CODEWITHSUNDEEP
javascriptregex
dvlden
dvlden

Reputation: 2462

Require both letters and numbers - regExp

I am trying to figure how to require both letters and numbers only without any other characters. So literally [a-z] and ( \d or [0-9] ) depending what is better way of doing it for numbers.

So if I had a string that requires validation:

$toValidate = 'Q23AS9D0APQQ2'; // It may start with letter or number, both cases possible.

And then if I had validation for it:

return /([a-z].*[0-9])|([0-9].*[a-z])/i.test($toValidate);

I used an i flag here because it could be that user enters it lowercase or uppercase, it's user preference... So that regex fails... It accepts special characters also, so that is not desired effect.

With the validation above, this passes as well:

$toValidate = 'asdas12312...1231@asda___213-1';

Then I tried something crazy and I don't even know what I have done, so if anyone could tell me beside the correct answer, I'll truly appreciate.

return /([a-z].*\d)+$|(\d.*[a-z])+$/i.test($toValidate);

This seemed to work great. But then when I tried to continue typing letters or numbers after an special character it still validates as true.

Example:

$toValidate = 'q2IasK231@!@!#_+123';

So please help me understand regularExpressions better and tell me what is the way to validate the string at the beginning of my question. Letters and numbers expected in the string.

Upvotes: 5

Views: 18075

Answers (3)

anubhava
anubhava

Reputation: 785146

To allow only letters and digits with at least one letter and at least one digit use:

/^(?=.*?\d)(?=.*?[a-zA-Z])[a-zA-Z\d]+$/

Regex breakdown:

^               # start of input
(?=.*?\d)       # lookahead to make sure at least one digit is there
(?=.*?[a-zA-Z]) # lookahead to make sure at least one letter is there
[a-zA-Z\d]+     # regex to match 1 or more of digit or letters
$               # end of input

RegEx Demo

You should not use .* in your regex otherwise it will allow any character in the input.

Upvotes: 14

Poyke
Poyke

Reputation: 655

what you are looking for in pseudo is: START-ANCOR [a-zA-Z0-9] 0->Inf times , ([a-zA-Z][0-9] OR [0-9][a-zA-Z]), [a-zA-Z0-9] 0->Inf times END-ANCOR

in words, start with anything from your lang, end with anything from your lang and contain a seam between letters and digits or the other way around

Should be like this:

/^([a-z0-9]* (([a-z][0-9]) | ([0-9][a-z])) [a-z0-9]*)$/i.

Upvotes: 1

BurningLights
BurningLights

Reputation: 2397

You can combine more than one type of character within the brackets. So, the following regex should work:

/^([a-z0-9]+)$/i.

The ^ and $ match the start and end of the string, so the whole string will be tested for the conditions. The [a-z0-9] makes it match only letters and numbers. The + makes it match one or more character. And the "i" at the end, as you know, makes it case insensitive.

Upvotes: 0

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