the addresses provided above, especially the first one with the dash, since they aren't very standard. you can modify my above code to allow for them if you wish--you could add
(-?)
to allow for a dash but not require one.
In addition, http://rubular.com/ is a quick and interactive way to learn regex. Try it out with the addresses above.
In case if you don't have a fixed format for the address as mentioned above, I would use regex expression just to eliminate the symbols which are not used in the address (like specialized sybmols - &(%#$^). Result would be: [A-Za-z0-9'\.\-\s\,]
Just to add to Serzas' answer(since don't have enough reps. to comment). alphabets and numbers can effectively be replaced by \w for words. Additionally apostrophe,comma,period and hyphen doesn't necessarily need a backslash. My requirement also involved front and back slashes so \/ and finally whitespaces with \s. The working regex for me ,as such was : pattern: "[\w',-\\/.\s]"
Regular expression for simple address validation ^[#.0-9a-zA-Z\s,-]+$ E.g. for Address match case #1, North Street, Chennai - 11 E.g. for Address not match case $1, North Street, Chennai # 11
I have succesfully used ;
Dim regexString = New stringbuilder
With regexString
.Append("(?<h>^[\d]+[ ])(?<s>.+$)|") 'find the 2013 1st ambonstreet
.Append("(?<s>^.*?)(?<h>[ ][\d]+[ ])(?<e>[\D]+$)|") 'find the 1-7-4 Dual Ampstreet 130 A
.Append("(?<s>^[\D]+[ ])(?<h>[\d]+)(?<e>.*?$)|") 'find the Terheydenlaan 320 B3
.Append("(?<s>^.*?)(?<h>\d*?$)") 'find the 245e oosterkade 9
End With
Dim Address As Match = Regex.Match(DataRow("customerAddressLine1"), regexString.ToString(), RegexOptions.Multiline)
If Not String.IsNullOrEmpty(Address.Groups("s").Value) Then StreetName = Address.Groups("s").Value
If Not String.IsNullOrEmpty(Address.Groups("h").Value) Then HouseNumber = Address.Groups("h").Value
If Not String.IsNullOrEmpty(Address.Groups("e").Value) Then Extension = Address.Groups("e").Value
The regex will attempt to find a result, if there is none, it move to the next alternative. If no result is found, none of the 4 formats where present.
This one worked for me: \d+[ ](?:[A-Za-z0-9.-]+[ ]?)+(?:Avenue|Lane|Road|Boulevard|Drive|Street|Ave|Dr|Rd|Blvd|Ln|St)\.? The source: https://www.codeproject.com/Tips/989012/Validate-and-Find-Addresses-with-RegEx
Regex is a very bad choice for this kind of task. Try to find a web service or an address database or a product which can clean address data instead. Related: Address validation using Google Maps API
As a simple one line expression recommend this,
^([a-zA-z0-9/\\''(),-\s]{2,255})$
I needed
STREET # | STREET | CITY | STATE | ZIP
So I wrote the following regex
[0-9]{1,5}( [a-zA-Z.]*){1,4},?( [a-zA-Z]*){1,3},? [a-zA-Z]{2},? [0-9]{5}
This allows
1-5 Street #s
1-4 Street description words
1-3 City words
2 Char State
5 Char Zip code
I also added option , for separating street, city, state, zip
Here is the approach I have taken to finding addresses using regular expressions:
A set of patterns is useful to find many forms that we might expect from an address starting with simply a number followed by set of strings (ex. 1 Basic Road) and then getting more specific such as looking for "P.O. Box", "c/o", "attn:", etc.
Below is a simple test in python. The test will find all the addresses but not the last 4 items which are company names. This example is not comprehensive, but can be altered to suit your needs and catch examples you find in your data.
