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Replace with a regex formula prop has quotation mark issue [duplicate]

This question already has answers here:
How do you use a variable in a regular expression?
(27 answers)
Closed 2 years ago.
So for example:
function(input){
var testVar = input;
string = ...
string.replace(/ReGeX + testVar + ReGeX/, "replacement")
}
But this is of course not working :)
Is there any way to do this?

const regex = new RegExp(`ReGeX${testVar}ReGeX`);
...
string.replace(regex, "replacement");
Update
Per some of the comments, it's important to note that you may want to escape the variable if there is potential for malicious content (e.g. the variable comes from user input)
ES6 Update
In 2019, this would usually be written using a template string, and the above code has been updated. The original answer was:
var regex = new RegExp("ReGeX" + testVar + "ReGeX");
...
string.replace(regex, "replacement");

You can use the RegExp object:
var regexstring = "whatever";
var regexp = new RegExp(regexstring, "gi");
var str = "whateverTest";
var str2 = str.replace(regexp, "other");
document.write(str2);
Then you can construct regexstring in any way you want.
You can read more about it here.

To build a regular expression from a variable in JavaScript, you'll need to use the RegExp constructor with a string parameter.
function reg(input) {
var flags;
//could be any combination of 'g', 'i', and 'm'
flags = 'g';
return new RegExp('ReGeX' + input + 'ReGeX', flags);
}
of course, this is a very naive example. It assumes that input is has been properly escaped for a regular expression. If you're dealing with user-input, or simply want to make it more convenient to match special characters, you'll need to escape special characters:
function regexEscape(str) {
return str.replace(/[-\/\\^$*+?.()|[\]{}]/g, '\\$&')
}
function reg(input) {
var flags;
//could be any combination of 'g', 'i', and 'm'
flags = 'g';
input = regexEscape(input);
return new RegExp('ReGeX' + input + 'ReGeX', flags);
}

You can create regular expressions in JS in one of two ways:
Using regular expression literal - /ab{2}/g
Using the regular expression constructor - new RegExp("ab{2}", "g") .
Regular expression literals are constant, and can not be used with variables. This could be achieved using the constructor. The stracture of the RegEx constructor is
new RegExp(regularExpressionString, modifiersString)
You can embed variables as part of the regularExpressionString. For example,
var pattern="cd"
var repeats=3
new RegExp(`${pattern}{${repeats}}`, "g")
This will match any appearance of the pattern cdcdcd.

if you're using es6 template literals are an option...
string.replace(new RegExp(`ReGeX${testVar}ReGeX`), "replacement")

You can always give regular expression as string, i.e. "ReGeX" + testVar + "ReGeX". You'll possibly have to escape some characters inside your string (e.g., double quote), but for most cases it's equivalent.
You can also use RegExp constructor to pass flags in (see the docs).

It's only necessary to prepare the string variable first and then convert it to the RegEx.
for example:
You want to add minLength and MaxLength with the variable to RegEx:
function getRegEx() {
const minLength = "5"; // for exapmle: min is 5
const maxLength = "12"; // for exapmle: man is 12
var regEx = "^.{" + minLength + ","+ maxLength +"}$"; // first we make a String variable of our RegEx
regEx = new RegExp(regEx, "g"); // now we convert it to RegEx
return regEx; // In the end, we return the RegEx
}
now if you change value of MaxLength or MinLength, It will change in all RegExs.
Hope to be useful. Also sorry about my English.

Here's an pretty useless function that return values wrapped by specific characters. :)
jsfiddle: https://jsfiddle.net/squadjot/43agwo6x/
function getValsWrappedIn(str,c1,c2){
var rg = new RegExp("(?<=\\"+c1+")(.*?)(?=\\"+c2+")","g");
return str.match(rg);
}
var exampleStr = "Something (5) or some time (19) or maybe a (thingy)";
var results = getValsWrappedIn(exampleStr,"(",")")
// Will return array ["5","19","thingy"]
console.log(results)

accepted answer doesn't work for me and doesn't follow MDN examples
see the 'Description' section in above link
I'd go with the following it's working for me:
let stringThatIsGoingToChange = 'findMe';
let flagsYouWant = 'gi' //simple string with flags
let dynamicRegExp = new RegExp(`${stringThatIsGoingToChange}`, flagsYouWant)
// that makes dynamicRegExp = /findMe/gi

