Regular Expression Pattern matching issue

Question

Not a lot of experience in RegEx stuff.

I have the following in java script which works perfectly fine.

The following pattern is used allow only alpha numeric

var valid = /^[A-Za-z0-9]+$/.test("a"); // returns true
var valid = /^[A-Za-z0-9]+$/.test("@"); // returns false

I am using the pattern part "^[A-Za-z0-9]" in some other places of the code and was asked to use the part "^[A-Za-z0-9]" in a variable and use it so that it is not repetitive. The following is a modification to the above:

var regExPart= "^[A-Za-z0-9]";
var regExString = ("/" + regExPart+ "+$/".replace(/\"/g, "")); // replacing the quotes
var regExp = new RegExp(regExString); // results in /^[A-Za-z0-9]+$/
var valid = regExp.test(charValue); // charValue could be any keyvalue "a" or "@"

//the above returns false for "a" 
//the above returns false for "@"

I am writing this in a keypress event to allow only alpha numeric

keypressValidation: function (e) {
var charCode = (e.which) ? e.which: event.keyCode;
var charValue = String.fromCharCode(charCode);
var valid = return /^[A-Za-z0-9]+$/.test(charValue);
if (!valid) 
{
    //prevent default (don't allow/ enter the value)
}

Not sure why. What am I missing in this. Need to return true for "a" and false for "@" for both the approaches. Any help/ suggestion would be of great help. Thank in advance.

I usually like DRY-ness, but IMO it's not something to try for across multiple separate regular expressions unless the repeated pattern is long. (also note that your character set simplifies to [a-z\d] with the i flag) — CertainPerformance
– CertainPerformance, Commented Oct 25, 2018 at 2:42
Also, do you only need to check if your character is a ? Why use a regex in that case ? — Nicolas
– Nicolas, Commented Oct 25, 2018 at 2:44
@Nicolas I am writing this as a part of keypress validation in java script to now allow user to enter only alpha numeric values. Updated my post. — Kishore
– Kishore, Commented Oct 25, 2018 at 2:49

Pablo · Accepted Answer · 2018-10-25 02:56:25Z

1

For the RegExp class constructor, you do not need to specify forward slashes /.

var regExPart= "^[A-Za-z0-9]";
var regExp = new RegExp(regExPart + "+$"); // results in /^[A-Za-z0-9]+$/

console.log('a', regExp.test('a'))
console.log('@', regExp.test('@'))

answered Oct 25, 2018 at 2:56

Pablo

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suertang · Accepted Answer · 2018-10-25 02:57:09Z

1

It is not a must to contain '/'s in regexp

new RegExp("^[0-9a-zA-Z]$").test('a')

return true

new RegExp("^[0-9a-zA-Z]$").test('@')

return false

So just do

var rex="^[0-9a-zA-Z]$"

And you can use it anywhere. Tested in Chrome console.

answered Oct 25, 2018 at 2:57

suertang

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Nicolas · Accepted Answer · 2018-10-25 03:02:56Z

0

I've made an example using your regex of what it should do, i think the way you were building your regex was not helping. You don't need to create a string and then create a new regex object , you can use /regex part/. Anyways here is a working example.

function keypress(e) {
  // Get the current typed key
  var keynum = e.key;
  // this regex only allow character between a and z and 0 and 9
  var regex = /^[a-zA-Z0-9]+$/;
  // we check if the current key matches our regex
  if(!keynum.match(regex) ) {
    // it doesn't ? well we stop the event from happening
    e.preventDefault();
  }
}

<input type="text" onkeypress="keypress(event)">

answered Oct 25, 2018 at 3:02

Nicolas

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