import re
strings = [
'701 FIFTH AVE',
'2157 Henderson Highway',
'Attn: Patent Docketing',
'HOLLYWOOD, FL 33022-2480',
'1940 DUKE STREET',
'111 MONUMENT CIRCLE, SUITE 3700',
'c/o Armstrong Teasdale LLP',
'1 Almaden Boulevard',
'999 Peachtree Street NE',
'P.O. BOX 2903',
'2040 MAIN STREET',
'300 North Meridian Street',
'465 Columbus Avenue',
'1441 SEAMIST DR.',
'2000 PENNSYLVANIA AVENUE, N.W.',
'465 Columbus Avenue',
'28 STATE STREET',
'P.O, Drawer 800889.',
'2200 CLARENDON BLVD.',
'840 NORTH PLANKINTON AVENUE',
'1025 Connecticut Avenue, NW',
'340 Commercial Street',
'799 Ninth Street, NW',
'11318 Lazarro Ln',
'P.O, Box 65745',
'c/o Ballard Spahr LLP',
'8210 SOUTHPARK TERRACE',
'1130 Connecticut Ave., NW, Suite 420',
'465 Columbus Avenue',
"BANNER & WITCOFF , LTD",
"CHIP LAW GROUP",
"HAMMER & ASSOCIATES, P.C.",
"MH2 TECHNOLOGY LAW GROUP, LLP",
]
patterns = [
"c\/o [\w ]{2,}",
"C\/O [\w ]{2,}",
"P.O\. [\w ]{2,}",
"P.O\, [\w ]{2,}",
"[\w\.]{2,5} BOX [\d]{2,8}",
"^[#\d]{1,7} [\w ]{2,}",
"[A-Z]{2,2} [\d]{5,5}",
"Attn: [\w]{2,}",
"ATTN: [\w]{2,}",
"Attention: [\w]{2,}",
"ATTENTION: [\w]{2,}"
]
contact_list = []
total_count = len(strings)
found_count = 0
for string in strings:
pat_no = 1
for pattern in patterns:
match = re.search(pattern, string.strip())
if match:
print("Item found: " + match.group(0) + " | Pattern no: " + str(pat_no))
found_count += 1
pat_no += 1
print("-- Total: " + str(total_count) + " Found: " + str(found_count))
UiPath Academy training video lists this RegEx for US addresses (and it works fine for me):
\b\d{1,8}(-)?[a-z]?\W[a-z|\W|\.]{1,}\W(road|drive|avenue|boulevard|circle|street|lane|waylrd\.|st\.|dr\.|ave\.|blvd\.|cir\.|In\.|rd|dr|ave|blvd|cir|ln)
I had a different use case - find any addresses in logs and scold application developers (favourite part of a devops job). I had the advantage of having the word "address" in the pattern but should work without that if you have specific field to scan
\baddress.[0-9\\\/# ,a-zA-Z]+[ ,]+[0-9\\\/#, a-zA-Z]{1,}
Look for the word "address" - skip this if not applicable
Look for first part numbers, letters, #, space - Unit Number / street number/suite number/door number
Separated by a space or comma
Look for one or more of rest of address numbers, letters, #, space
Tested against :
1 Sleepy Boulevard PO, Box 65745
Suite #100 /98,North St,Snoozepura
Ave., New Jersey,
Suite 420 1130 Connect Ave., NW,
Suite 420 19 / 21 Old Avenue,
Suite 12, Springfield, VIC 3001
Suite#100/98 North St Snoozepura
This worked for me when there were street addresses with unit/suite numbers, zip codes, only street. It also didn't match IP addresses or mac addresses. Worked with extra spaces.
This assumes users are normal people separate elements of a street address with a comma, hash sign, or space and not psychopaths who use characters like "|" or ":"!
For French address and some international address too, I use it. [\\D+ || \\d]+\\d+[ ||,||[A-Za-z0-9.-]]+(?:[Rue|Avenue|Lane|... etcd|Ln|St]+[ ]?)+(?:[A-Za-z0-9.-](.*)]?)
I was inspired from the responses given here and came with those 2 solutions support optional uppercase support french also regex structure numbers (required) letters, chars and spaces at least one common address keyword (required) as many chars you want before the line break definitions: accuracy capacity of detecting addresses and not something that looks like an address which is not. range capacity to detect uncommon addresses. Regex 1: high accuracy low range /[0-9]+[ |[a-zà-ú.,-]* ((highway)|(autoroute)|(north)|(nord)|(south)|(sud)|(east)|(est)|(west)|(ouest)|(avenue)|(lane)|(voie)|(ruelle)|(road)|(rue)|(route)|(drive)|(boulevard)|(circle)|(cercle)|(street)|(cer\.)|(cir\.)|(blvd\.)|(hway\.)|(st\.)|(aut\.)|(ave\.)|(ln\.)|(rd\.)|(hw\.)|(dr\.)|(a\.))([ .,-]*[a-zà-ú0-9]*)*/i regex 2: low accuracy high range /[0-9]*[ |[a-zà-ú.,-]* ((highway)|(autoroute)|(north)|(nord)|(south)|(sud)|(east)|(est)|(west)|(ouest)|(avenue)|(lane)|(voie)|(ruelle)|(road)|(rue)|(route)|(drive)|(boulevard)|(circle)|(cercle)|(street)|(cer\.?)|(cir\.?)|(blvd\.?)|(hway\.?)|(st\.?)|(aut\.?)|(ave\.?)|(ln\.?)|(rd\.?)|(hw\.?)|(dr\.?)|(a\.))([ .,-]*[a-zà-ú0-9]*)*/i
This one works well for me ^(\d+) ?([A-Za-z](?= ))? (.*?) ([^ ]+?) ?((?<= )APT)? ?((?<= )\d*)?$ Source : https://community.alteryx.com/t5/Alteryx-Designer-Discussions/RegEx-Addresses-different-formats-and-headaches/td-p/360147
Here is my RegEx for address, city & postal validation rules
validation rules:
address -
1 - 40 characters length.