Javascript regexp constructor with backtick [duplicate]

This question already has answers here:
How do you use a variable in a regular expression?
(27 answers)
Closed 2 years ago.
So for example:
function(input){
var testVar = input;
string = ...
string.replace(/ReGeX + testVar + ReGeX/, "replacement")
}
But this is of course not working :)
Is there any way to do this?

const regex = new RegExp(`ReGeX${testVar}ReGeX`);
...
string.replace(regex, "replacement");
Update
Per some of the comments, it's important to note that you may want to escape the variable if there is potential for malicious content (e.g. the variable comes from user input)
ES6 Update
In 2019, this would usually be written using a template string, and the above code has been updated. The original answer was:
var regex = new RegExp("ReGeX" + testVar + "ReGeX");
...
string.replace(regex, "replacement");

You can use the RegExp object:
var regexstring = "whatever";
var regexp = new RegExp(regexstring, "gi");
var str = "whateverTest";
var str2 = str.replace(regexp, "other");
document.write(str2);
Then you can construct regexstring in any way you want.
You can read more about it here.

To build a regular expression from a variable in JavaScript, you'll need to use the RegExp constructor with a string parameter.
function reg(input) {
var flags;
//could be any combination of 'g', 'i', and 'm'
flags = 'g';
return new RegExp('ReGeX' + input + 'ReGeX', flags);
}
of course, this is a very naive example. It assumes that input is has been properly escaped for a regular expression. If you're dealing with user-input, or simply want to make it more convenient to match special characters, you'll need to escape special characters:
function regexEscape(str) {
return str.replace(/[-\/\\^$*+?.()|[\]{}]/g, '\\$&')
}
function reg(input) {
var flags;
//could be any combination of 'g', 'i', and 'm'
flags = 'g';
input = regexEscape(input);
return new RegExp('ReGeX' + input + 'ReGeX', flags);
}

You can create regular expressions in JS in one of two ways:
Using regular expression literal - /ab{2}/g
Using the regular expression constructor - new RegExp("ab{2}", "g") .
Regular expression literals are constant, and can not be used with variables. This could be achieved using the constructor. The stracture of the RegEx constructor is
new RegExp(regularExpressionString, modifiersString)
You can embed variables as part of the regularExpressionString. For example,
var pattern="cd"
var repeats=3
new RegExp(`${pattern}{${repeats}}`, "g")
This will match any appearance of the pattern cdcdcd.

if you're using es6 template literals are an option...
string.replace(new RegExp(`ReGeX${testVar}ReGeX`), "replacement")

You can always give regular expression as string, i.e. "ReGeX" + testVar + "ReGeX". You'll possibly have to escape some characters inside your string (e.g., double quote), but for most cases it's equivalent.
You can also use RegExp constructor to pass flags in (see the docs).

It's only necessary to prepare the string variable first and then convert it to the RegEx.
for example:
You want to add minLength and MaxLength with the variable to RegEx:
function getRegEx() {
const minLength = "5"; // for exapmle: min is 5
const maxLength = "12"; // for exapmle: man is 12
var regEx = "^.{" + minLength + ","+ maxLength +"}$"; // first we make a String variable of our RegEx
regEx = new RegExp(regEx, "g"); // now we convert it to RegEx
return regEx; // In the end, we return the RegEx
}
now if you change value of MaxLength or MinLength, It will change in all RegExs.
Hope to be useful. Also sorry about my English.

Here's an pretty useless function that return values wrapped by specific characters. :)
jsfiddle: https://jsfiddle.net/squadjot/43agwo6x/
function getValsWrappedIn(str,c1,c2){
var rg = new RegExp("(?<=\\"+c1+")(.*?)(?=\\"+c2+")","g");
return str.match(rg);
}
var exampleStr = "Something (5) or some time (19) or maybe a (thingy)";
var results = getValsWrappedIn(exampleStr,"(",")")
// Will return array ["5","19","thingy"]
console.log(results)

accepted answer doesn't work for me and doesn't follow MDN examples
see the 'Description' section in above link
I'd go with the following it's working for me:
let stringThatIsGoingToChange = 'findMe';
let flagsYouWant = 'gi' //simple string with flags
let dynamicRegExp = new RegExp(`${stringThatIsGoingToChange}`, flagsYouWant)
// that makes dynamicRegExp = /findMe/gi

Use dynamic (variable) string as regex pattern in JavaScript

I want to add a (variable) tag to values with regex, the pattern works fine with PHP but I have troubles implementing it into JavaScript.
The pattern is (value is the variable):
/(?!(?:[^<]+>|[^>]+<\/a>))\b(value)\b/is
I escaped the backslashes:
var str = $("#div").html();
var regex = "/(?!(?:[^<]+>|[^>]+<\\/a>))\\b(" + value + ")\\b/is";
$("#div").html(str.replace(regex, "" + value + ""));
But this seem not to be right, I logged the pattern and its exactly what it should be.
Any ideas?

