How to create regex pattern which is concatenate with variable, something like this:
var test ="52";
var re = new RegExp("/\b"+test+"\b/");
alert('51,52,53'.match(re));
Thanks
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3 Answers
var re = new RegExp("/\b"+test+"\b/");
\b in a string literal is a backspace character. When putting a regex in a string literal you need one more round of escaping:
var re = new RegExp("\\b"+test+"\\b");
(You also don't need the // in this context.)
7 Comments
bobince
There are many places where the constructor-function of a built-in type may be used both with or without
new. However, for consistency with other objects where this may not hold true, and clarity in general, I would always use new.
IQAndreas
Another way is to use single-quotes for clarity so you don't need to escape the back-slashes:
new RegExp('\b'+test+'\b');bobince
You still need to escape the backslashes. Single quotes don't have different escaping rules to double quotes in JavaScript (unlike, say, PHP).
HoldOffHunger
This answer also solves the same question when trying to use the border-operators on a variable with the .match() and .replace() JS regexing functions.
Bas Peeters
Just what I was looking for. The part about
When putting a regex in a string literal you need one more round of escaping; is there any background information about (the reasons of) this? |
With ES2015 (aka ES6) you can use template literals when constructing RegExp:
let test = '53'
const regexp = new RegExp(`\\b${test}\\b`, 'gi') // showing how to pass optional flags
console.log('51, 52, 53, 54'.match(regexp))
1 Comment
NFpeter
typo: Regexp must be RegExp
you can use
/(^|,)52(,|$)/.test('51,52,53')
but i suggest to use
var list = '51,52,53';
function test2(list, test){
return !((","+list+",").indexOf(","+test+",") === -1)
}
alert( test2(list,52) )