Letters, numbers, space and . , : ' #
city -
1 - 19 characters length
Only Alpha characters are allowed
Spaces are allowed
postalCode -
The USA zip must meet the following criteria and is required:
Minimum of 5 digits (9 digits if zip + 4 is provided)
Numeric only
A Canadian postal code is a six-character string.
in the format A1A 1A1, where A is a letter and 1 is a digit.
a space separates the third and fourth characters.
do not include the letters D, F, I, O, Q or U.
the first position does not make use of the letters W or Z.
address: ^[a-zA-Z0-9 .,#;:'-]{1,40}$
city: ^[a-zA-Z ]{1,19}$
usaPostal: ^([0-9]{5})(?:[-]?([0-9]{4}))?$
canadaPostal : ^(?!.*[DFIOQU])[A-VXY][0-9][A-Z] ?[0-9][A-Z][0-9]$
\b(\d{1,8}[a-z]?[0-9\/#- ,a-zA-Z]+[ ,]+[.0-9\/#, a-zA-Z]{1,})\n
A more dynamic approach to #micah would be the following: (?'Address'(?'Street'[0-9][a-zA-Z\s]),?\s*(?'City'[A-Za-z\s]),?\s(?'Country'[A-Za-z])\s(?'Zipcode'[0-9]-?[0-9])) It won't care about individual lengths of segments of code. https://regex101.com/r/nuy7hB/1
RegEx: Uk Landlines, Mobile phone numbers
I've been struggling with finding a suitable solution :-
I need an regex expression that will match all UK phone numbers and mobile phones.
So far this one appears to cover most of the UK numbers:
^0\d{2,4}[ -]{1}[\d]{3}[\d -]{1}[\d -]{1}[\d]{1,4}$
However mobile numbers do not work with this regex expression or phone-numbers written in a single solid block such as 01234567890.
Could anyone help me create the required regex expression?
[\d -]{1}
is blatently incorrect: a digit OR a space OR a hyphen.
01000 123456
01000 is not a valid UK area code. 123456 is not a valid local number.
It is important that test data be real area codes and real number ranges.
^\s*(?(020[7,8]{1})?[ ]?[1-9]{1}[0-9{2}[ ]?[0-9]{4})|(0[1-8]{1}[0-9]{3})?[ ]?[1-9]{1}[0-9]{2}[ ]?[0-9]{3})\s*|[0-9]+[ ]?[0-9]+$
The above pattern is garbage for many different reasons.
[7,8] matches 7 or comma or 8. You don't need to match a comma.
London numbers also begin with 3 not just 7 or 8.
London 020 numbers aren't the only 2+8 format numbers; see also 023, 024, 028 and 029.
[1-9]{1} simplifies to [1-9]
[ ]? simplifies to \s?
Having found the intial 0 once, why keep searching for it again and again?
^(0....|0....|0....|0....)$ simplifies to ^0(....|....|....|....)$
Seriously. ([1]|[2]|[3]|[7]){1} simplifies to [1237] here.
UK phone numbers use a variety of formats: 2+8, 3+7, 3+6, 4+6, 4+5, 5+5, 5+4. Some users don't know which format goes with which number range and might use the wrong one on input. Let them do that; you're interested in the DIGITS.
Step 1: Check the input format looks valid
Make sure that the input looks like a UK phone number. Accept various dial prefixes, +44, 011 44, 00 44 with or without parentheses, hyphens or spaces; or national format with a leading 0. Let the user use any format they want for the remainder of the number: (020) 3555 7788 or 00 (44) 203 555 7788 or 02035-557-788 even if it is the wrong format for that particular number. Don't worry about unbalanced parentheses. The important part of the input is making sure it's the correct number of digits. Punctuation and spaces don't matter.
