Regex, two ways, different results.

Question

Here goes two same ways to achive something. They both should do the same thing, but they don't.

a1 = function(str) {
    return str.replace(new RegExp("[^a-zA-Z0-9\s]", "gi"), "");
}

a2 = function(str) {
    return str.replace(/[^a-zA-Z0-9\s]/gi, "");
}

var string = "abc abc";
console.log(a1(string));
console.log(a2(string));

output:

abcabc
abc abc

Please, tell me why? Code is the same, except that first function creates regex, by new operator, second not.

this relates to a question I asked before, here you go: http://stackoverflow.com/questions/28910934/why-does-regex-test-changes-the-result-in-subsequent-calls — Edwin Reynoso, Mar 14 '15 at 17:22

score 2 · Accepted Answer · answered Mar 14 '15 at 17:22

2

You need to use \\s while using RegExp object:

a1 = function(str) {
    return str.replace(new RegExp("[^a-zA-Z0-9\\s]", "gi"), "");
}

Otherwise \s is interpreted as literal s in RegExp object.

answered Mar 14 '15 at 17:22

anubhava

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Woohoo! I will mark this post as an answer in next 9 minutes. Thanks! – ElSajko Mar 14 '15 at 17:26

score 2 · Answer 2 · answered Mar 14 '15 at 17:22

> console.log("[^a-zA-Z0-9\s]");
"[^a-zA-Z0-9s]"

String literals also use backslashes as escapes, and unrecognized sequences have their backslashes ignored. The escape never makes it to the RegExp constructor.

The correct way to do things, of course, is to always use regular expression literals. If you need a literal backslash in any string, though, it can be escaped too.

new RegExp("[^a-zA-Z0-9\\s]", "gi")

Regex, two ways, different results.

2 Answers2