Get JavaScript to ignore everything that is not a letter

Question

How would I get JS to ignore everything that is not a letter (e.g. abc...) without using regex.

Examples
match("abcdef","@C2D!") returns true
match("abcdef","CAfe") returns true
match("abcdef","CG") returns false

This is what I've done so far...

function match(string, pattern) {
    string = string.toLowerCase();
    pattern = pattern.toLowerCase();
    for (var i = 0, l = string.length; i < l; ++i) {
        if(pattern.indexOf(string[i]) === -1) return false;
    }
    return true;
}
alert(match("abcdef", "@C2D!"));

Fiddle here: http://jsfiddle.net/5UCwW/

http://stackoverflow.com/q/3073176/1636522, in your case : `!/^[a-zA-Z]+$/.test('@C2D!')`. — , Nov 09 '13 at 17:50
I find it really hard to understand the requirement. The string value `abcdef` always consists of just letters and it seems to match none of the pattern you pass in. Can you elaborate on the requirement a bit? — Frank van Puffelen, Nov 09 '13 at 17:50
If you don't want to use a part of the language, at least try to explain why so. — raina77ow, Nov 09 '13 at 18:06

Hugo Tunius · Accepted Answer · 2013-11-09T18:32:12.753

1

Use a regular expression like the following [^A-Za-z].

var re = /[^A-Za-z]/g;

function match(string, pattern) {
    string = string.toLowerCase();
    pattern = pattern.toLowerCase().replace(re, "");
    for (var i = 0, l = string.length; i < l; ++i) {
        if(pattern.indexOf(string[i]) === -1) return false;
    }
    return true;
}

As a function without reg ex

var sanitize = function (str) { 
  var newStr = "", i, charCode;

  for (i = 0; i < str.length; ++i) {
    charCode = str.charCodeAt(i);  
    if (charCode >= 65 && charCode <= 90 ||
        charCode >= 97 && charCode <= 122)  {
      newStr += str[i];
    } 

  return newStr;
}

EDIT: A-z uses the range of the ascii values from capital a to small case z. Thus this includes ], \, [, ^, _ and `. I was not aware of this, [A-Za-z] is the correct pattern.

edited Nov 09 '13 at 18:32

answered Nov 09 '13 at 17:49

Hugo Tunius

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Can it be placed within a function? – theshizy Nov 09 '13 at 17:50
`\w` includes digits and underscore; nowhere in the question the OP mentions these shouldn't be ignored. Also, `[^\w]` is the same as just `\W`. – raina77ow Nov 09 '13 at 17:51
Yupp you guys are corret, fixed my answer to reflect it. – Hugo Tunius Nov 09 '13 at 17:55
`[A-z]` is not the same as `[A-Za-z]`. – raina77ow Nov 09 '13 at 18:05
@raina77ow I figured that out, thanks for telling me though. I didn't know that. I guess I am one of todays luck 10000 :) – Hugo Tunius Nov 09 '13 at 18:07
Of course it can, but why would you want to do that? The simple solution is to just loop over all characters in the string and check their ascii values. – Hugo Tunius Nov 09 '13 at 18:14
@HugoTunius sounds like what I need, how would you go about doing that? – theshizy Nov 09 '13 at 18:16
Loop through the string and check if the ascii value is in the range 65-90 or 97-122. TO get the ascii value look here http://stackoverflow.com/questions/10879244/how-to-get-the-ascii-value-in-javascript-for-the-characters – Hugo Tunius Nov 09 '13 at 18:17
Thanks - can the answer take the form `function match(string, pattern)` – theshizy Nov 09 '13 at 18:34
Just use the sanitize after you have made the pattern lowercase. – Hugo Tunius Nov 09 '13 at 18:36
let us [continue this discussion in chat](http://chat.stackoverflow.com/rooms/40903/discussion-between-hugo-tunius-and-antoine-laurent-lavoisier) – Hugo Tunius Nov 10 '13 at 14:09

score 1 · Answer 2 · answered Nov 09 '13 at 17:54

Use Javascript's built-in match method on strings:

 var mystring = '@C2D!';
 mystring.match(/[^a-zA-Z]/gi).length returns true

and

 var mystring2 = 'abcdef';
 mystring2.match(/[^a-zA-Z]/gi).length returns false

I'm assuming you're looking for lower and upper-case letters. If just lower, use [^a-z].

For more info on the match method, go here: http://www.w3schools.com/jsref/jsref_match.asp

Get JavaScript to ignore everything that is not a letter

2 Answers2