^\(?(?:(?:0(?:0|11)\)?[\s-]?\(?|\+)44\)?[\s-]?\(?(?:0\)?[\s-]?\(?)?|0)(?:\d{5}\)?[\s-]?\d{4,5}|\d{4}\)?[\s-]?(?:\d{5}|\d{3}[\s-]?\d{3})|\d{3}\)?[\s-]?\d{3}[\s-]?\d{3,4}|\d{2}\)?[\s-]?\d{4}[\s-]?\d{4}|8(?:00[\s-]?11[\s-]?11|45[\s-]?46[\s-]?4\d))(?:(?:[\s-]?(?:x|ext\.?\s?|\#)\d+)?)$
The above pattern matches optional opening parentheses, followed by 00 or 011 and optional closing parentheses, followed by an optional space or hyphen, followed by optional opening parentheses. Alternatively, the initial opening parentheses are followed by a literal + without a following space or hyphen. Any of the previous two options are then followed by 44 with optional closing parentheses, followed by optional space or hyphen, followed by optional 0 in optional parentheses, followed by optional space or hyphen, followed by optional opening parentheses (international format). Alternatively, the pattern matches optional initial opening parentheses followed by the 0 trunk code (national format).
The previous part is then followed by the NDC (area code) and the subscriber phone number in 2+8, 3+7, 3+6, 4+6, 4+5, 5+5 or 5+4 format with or without spaces and/or hyphens. This also includes provision for optional closing parentheses and/or optional space or hyphen after where the user thinks the area code ends and the local subscriber number begins. The pattern allows any format to be used with any GB number. The display format must be corrected by later logic if the wrong format for this number has been used by the user on input.
The pattern ends with an optional extension number arranged as an optional space or hyphen followed by x, ext and optional period, or #, followed by the extension number digits. The entire pattern does not bother to check for balanced parentheses as these will be removed from the number in the next step.
At this point you don't care whether the number begins 01 or 07 or something else. You don't care whether it's a valid area code. Later steps will deal with those issues.
Step 2: Extract the NSN so it can be checked in more detail for length and range
After checking the input looks like a GB telephone number using the pattern above, the next step is to extract the NSN part so that it can be checked in greater detail for validity and then formatted in the right way for the applicable number range.
^\(?(?:(?:0(?:0|11)\)?[\s-]?\(?|\+)(44)\)?[\s-]?\(?(?:0\)?[\s-]?\(?)?|0)([1-9]\d{1,4}\)?[\s\d-]+)(?:((?:x|ext\.?\s?|\#)\d+)?)$
Use the above pattern to extract the '44' from $1 to know that international format was used, otherwise assume national format if $1 is null.
Extract the optional extension number details from $3 and store them for later use.
Extract the NSN (including spaces, hyphens and parentheses) from $2.
Step 3: Validate the NSN
Remove the spaces, hyphens and parentheses from $2 and use further RegEx patterns to check the length and range and identify the number type.
These patterns will be much simpler, since they will not have to deal with various dial prefixes or country codes.
The pattern to match valid mobile numbers is therefore as simple as
^7([45789]\d{2}|624)\d{6}$
Premium rate is
^9[018]\d{8}$
There will be a number of other patterns for each number type: landlines, business rate, non-geographic, VoIP, etc.
By breaking the problem into several steps, a very wide range of input formats can be allowed, and the number range and length for the NSN checked in very great detail.
Step 4: Store the number
Once the NSN has been extracted and validated, store the number with country code and all the other digits with no spaces or punctuation, e.g. 442035557788.
Step 5: Format the number for display
Another set of simple rules can be used to format the number with the requisite +44 or 0 added at the beginning.
The rule for numbers beginning 03 is
^44(3\d{2})(\d{3])(\d{4})$
formatted as
0$1 $2 $3 or as +44 $1 $2 $3
and for numbers beginning 02 is
^44(2\d)(\d{4})(\d{4})$
formatted as
(0$1) $2 $3 or as +44 $1 $2 $3
The full list is quite long. I could copy and paste it all into this thread, but it would be hard to maintain that information in multiple places over time. For the present the complete list can be found at: http://aa-asterisk.org.uk/index.php/Regular_Expressions_for_Validating_and_Formatting_GB_Telephone_Numbers
Given that people sometimes write their numbers with spaces in random places, you might be better off ignoring the spaces all together - you could use a regex as simple as this then:
^0(\d ?){10}$
This matches:
01234567890
01234 234567
0121 3423 456
01213 423456
01000 123456
But it would also match:
01 2 3 4 5 6 7 8 9 0
So you may not like it, but it's certainly simpler.