To create the regex from a string, you have to use JavaScript's RegExp object.
If you also want to match/replace more than one time, then you must add the g (global match) flag. Here's an example:
var stringToGoIntoTheRegex = "abc";
var regex = new RegExp("#" + stringToGoIntoTheRegex + "#", "g");
// at this point, the line above is the same as: var regex = /#abc#/g;
var input = "Hello this is #abc# some #abc# stuff.";
var output = input.replace(regex, "!!");
alert(output); // Hello this is !! some !! stuff.
JSFiddle demo here.
In the general case, escape the string before using as regex:
Not every string is a valid regex, though: there are some speciall characters, like ( or [. To work around this issue, simply escape the string before turning it into a regex. A utility function for that goes in the sample below:
function escapeRegExp(stringToGoIntoTheRegex) {
return stringToGoIntoTheRegex.replace(/[-\/\\^$*+?.()|[\]{}]/g, '\\$&');
}
var stringToGoIntoTheRegex = escapeRegExp("abc"); // this is the only change from above
var regex = new RegExp("#" + stringToGoIntoTheRegex + "#", "g");
// at this point, the line above is the same as: var regex = /#abc#/g;
var input = "Hello this is #abc# some #abc# stuff.";
var output = input.replace(regex, "!!");
alert(output); // Hello this is !! some !! stuff.
JSFiddle demo here.
Note: the regex in the question uses the s modifier, which didn't exist at the time of the question, but does exist -- a s (dotall) flag/modifier in JavaScript -- today.

If you are trying to use a variable value in the expression, you must use the RegExp "constructor".
var regex = "(?!(?:[^<]+>|[^>]+<\/a>))\b(" + value + ")\b";
new RegExp(regex, "is")

I found I had to double slash the \b to get it working. For example to remove "1x" words from a string using a variable, I needed to use:
str = "1x";
var regex = new RegExp("\\b"+str+"\\b","g"); // same as inv.replace(/\b1x\b/g, "")
inv=inv.replace(regex, "");

You don't need the " to define a regular expression so just:
var regex = /(?!(?:[^<]+>|[^>]+<\/a>))\b(value)\b/is; // this is valid syntax
If value is a variable and you want a dynamic regular expression then you can't use this notation; use the alternative notation.
String.replace also accepts strings as input, so you can do "fox".replace("fox", "bear");
Alternative:
var regex = new RegExp("/(?!(?:[^<]+>|[^>]+<\/a>))\b(value)\b/", "is");
var regex = new RegExp("/(?!(?:[^<]+>|[^>]+<\/a>))\b(" + value + ")\b/", "is");
var regex = new RegExp("/(?!(?:[^<]+>|[^>]+<\/a>))\b(.*?)\b/", "is");
Keep in mind that if value contains regular expressions characters like (, [ and ? you will need to escape them.

I found this thread useful - so I thought I would add the answer to my own problem.
I wanted to edit a database configuration file (datastax cassandra) from a node application in javascript and for one of the settings in the file I needed to match on a string and then replace the line following it.
This was my solution.
dse_cassandra_yaml='/etc/dse/cassandra/cassandra.yaml'
// a) find the searchString and grab all text on the following line to it
// b) replace all next line text with a newString supplied to function
// note - leaves searchString text untouched
function replaceStringNextLine(file, searchString, newString) {
fs.readFile(file, 'utf-8', function(err, data){
if (err) throw err;
// need to use double escape '\\' when putting regex in strings !
var re = "\\s+(\\-\\s(.*)?)(?:\\s|$)";
var myRegExp = new RegExp(searchString + re, "g");
var match = myRegExp.exec(data);
var replaceThis = match[1];
var writeString = data.replace(replaceThis, newString);
fs.writeFile(file, writeString, 'utf-8', function (err) {
if (err) throw err;
console.log(file + ' updated');
});
});
}
searchString = "data_file_directories:"
newString = "- /mnt/cassandra/data"
replaceStringNextLine(dse_cassandra_yaml, searchString, newString );
After running, it will change the existing data directory setting to the new one:
config file before:
data_file_directories:
- /var/lib/cassandra/data
config file after:
data_file_directories:
- /mnt/cassandra/data

