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RegEx for Javascript to allow only alphanumeric

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轻奢々
轻奢々 2020-11-22 15:13

I need to find a reg ex that only allows alphanumeric. So far, everyone I try only works if the string is alphanumeric, meaning contains both a letter and a number. I just w

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  • 长发绾君心
    2020-11-22 15:38

    Use the word character class. The following is equivalent to a ^[a-zA-Z0-9_]+$:

    ^\w+$

    Explanation:

    • ^ start of string
    • \w any word character (A-Z, a-z, 0-9, _).
    • $ end of string

    Use /[^\w]|_/g if you don't want to match the underscore.

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  • 北荒
    2020-11-22 15:39

    Instead of checking for a valid alphanumeric string, you can achieve this indirectly by checking the string for any invalid characters. Do so by checking for anything that matches the complement of the valid alphanumeric string.

    /[^a-z\d]/i

    Here is an example:

    var alphanumeric = "someStringHere";
var myRegEx  = /[^a-z\d]/i;
var isValid = !(myRegEx.test(alphanumeric));

    Notice the logical not operator at isValid, since I'm testing whether the string is false, not whether it's valid.

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  • 伪装坚强ぢ
    2020-11-22 15:42

    Try this... Replace you field ID with #name... a-z(a to z), A-Z(A to Z), 0-9(0 to 9)

    jQuery(document).ready(function($){
    $('#name').keypress(function (e) {
        var regex = new RegExp("^[a-zA-Z0-9\s]+$");
        var str = String.fromCharCode(!e.charCode ? e.which : e.charCode);
        if (regex.test(str)) {
            return true;
        }
        e.preventDefault();
        return false;
    });
});
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  • 囚心锁ツ
    2020-11-22 15:43 
    ^\s*([0-9a-zA-Z]*)\s*$

    or, if you want a minimum of one character:

    ^\s*([0-9a-zA-Z]+)\s*$

    Square brackets indicate a set of characters. ^ is start of input. $ is end of input (or newline, depending on your options). \s is whitespace.

    The whitespace before and after is optional.

    The parentheses are the grouping operator to allow you to extract the information you want.

    EDIT: removed my erroneous use of the \w character set.

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  • 醉酒成梦
    2020-11-22 15:45

    JAVASCRIPT to accept only NUMBERS, ALPHABETS and SPECIAL CHARECTERS

    document.getElementById("onlynumbers").onkeypress = function (e) {
	onlyNumbers(e.key, e)
};

document.getElementById("onlyalpha").onkeypress = function (e) {
	onlyAlpha(e.key, e)
};

document.getElementById("speclchar").onkeypress = function (e) {
	speclChar(e.key, e)
};

function onlyNumbers(key, e) {
	var letters = /^[0-9]/g; //g means global
	if (!(key).match(letters)) e.preventDefault();
}

function onlyAlpha(key, e) {
	var letters = /^[a-z]/gi; //i means ignorecase
	if (!(key).match(letters)) e.preventDefault();
}

function speclChar(key, e) {
	var letters = /^[0-9a-z]/gi;
	if ((key).match(letters)) e.preventDefault();
}
    <html>
   <head></head>
   <body>
      Enter Only Numbers: 
      <input id="onlynumbers" type="text">
      <br><br>
      Enter Only Alphabets: 
      <input id="onlyalpha" type="text" >
      <br><br>
      Enter other than Alphabets and numbers like special characters: 
      <input id="speclchar" type="text" >
   </body>
</html>

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  • 死守一世寂寞
    2020-11-22 15:47

    If you wanted to return a replaced result, then this would work:

    var a = 'Test123*** TEST';
var b = a.replace(/[^a-z0-9]/gi,'');
console.log(b);

    This would return:

    Test123TEST

    Note that the gi is necessary because it means global (not just on the first match), and case-insensitive, which is why I have a-z instead of a-zA-Z. And the ^ inside the brackets means "anything not in these brackets".

    WARNING: Alphanumeric is great if that's exactly what you want. But if you're using this in an international market on like a person's name or geographical area, then you need to account for unicode characters, which this won't do. For instance, if you have a name like "Âlvarö", it would make it "lvar".

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