Would this regex do?
// using System.Text.RegularExpressions;
/// <summary>
/// Regular expression built for C# on: Wed, Sep 8, 2010, 06:38:28
/// Using Expresso Version: 3.0.2766, http://www.ultrapico.com
///
/// A description of the regular expression:
///
/// [1]: A numbered capture group. [\+44], zero or one repetitions
/// \+44
/// Literal +
/// 44
/// [2]: A numbered capture group. [\s+], zero or one repetitions
/// Whitespace, one or more repetitions
/// [3]: A numbered capture group. [\(?]
/// Literal (, zero or one repetitions
/// [area_code]: A named capture group. [(\d{1,5}|\d{4}\s+?\d{1,2})]
/// [4]: A numbered capture group. [\d{1,5}|\d{4}\s+?\d{1,2}]
/// Select from 2 alternatives
/// Any digit, between 1 and 5 repetitions
/// \d{4}\s+?\d{1,2}
/// Any digit, exactly 4 repetitions
/// Whitespace, one or more repetitions, as few as possible
/// Any digit, between 1 and 2 repetitions
/// [5]: A numbered capture group. [\)?]
/// Literal ), zero or one repetitions
/// [6]: A numbered capture group. [\s+|-], zero or one repetitions
/// Select from 2 alternatives
/// Whitespace, one or more repetitions
/// -
/// [tel_no]: A named capture group. [(\d{1,4}(\s+|-)?\d{1,4}|(\d{6}))]
/// [7]: A numbered capture group. [\d{1,4}(\s+|-)?\d{1,4}|(\d{6})]
/// Select from 2 alternatives
/// \d{1,4}(\s+|-)?\d{1,4}
/// Any digit, between 1 and 4 repetitions
/// [8]: A numbered capture group. [\s+|-], zero or one repetitions
/// Select from 2 alternatives
/// Whitespace, one or more repetitions
/// -
/// Any digit, between 1 and 4 repetitions
/// [9]: A numbered capture group. [\d{6}]
/// Any digit, exactly 6 repetitions
///
///
/// </summary>
public Regex MyRegex = new Regex(
"(\\+44)?\r\n(\\s+)?\r\n(\\(?)\r\n(?<area_code>(\\d{1,5}|\\d{4}\\s+"+
"?\\d{1,2}))(\\)?)\r\n(\\s+|-)?\r\n(?<tel_no>\r\n(\\d{1,4}\r\n(\\s+|-"+
")?\\d{1,4}\r\n|(\\d{6})\r\n))",
RegexOptions.IgnoreCase
| RegexOptions.Singleline
| RegexOptions.ExplicitCapture
| RegexOptions.CultureInvariant
| RegexOptions.IgnorePatternWhitespace
| RegexOptions.Compiled
);
//// Replace the matched text in the InputText using the replacement pattern
// string result = MyRegex.Replace(InputText,MyRegexReplace);
//// Split the InputText wherever the regex matches
// string[] results = MyRegex.Split(InputText);
//// Capture the first Match, if any, in the InputText
// Match m = MyRegex.Match(InputText);
//// Capture all Matches in the InputText
// MatchCollection ms = MyRegex.Matches(InputText);
//// Test to see if there is a match in the InputText
// bool IsMatch = MyRegex.IsMatch(InputText);
//// Get the names of all the named and numbered capture groups
// string[] GroupNames = MyRegex.GetGroupNames();
//// Get the numbers of all the named and numbered capture groups
// int[] GroupNumbers = MyRegex.GetGroupNumbers();
Notice how the spaces and dashes are optional and can be part of it.. also it is now divided into two capture groups called area_code and tel_no to break it down and easier to extract.
Strip all whitespace and non-numeric characters and then do the test. It'll be musch , much easier than trying to account for all the possible options around brackets, spaces, etc.
Try the following:
#"^(([0]{1})|([\+][4]{2}))([1]|[2]|[3]|[7]){1}\d{8,9}$"
Starts with 0 or +44 (for international) - I;m sure you could add 0044 if you wanted.
It then has a 1, 2, 3 or 7.
It then has either 8 or 9 digits.