Much easier way: use template literals.
var variable = 'foo'
var expression = `.*${variable}.*`
var re = new RegExp(expression, 'g')
re.test('fdjklsffoodjkslfd') // true
re.test('fdjklsfdjkslfd') // false

Using string variable(s) content as part of a more complex composed regex expression (es6|ts)
This example will replace all urls using my-domain.com to my-other-domain (both are variables).
You can do dynamic regexs by combining string values and other regex expressions within a raw string template. Using String.raw will prevent javascript from escaping any character within your string values.
// Strings with some data
const domainStr = 'my-domain.com'
const newDomain = 'my-other-domain.com'
// Make sure your string is regex friendly
// This will replace dots for '\'.
const regexUrl = /\./gm;
const substr = `\\\.`;
const domain = domainStr.replace(regexUrl, substr);
// domain is a regex friendly string: 'my-domain\.com'
console.log('Regex expresion for domain', domain)
// HERE!!! You can 'assemble a complex regex using string pieces.
const re = new RegExp( String.raw `([\'|\"]https:\/\/)(${domain})(\S+[\'|\"])`, 'gm');
// now I'll use the regex expression groups to replace the domain
const domainSubst = `$1${newDomain}$3`;
// const page contains all the html text
const result = page.replace(re, domainSubst);
note: Don't forget to use regex101.com to create, test and export REGEX code.

var string = "Hi welcome to stack overflow"
var toSearch = "stack"
//case insensitive search
var result = string.search(new RegExp(toSearch, "i")) > 0 ? 'Matched' : 'notMatched'
https://jsfiddle.net/9f0mb6Lz/
Hope this helps

In Javascript, how can I perform a global replace on string with a variable inside '/' and '/g'?

I want to perform a global replace of string using String.replace in Javascript.
In the documentation I read that I can do this with /g, i.e. for example;
var mystring = mystring.replace(/test/g, mystring);
and this will replace all occurrences inside mystring. No quotes for the expression.
But if I have a variable to find, how can I do this without quotes?
I've tried something like this:
var stringToFind = "test";
//first try
mystring = mystring.replace('/' + stringToFind + '/g', mystring);
//second try, not much sense at all
mystring = mystring.replace(/stringToFind/g, mystring);
but they don't work. Any ideas?

var mystring = "hello world test world";
var find = "world";
var regex = new RegExp(find, "g");
alert(mystring.replace(regex, "yay")); // alerts "hello yay test yay"
In case you need this into a function
replaceGlobally(original, searchTxt, replaceTxt) {
const regex = new RegExp(searchTxt, 'g');
return original.replace(regex, replaceTxt) ;
}

For regex, new RegExp(stringtofind, 'g');. BUT. If ‘find’ contains characters that are special in regex, they will have their regexy meaning. So if you tried to replace the '.' in 'abc.def' with 'x', you'd get 'xxxxxxx' — whoops.
If all you want is a simple string replacement, there is no need for regular expressions! Here is the plain string replace idiom:
mystring= mystring.split(stringtofind).join(replacementstring);

Regular expressions are much slower then string search. So, creating regex with escaped search string is not an optimal way. Even looping though the string would be faster, but I suggest using built-in pre-compiled methods.
Here is a fast and clean way of doing fast global string replace:
sMyString.split(sSearch).join(sReplace);
And that's it.

