Regex to accept one or two words

Question

I need a RegEx to match the following condition:

1 or 2 words
Total minimum length is 3
Total maximum length is 50
Accept accented characters
No numbers is a plus

Examples of accepted strings:

Karmen Garcia
João Pedro
Vitor
maxiliamum LóPezz
Nadál
loriPsium PsisVõèrbizbbbbbbbbbb

Examples of non accepted string:

Jony Cam Vanis
9lucas P.us-kas1 Li1

What I tried:

^(?!.*?\s{2})[A-Za-z ]{3,50}$
^([a-zA-Z][a-zA-Z]){3,50}$

you can look JavaScript RegExp Reference – Sercan Elvanağacı Jan 13 '17 at 14:41 — Sercan Elvanağacı, Jan 13 '17 at 14:41
Does the minimum and maximum values include any space? – SamWhan Jan 13 '17 at 15:00 — SamWhan, Jan 13 '17 at 15:00
@ClasG Yes, includes all – Fábio Silva Jan 13 '17 at 15:10 — Fábio Silva, Jan 13 '17 at 15:10

score 1 · Answer 1 · edited May 23 '17 at 12:33

1

Here you go:

(?=^.{3,50}$)(\s*\p{L}+\s*){1,2}

Or if you don't want any leading whitespace, trailing whitespace, or more than one space between the words:

(?=^.{3,50}$)\p{L}+(\s\p{L}+)?

EDIT:

As this other SO post shows, Javascript has a problem with Unicode character classes. So, \p{L} won't work in Javascript.

What does that mean for you? Well, that other post shows three different solutions. Which solution is right for you depends on whether or not you know in advance exactly which accented characters or non-word (e.g. punctuation) characters might entered.

One possible approach is to list out the valid accented characters then concatenate it in to the regex:

var validWordCharacters = "[" + "a-z" + "A-Z" + "àèìòùÀÈÌÒÙáéíóúýÁÉÍÓÚÝâêîôûÂÊÎÔÛãñõÃÑÕäëïöüÿÄËÏÖÜŸçÇßØøÅåÆæœ" + "]";

var regex = "(?=^.{3,50}$)" + validWordCharacters + "+(\s" + validWordCharacters + "+)?";

regexCompiled = new RegExp(regex);

Another possible (and more concise) solution is to use a range of code points:

var validWordCharacters = "[" + "a-z" + "A-Z" + "\u00C0-\u024F" + "]";

var regex = "(?=^.{3,50}$)" + validWordCharacters + "+(\s" + validWordCharacters + "+)?";

regexCompiled = new RegExp(regex);

edited May 23 '17 at 12:33

Community

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answered Jan 13 '17 at 14:56

Travis

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assuming there is no space starting or ending i guess (like sample show so seems OK) – NeronLeVelu Jan 13 '17 at 14:58
something wrong. It accepts 'first ' but wont accept 'first second' – Fábio Silva Jan 13 '17 at 14:59
@FábioSilva Try it now. I corrected the multiplicity on the whitespace character class. – Travis Jan 13 '17 at 15:01
@Travis Well now it accepts 'first second' but is struggling with accented characters – Fábio Silva Jan 13 '17 at 15:05
@FábioSilva I changed `\w` to `\p{L}`, which should now accept accented letters. Sorry about that... I was previously unaware that `\w` did not match accented "word" characters. – Travis Jan 13 '17 at 15:09
@FábioSilva See my edit about a workaround for the unicode letter class in Javascript. – Travis Jan 13 '17 at 15:28
@Travis thank you, ^(?=.{3,50}$)[^\s\d]*(\s[^\s\d]*)?$ is doing ok for what I need. – Fábio Silva Jan 13 '17 at 15:43

SamWhan · Accepted Answer · 2017-01-13T15:28:03.527

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You failed to tag what regex flavor/programming language/tool. But here's one way of doing it:

^(?=.{3,50}$)\p{L}*(\s\p{L}*)?$

It uses a positive look-ahead to make sure it's between 3 and 50 characters. Follwing that it simple checks for unicode letter class characters, optionally followed by a space and more letters, i.e. one or two words.

See it here at regex101.

Edit

OK - so for javascript you could try

^(?=.{3,50}$)[^\s\d]*(\s[^\s\d]*)?$

It's basically the same, only instead of matching the unicode class, it matches non space and digits. That's not perfect, but a simple solution that'll work in most (latin) cases.

See this here at regex101.

edited Jan 13 '17 at 15:28

answered Jan 13 '17 at 15:04

SamWhan

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I added the flavor, javascript. It seems not to work – Fábio Silva Jan 13 '17 at 15:12
For the Javascript solution, what about punctuation? `[^\s\d]` would still match any of "!@#$" for example. – Travis Jan 13 '17 at 15:20
@Travis that seems to be very close to what I asked for, but I think it isn't validating for maximum length of 50, do you understand why? – Fábio Silva Jan 13 '17 at 15:26
@Travis "That's not perfect, but a simple solution that'll work in most (latin) cases.", Just add unwanted characters as needed ;) – SamWhan Jan 13 '17 at 15:26
That's my bad - fixed in answer and changed second example. – SamWhan Jan 13 '17 at 15:29
^(?=.{3,50}$)[^\s\d]*(\s[^\s\d]*)?$ Seems to be working like a charm, thank you. – Fábio Silva Jan 13 '17 at 15:35

score 0 · Answer 3 · answered Jan 13 '17 at 14:50

0

I think you may use next. It check minimum length but not check maximum. You may ensure it in different check

^([A-Za-z]{3,50}|([A-Za-z]+\s[A-Za-z]+))$

answered Jan 13 '17 at 14:50

Michael Piankov

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Well thats fine. Do you know how to add accented characters? 'Joao Pedro' is OK but 'João Pedro' is not passing. Thank you – Fábio Silva Jan 13 '17 at 14:55
does not meet the request, you could have a word of 1 char with a second of 2 that is good and dont pass or second word of 150 char withfirst of 10 that is bad but pass – NeronLeVelu Jan 13 '17 at 14:56
@FábioSilva well in truth it you may add the first and the last character as A-Z. I'm sorry | i haven't your alphabet and I just copy some letters from post (ã-õ). – Michael Piankov Jan 13 '17 at 15:00
@NeronLeVelu, thank you for your comment. But it not clear. In my two words first with 1 char and next with 2 (`'a bc'`) will pass with right part of regex `([A-Za-z]+\s[A-Za-z]+)`. And two words for 161 chars are passed to and I write that I dont check the whole length in that case. Am I ask for your question? – Michael Piankov Jan 13 '17 at 15:06

Regex to accept one or two words

3 Answers3