What is the difference between
var regEx = /\d/;
and
var regEx = new RegEx("\d");
Bob
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1 Answer
Both evaluate to be the same exact regex, but the first is a literal, meaning you cannot use any variables inside it, you cannot dynamically generate a regex.
The second uses the constructor explicitly and can be used to create dynamic regular expressions.
var x = '3', r = ( new RegExp( x + '\d' ) ); r.test('3d')
The above is an example of dynamically constructing the regex with the constructor, which you can not do in literal form.
In 99% of cases you will just rely on the first version ( literal ) for all your regexpes in JS. In an advanced scenario where you need say, user input to dynamically construct a regex then that's when you'll need the second form.
EDIT #1: The first matches a digit, the second just matches the letter d. You have to double-escape the second one in order for it to equal the first one, which I assume you meant to do. Just remember the advice I typed above is accurate if the second example is new RegExp('\\d').
/\d/.test('3') // true
( new RegExp('\d') ).test('3') // false
( new RegExp('\\d') ).test('3') // true
2 Comments
bob
Thanks for the explaination, the reason I asked was I had this expression /^ +|[\\/#;]| +$/gi (1 or more spaces OR contains \ / # ; OR ends with one or more space). When using the non-constructor every odd time the result resulted in false. Eg. the result was alternating. When using the constructor everything worked as expected every time. Strange!
Felix Kling
There is yet another difference: The literal only creates one object at parse time whereas the constructor function always creates a new object. This will be changed though in ES5 and the literal will also always return a new object.