Validate Indian Phone Number With Optional +91 or 0 Preceeding 10 Digits

Validate Indian Phone number with optional +91 or 0 preceeding 10 digits

You can use ^([0]|\+91)?\d{10} as starts with 0 or +91 and 10 digits after that.

..but as you must have seen number starts with 7,8 or 9

Then, you should use ^([0]|\+91)?[789]\d{9}$ which means starts with 7, 8 or 9 and follows 9 digits after that.

Some random matches;

+919802422462
08150166859

Regular Expression Validation For Indian Phone Number and Mobile number

For land Line Number

03595-259506
03592 245902
03598245785

you can use this

\d{5}([- ]*)\d{6}

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NEW for all ;)

OLD: ((\+*)(0*|(0 )*|(0-)*|(91 )*)(\d{12}+|\d{10}+))|\d{5}([- ]*)\d{6}
NEW: ((\+*)((0[ -]*)*|((91 )*))((\d{12})+|(\d{10})+))|\d{5}([- ]*)\d{6}

9775876662
0 9754845789
0-9778545896
+91 9456211568
91 9857842356
919578965389

03595-259506
03592 245902
03598245785

this site is useful for me, and maby for you .;)http://gskinner.com/RegExr/

regular expression for Indian mobile numbers

^[789]\d{9}$ should do the trick.

^     #Match the beginning of the string
[789] #Match a 7, 8 or 9
\d    #Match a digit (0-9 and anything else that is a "digit" in the regex engine)
{9}   #Repeat the previous "\d" 9 times (9 digits)
$     #Match the end of the string

UPDATE

Some Telecom operators in india introduced new mobile number series which starts with digit 6.

for that use:

^[6-9]\d{9}$

Regular expression for Indian mobile phone numbers?

Try something like this:

^9\d{9}$

regular expression for indian phone number

You can try

^\+?[0-9-]+$

See it here on Regexr

The important parts are the anchors ^ and $ for the start and the end of the string. I added also \+? at the start, to match an optional +. The + needs to be escaped since it is a special character in regex, the ? after it makes it optional.

Of course this is a very simple pattern, be aware that e.g. "-----" would also be valid.

HTML5 phone number validation with pattern

How about this? /(7|8|9)\d{9}/

It starts by either looking for 7 or 8 or 9, and then followed by 9 digits.

Validate phone number with JavaScript

First off, your format validator is obviously only appropriate for NANP (country code +1) numbers. Will your application be used by someone with a phone number from outside North America? If so, you don't want to prevent those people from entering a perfectly valid [international] number.

Secondly, your validation is incorrect. NANP numbers take the form NXX NXX XXXX where N is a digit 2-9 and X is a digit 0-9. Additionally, area codes and exchanges may not take the form N11 (end with two ones) to avoid confusion with special services except numbers in a non-geographic area code (800, 888, 877, 866, 855, 900) may have a N11 exchange.

So, your regex will pass the number (123) 123 4566 even though that is not a valid phone number. You can fix that by replacing \d{3} with [2-9]{1}\d{2}.

Finally, I get the feeling you're validating user input in a web browser. Remember that client-side validation is only a convenience you provide to the user; you still need to validate all input (again) on the server.

TL;DR don't use a regular expression to validate complex real-world data like phone numbers or URLs. Use a specialized library.

Regex for Mobile Number Validation

Satisfies all your requirements if you use the trick told below

Regex: /^(\+\d{1,3}[- ]?)?\d{10}$/

1. ^ start of line
2. A + followed by \d+ followed by a or - which are optional.
3. Whole point two is optional.
4. Negative lookahead to make sure 0s do not follow.
5. Match \d+ 10 times.
6. Line end.

DEMO ^{Added multiline flag in demo to check for all cases}

P.S. You really need to specify which language you use so as to use an if condition something like below:

// true if above regex is satisfied and (&&) it does not (`!`) match `0`s `5` or more times

if(number.match(/^(\+\d{1,3}[- ]?)?\d{10}$/) && ! (number.match(/0{5,}/)) )

Validate Bangladeshi phone number with optional +88 or 01 preceeding 11 digits

I see you have tried but your regex is not accurate.

1. You do not use an end of string $ anchor (thus, even "abc" at the end will not prevent the IsMatch from returning true)
2. You are using 01 inside square brackets thus creating a character class, meaning either 0 or 1.
3. No need in a capturing group here, a non-capturing is best for optional subpatterns.
4. As has been pointed out in the follow up answers, the regex you are using is not actually meeting all the requirements for the Bangladeshi phone number, see Kobi's ^(?:\+?88|0088)?01[15-9]\d{8}$ answer.

In order to create a regex that will validate a string that has "optional +88, 0088 or 01 preceeding 11 digits", you need something like:

@"^(?:(?:\+|00)88|01)?\d{11}$"

See RegexStorm demo

UPDATE

If you want to validate Bangladeshi phone numbers with this regex, nothing changes in the pattern (only \r? is totally redundant), but if you plan to allow 13 or 11 digits after +88 or 01, you need to use an alternation:

ng-pattern="/^(?:(?:\+|00)88|01)?\d{11}$/"

See demo

In Angular:

Validators.pattern('(?:(?:\\+|00)88|01)?\\d{11}')
// or
Validators.pattern(/^(?:(?:\+|00)88|01)?\d{11}$/)

Regex for Mobile Number with or without Country Code

From what I can see, this should work. The prefix is optional and is stored into the first match group, with the main number going into the second group.

^([0|\+[0-9]{1,5})?([7-9][0-9]{9})$

But if you can give us some test cases for each, it'd help us in giving you a working regex for what you want.

Props to SchlaWiener in the comments for the correct limit on the country code length.

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