JavaScript and regular expressions: literal syntax Vs. RegExp object

Question

I have some troubles with this small JavaScript code:

var text="Z Test Yeah ! Z";

// With literal syntax, it returns true: good!
alert(/(Z[\s\S]*?Z)/g.test(text));

// But not with the RegExp object O_o
var reg=new RegExp('Z[\s\S]*?Z','g');
alert(reg.test(text));

I don't understand why the literal syntax and the RegExp object don't give me the same result... The problem is that I have to use the RegExp object since I'll have some variables later.

Any ideas?

Thanks in advance :)

zzzzBov · Accepted Answer · 2011-08-10 17:03:42Z

24

You need to double escape \ characters in string literals, which is why the regex literal is typically preferred.

Try:

'Z[\\s\\S]*?Z'

answered Aug 10, 2011 at 17:03

zzzzBov

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3 Comments

KorHosik Over a year ago

It's correct, with backslashes it works...! Thank you so much, since we don't need them with literal syntax, I didn't think that they are mandatory for RegExp object. Thank you again!

Geek Over a year ago

@zzzzBov Why do we need to escape the `\` in the second case and not the first case?

zzzzBov Over a year ago

@Geek, the first regular expression used is a RegExp literal. The second regular expression used is a RegExp object constructor with a string literal as an argument. My answer specifically says you need to double escape \ characters in string literals. For string literals, \ is used to create escape sequences for values, whereas in a regular expression we want the escape sequence to be the value. "\n" creates a string with a value of newline, whereas "\\n" creates a string with a value of \n. The distinction is very important for regular expression patterns.

Benjamin Atkin · Accepted Answer · 2011-08-10 17:04:26Z

3

I think it's because you have to escape your backslashes, even when using single quotes. Try this:

new RegExp('Z[\\s\\S]*?Z','g')

answered Aug 10, 2011 at 17:04

Benjamin Atkin

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