Regular Expression for alphanumeric and underscore

问题:

I would like to have a regular expression that checks if a string contains only upper and lowercase letters, numbers, and underscores.

回答1:

To match a string that contains only those characters (or an empty string), try

\"^[a-zA-Z0-9_]*$\"

This works for .NET regular expressions, and probably a lot of other languages as well.

Breaking it down:

^ : start of string
[ : beginning of character group
a-z : any lowercase letter
A-Z : any uppercase letter
0-9 : any digit
_ : underscore
] : end of character group
* : zero or more of the given characters
$ : end of string

If you don\'t want to allow empty strings, use + instead of *.

EDIT As others have pointed out, some regex languages have a shorthand form for [a-zA-Z0-9_]. In the .NET regex language, you can turn on ECMAScript behavior and use \\w as a shorthand (yielding ^\\w*$ or ^\\w+$). Note that in other languages, and by default in .NET, \\w is somewhat broader, and will match other sorts of unicode characters as well (thanks to Jan for pointing this out). So if you\'re really intending to match only those characters, using the explicit (longer) form is probably best.

回答2:

There\'s a lot of verbosity in here, and I\'m deeply against it, so, my conclusive answer would be:

/^\\w+$/

\\w is equivalent to [A-Za-z0-9_], which is pretty much what you want. (unless we introduce unicode to the mix)

Using the + quantifier you\'ll match one or more characters. If you want to accept an empty string too, use * instead.

回答3:

You want to check that each character matches your requirements, which is why we use:

[A-Za-z0-9_]

And you can even use the shorthand version:

\\w

Which is equivalent (in some regex flavors, so make sure you check before you use it). Then to indicate that the entire string must match, you use:

^

To indicate the string must start with that character, then use

$

To indicate the string must end with that character. Then use

\\w+ or \\w*

To indicate \"1 or more\", or \"0 or more\". Putting it all together, we have:

^\\w*$

回答4:

Although it\'s more verbose than \\w, I personally appreciate the readability of the full POSIX character class names ( http://www.zytrax.com/tech/web/regex.htm#special ), so I\'d say:

^[[:alnum:]_]+$

However, while the documentation at the above links states that \\w will \"Match any character in the range 0 - 9, A - Z and a - z (equivalent of POSIX [:alnum:])\", I have not found this to be true. Not with grep -P anyway. You need to explicitly include the underscore if you use [:alnum:] but not if you use \\w. You can\'t beat the following for short and sweet:

^\\w+$

Along with readability, using the POSIX character classes (http://www.regular-expressions.info/posixbrackets.html) means that your regex can work on non ASCII strings, which the range based regexes won\'t do since they rely on the underlying ordering of the ASCII characters which may be different from other character sets and will therefore exclude some non-ASCII characters (letters such as œ) which you might want to capture.

回答5:

Um...question: Does it need to have at least one character or no? Can it be an empty string?

^[A-Za-z0-9_]+$

Will do at least one upper or lower case alphanumeric or underscore. If it can be zero length, then just substitute the + for *

^[A-Za-z0-9_]*$

Edit:

If diacritics need to be included (such as cedilla - ç) then you would need to use the word character which does the same as the above, but includes the diacritic characters:

^\\w+$

Or

^\\w*$

回答6:

In Computer Science, an Alphanumeric value often means the first character is not a number but is an alphabet or underscore. Thereafter the character can be 0-9, A-Z, a-z, or underscore (_).

Here is how you would do that:

Tested under php:

$regex = \'/^[A-Za-z_][A-Za-z\\d_]*$/\'

or take this

^[A-Za-z_][A-Za-z\\d_]*$

and place it in your development language.

回答7:

How about:

^([A-Za-z]|[0-9]|_)+$

...if you want to be explicit, or:

^\\w+$

...if you prefer concise (Perl syntax).

回答8:

use lookaheads to do the \"at least one\" stuff. Trust me it\'s much easier.