If you want to be even smarter, the following may be a useful reference: http://en.wikipedia.org/wiki/Telephone_numbers_in_the_United_Kingdom
It's not a single regex, but there's sample code from Braemoor Software that is simple to follow and fairly thorough. The JS version is probably easiest to read. It strips out spaces and hyphens (which I realise you said you can't do) then applies a number of positive and negative regexp checks.
Start by stripping the non-numerics, excepting a + as the first character.
(Javascript)
var tel=document.getElementById("tel").value;
tel.substr(0,1).replace(/[^+0-9]/g,'')+tel.substr(1).replace(/[^0-9]/g,'')
The regex below allows, after the international indicator +, any combination of between 7 and 15 digits (the ITU maximum) UNLESS the code is +44 (UK). Otherwise if the string either begins with +44, +440 or 0, it is followed by 2 or 7 and then by nine of any digit, or it is followed by 1, then any digit except 0, then either seven or eight of any digit. (So 0203 is valid, 0703 is valid but 0103 is not valid). There is currently no such code as 025 (or in London 0205), but those could one day be allocated.
/(^\+(?!44)[0-9]{7,15}$)|(^(\+440?|0)(([27][0-9]{9}$)|(1[1-9][0-9]{7,8}$)))/
Its primary purpose is to identify a correct starting digit for a non-corporate number, followed by the correct number of digits to follow. It doesn't deduce if the subscriber's local number is 5, 6, 7 or 8 digits. It does not enforce the prohibition on initial '1' or '0' in the subscriber number, about which I can't find any information as to whether those old rules are still enforced. UK phone rules are not enforced on properly formatted international phone numbers from outside the UK.
After a long search for valid regexen to cover UK cases, I found that the best way (if you're using client side javascript) to validate UK phone numbers is to use libphonenumber-js along with custom config to reduce bundle size:
If you're using NodeJS, generate UK metadata by running:
npx libphonenumber-metadata-generator metadata.custom.json --countries GB --extended
then import and use the metadata with libphonenumber-js/core:
import { isValidPhoneNumber } from "libphonenumber-js/core";
import data from "./metadata.custom.json";
isValidPhoneNumber("01234567890", "GB", data);
CodeSandbox Example
Custom RegEx expression for validating different possibilities of phone number entries?
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.
Validation for a 10 digit phone number
I'm looking for a simple regex that will validate a 10 digit phone number. I'd like to make sure that the number is exactly 10 digits, no letters, hyphens or parens and that the first two digits do not start with 0 or 1. Can someone help out?
/[2-9]{2}\d{8}/
^[2-9]{2}[0-9]{8}$
I consider [0-9] to be better to read than \d, especially considering the preceding [2-9]
The ^ and $ ensure that the input string consists ONLY of those 8 characters - otherwise it is not guaranteed that the input string is not larger - i.e. "12345678901" would match the regex w/o those two characters - although it is 11 chars and starts with a 1!
As Randal pointed out, this question is not consistent with the way phone numbers are formatted in North America (even though the OP stated 'first two digits do not start with 0 or 1'). A better regex for North American phone numbers would be:
^[2-9]{1}[0-9]{9}$
For example, Washington DC's area code is (202). NYC has area code (212). Northern New Jersey has (201).
But more accurately, the NANP has a lot of rules as it relates to what is allowed in area code and exchange (first six digits). This regex should still cover most cases. https://en.wikipedia.org/wiki/North_American_Numbering_Plan
This regex script might help out. I essentially strips any "punctuation" characters, including a leading 1-, then validates it is 10 digits.
The extra part you probably don't need is the formatting to 000-000-0000
formatPhone = function() {
var phone = this.value;
phone = phone.replace(/^1(|-|\(|\)|\.| )*|-|\(|\)|\.| /g, '');
if(phone.length === 10) {
this.value = phone.slice(0,3) + '-' + phone.slice(3,6) + '-' + phone.slice(6,10);
}
}
The Phone Numbers will be of 10 digits, and it will start from 7,8 and 9
[RegularExpression("^([07][0-9][0-9][0-9][0-9][0-9][0-9][0-9][0-9][0-9] | 8[0-9][0-9][0-9][0-9][0-9][0-9][0-9][0-9][0-9] | 9[0-9][0-9][0-9][0-9][0-9][0-9][0-9][0-9][0-9])$", ErrorMessage = "Enter Valid Mobile Number")]
reference : http://www.regular-expressions.info/numericranges.html