Javascript: test single character is found in regex string

Question

I confess at the outset that I dislike and don't really understand regex properly. I want to check that a single character ch is one of a set of acceptable characters. I thought this should work, but it doesn't:

if (/aCcehIikmNnOoprSstxYy/.test(ch)) {

What am I doing wrong?

Thanks.

Your regex means you are expecting exactly the string aCcehIikmNnOoprSstxYy. — helpermethod
– helpermethod, Commented Jun 13, 2012 at 8:14

Petar Ivanov · Accepted Answer · 2012-06-13 08:13:31Z

3

you need to enclose the set of characters in [ ]:

if (/[aCcehIikmNnOoprSstxYy]/.test(ch)) {

Without that you are trying to match the whole string 'aCcehIikmNnOoprSstxYy'.

answered Jun 13, 2012 at 8:13

Petar Ivanov

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Comments

ioseb · Accepted Answer · 2012-06-13 08:21:42Z

2

I think this can be solved without regex:

var characters = "aCcehIikmNnOoprSstxYy";
var allowed    = characters.indexOf("C") != -1;

if (allowed) {
  // do something here
}

String.indexOf() returns -1 if character is not in string, otherwise positive number.

answered Jun 13, 2012 at 8:21

ioseb

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4 Comments

h2ooooooo Over a year ago

A much better solution. Indexof in this case is also much faster than regex. See a speed comparison here: jsperf.com/regexp-vs-indexof For example my Chrome 19's indexOf is about twice as fast as regex is.

h2ooooooo Over a year ago

I should add that if you prefer, this could also be made a one liner very easily. if (String("aCcehIikmNnOoprSstxYy").indexOf("C") != -1) {

ioseb Over a year ago

String(...) is not necessary, it's legitimate to use directly: "aCcehIikmNnOoprSstxYy".indexOf("C")

Nick Over a year ago

Thanks, everyone. I used if ("aCcehIikmNnOoprSstxYy".indexOf(ch) !== -1) {

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