Here\'s an example that would require 1-10 characters, containing at least one digit and one letter:

^(?=.*\\d)(?=.*[A-Za-z])[A-Za-z0-9]{1,10}$

NOTE: could have used \\w but then ECMA/Unicode considerations come into play increasing the character coverage of the \\w \"word character\".

回答9:

Try these multi-lingual extensions I have made for string.

IsAlphaNumeric - String must contain atleast 1 alpha (letter in Unicode range, specified in charSet) and atleast 1 number (specified in numSet). Also, the string should comprise only of alpha and numbers.

IsAlpha - String should contain atleast 1 alpha (in the language charSet specified) and comprise only of alpha.

IsNumeric - String should contain atleast 1 number (in the language numSet specified) and comprise only of numbers.

The charSet/numSet range for the desired language can be specified. The Unicode ranges are available on below link:

http://www.ssec.wisc.edu/~tomw/java/unicode.html

API :

    public static bool IsAlphaNumeric(this string stringToTest)
    {
        //English
        const string charSet = \"a-zA-Z\";
        const string numSet = @\"0-9\";

        //Greek
        //const string charSet = @\"\\u0388-\\u03EF\";            
        //const string numSet = @\"0-9\";

        //Bengali
        //const string charSet = @\"\\u0985-\\u09E3\";
        //const string numSet = @\"\\u09E6-\\u09EF\";

        //Hindi
        //const string charSet = @\"\\u0905-\\u0963\";
        //const string numSet = @\"\\u0966-\\u096F\";

        return Regex.Match(stringToTest, @\"^(?=[\" + numSet + @\"]*?[\" + charSet + @\"]+)(?=[\" + charSet + @\"]*?[\" + numSet + @\"]+)[\" + charSet + numSet +@\"]+$\").Success;
    }

    public static bool IsNumeric(this string stringToTest)
    {
        //English
        const string numSet = @\"0-9\";

        //Hindi
        //const string numSet = @\"\\u0966-\\u096F\";

        return Regex.Match(stringToTest, @\"^[\" + numSet + @\"]+$\").Success;
    }

    public static bool IsAlpha(this string stringToTest)
    {
        //English
        const string charSet = \"a-zA-Z\";

        return Regex.Match(stringToTest, @\"^[\" + charSet + @\"]+$\").Success;
    }

Usage :

        //English
        string test = \"AASD121asf\";

        //Greek
        //string test = \"Ϡϛβ123\";

        //Bengali
        //string test = \"শর৩৮\";

        //Hindi
        //string test = @\"क़लम३७ख़\";

        bool isAlphaNum = test.IsAlphaNumeric();

回答10:

The following regex matches alphanumeric characters and underscore:

^[a-zA-Z0-9_]+$

For example, in Perl:

#!/usr/bin/perl -w

my $arg1 = $ARGV[0];

# check that the string contains *only* one or more alphanumeric chars or underscores
if ($arg1 !~ /^[a-zA-Z0-9_]+$/) {
  print \"Failed.\\n\";
} else {
    print \"Success.\\n\";
}

回答11:

For me there was an issue in that I want to distinguish between alpha, numeric and alpha numeric, so to ensure an alphanumeric string contains at least one alpha and at least one numeric, I used :

^([a-zA-Z_]{1,}\\d{1,})+|(\\d{1,}[a-zA-Z_]{1,})+$

回答12:

For those of you looking for unicode alphanumeric matching, you might want to do something like:

^[\\p{L} \\p{Nd}_]+$

Further reading at http://unicode.org/reports/tr18/ and at http://www.regular-expressions.info/unicode.html

回答13:

Here is the regex for what you want with a quantifier to specify at least 1 character and no more than 255 characters

[^a-zA-Z0-9 _]{1,255}

回答14:

matching diacritics in a regexp opens a whole can of worms, especially when taking Unicode into consideration. You might want to read about Posix locales in particular.

回答15:

To check the entire string and not allow empty strings, try

^[A-Za-z0-9_]+$

回答16:

this works for me you can try [\\p{Alnum}_]

回答17:

^\\w*$ will work for below combination 1 123 1av pRo av1