JavaScript regular expression returning false?

Question

I'm learning javascript and this just doesn't make sense to me. test() returns true or false if a character is a match. So can someone please explain why my code below is not working?

var str = "hello*3";
var arr = str.split('');

for(var i in arr){
    re = new RegExp("/[a-z]/i");
    if(re.test(arr[i])) {
        console.log(String.fromCharCode(arr[i].charCodeAt() + 1));   
    }
}

re.test(arr[0]); //arr[0] == 'h'
// false

Updated example:

var re = new RegExp("/[a-z]/i");
re.test('h');
// returns 'false' (I'm expecting this to be true)

re = new RegExp("/[a-z]/i"); should be outside your for loop btw — Dalorzo
– Dalorzo, Commented May 30, 2014 at 17:35
oh no that was not the solution just an advice. @subhaze has your answer — Dalorzo
– Dalorzo, Commented May 30, 2014 at 17:37

subhaze · Accepted Answer · 2014-05-30 17:43:39Z

5

You don't need the / when using the RegExp and you pass flags as the second param

This should work re = new RegExp("[a-z]", "i");

or the literal form

re = /[a-z]/i;

More info can be found on MDN here https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/RegExp

Some info from the link above

Constructor Literal and constructor notations are possible:

/pattern/flags; 
new RegExp(pattern [, flags]);

pattern

The text of the regular expression.

flags

If specified, flags can have any combination of the following values:

g

global match

i

ignore case

m

multiline; treat beginning and end characters (^ and $) as working over multiple lines (i.e., match the beginning or end of each line (delimited by \n or \r), not only the very beginning or end of the whole input string)

y

sticky; matches only from the index indicated by the lastIndex property of this regular expression in the target string (and does not attempt to match from any later indexes). This allows the match-only-at-start capabilities of the character "^" to effectively be used at any location in a string by changing the value of the lastIndex property.

edited May 30, 2014 at 17:43

answered May 30, 2014 at 17:36

subhaze

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2 Comments

MrPizzaFace Over a year ago

Ah I see... So basically I can do this var re = new RegExp("[a-z]","i"); or var re = /[a-z]/i;

subhaze Over a year ago

Yup! I usually just go with the literal form instead of instantiating an object

hwnd · Accepted Answer · 2014-05-30 17:51:03Z

2

If you want to create a RegExp Object you must supply the correct syntax.

Syntax:

var re = new RegExp(pattern, modifiers);
var re = /pattern/modifiers;

So you could use either of the following here.

var re = new RegExp('[a-z]', 'i');
var re = /[a-z]/i;

edited May 30, 2014 at 17:51

answered May 30, 2014 at 17:40

hwnd

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1 Comment

MrPizzaFace Over a year ago

Yep, thanks.. just figured that out... originally had re = '/[a-z]/i'; which wasn't working.. which is why I opted to use new RegExp.. My syntax was wrong.. But since it was a Str I can see why it didn't throw a syntax error.. Well that was 10 minutes wasted. Thanks.

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