String.prototype.replaceAll = function (replaceThis, withThis) {
var re = new RegExp(RegExp.quote(replaceThis),"g");
return this.replace(re, withThis);
};
RegExp.quote = function(str) {
return str.replace(/([.?*+^$[\]\\(){}-])/g, "\\$1");
};
var aa = "qwerr.erer".replaceAll(".","A");
alert(aa);
silmiar post

You can use the following solution to perform a global replace on a string with a variable inside '/' and '/g':
myString.replace(new RegExp(strFind, 'g'), strReplace);

Thats a regular expression, not a string. Use the constructor for a RegExp object to dynamically create a regex.
var r = new RegExp(stringToFind, 'g');
mystring.replace(r, 'some replacement text');

Try:
var stringToFind = "test";
mystring = mystring.replace(new RegExp(stringToFind, "g"), mystring);

You can do using following method
see this function:
function SetValue()
{
var txt1='This is a blacK table with BLack pen with bLack lady';
alert(txt1);
var txt2= txt1.replace(/black/gi,'green');
alert(txt2);
}
syntax:
/search_string/{g|gi}
where
g is global case-sensitive replacement
gi is blobal case-insensitive replacement
You can check this JSBIN link
http://jsbin.com/nohuroboxa/edit?html,js,output

If you want variables interpolated, you need to use the RegExp object
https://developer.mozilla.org/en/Core_JavaScript_1.5_Guide/Regular_Expressions
Example:
var str = "This is my name";
var replace = "i";
var re = new RegExp(replace, 'g')
str = str.replace(re, 'p');
alert(str);

Dynamic global replace
I came to this thread looking for a slightly more complex solution which isn't answered here. I've now found the answer so I'm going to post it in case anyone else finds it useful.
I wanted to do a dynamic global replace, where the replacement strings are based on the original matches.
For example, to capitalise the first letter of all words (e.g. "cat sat mat" into "Cat Sat Mat") with a global find replace. Here's how to do that.
function capitaliseWords(aString) {
// Global match for lowercase letters following a word boundary
var letters = aString.match(/\b[a-z]/g), i, letterMatch;
// Loop over all matched letters
for( i = 0; i < letters.length; i++ ) {
// Replace the matched letters with upper case versions
letterMatch = new RegExp('\\b'+letters[i]); // EDIT - slight fix
aString = aString.replace(letterMatch, letters[i].toUpperCase());
}
// Return our newly capitalised string
return aString;
}
alert( capitaliseWords("cat sat mat") ); // Alerts "Cat Sat Mat"

WIth modern day linters, they prefer you to regEx literal, so rather than new RegExp it would just be `//
With an example:
'test'.replace(/ /gi, '_')
with the test you are looking to replace inside the regex or the /searchableText/ and then replace text in the second parameter. In my case I wanted to replace all spaces with underscores.

Can you use prototype.js? If so you could use String.gsub, like
var myStr = "a day in a life of a thing";
var replace = "a";
var resultString = myStr.gsub(replace, "g");
// resultString will be "g day in g life of g thing"
It will also take regular expressions. To me this is one of the more elegant ways to solve it. prototypejs gsub documentation

How can I concatenate regex literals in JavaScript?

Is it possible to do something like this?
var pattern = /some regex segment/ + /* comment here */
/another segment/;
Or do I have to use new RegExp() syntax and concatenate a string? I'd prefer to use the literal as the code is both more self-evident and concise.

Here is how to create a regular expression without using the regular expression literal syntax. This lets you do arbitary string manipulation before it becomes a regular expression object:
var segment_part = "some bit of the regexp";
var pattern = new RegExp("some regex segment" + /*comment here */
segment_part + /* that was defined just now */
"another segment");
If you have two regular expression literals, you can in fact concatenate them using this technique:
var regex1 = /foo/g;
var regex2 = /bar/y;
var flags = (regex1.flags + regex2.flags).split("").sort().join("").replace(/(.)(?=.*\1)/g, "");
var regex3 = new RegExp(expression_one.source + expression_two.source, flags);
// regex3 is now /foobar/gy
It's just more wordy than just having expression one and two being literal strings instead of literal regular expressions.

Just randomly concatenating regular expressions objects can have some adverse side effects. Use the RegExp.source instead:
var r1 = /abc/g;
var r2 = /def/;
var r3 = new RegExp(r1.source + r2.source,
(r1.global ? 'g' : '')
+ (r1.ignoreCase ? 'i' : '') +
(r1.multiline ? 'm' : ''));
console.log(r3);
var m = 'test that abcdef and abcdef has a match?'.match(r3);
console.log(m);
// m should contain 2 matches
This will also give you the ability to retain the regular expression flags from a previous RegExp using the standard RegExp flags.
jsFiddle

I don't quite agree with the "eval" option.
var xxx = /abcd/;
var yyy = /efgh/;
var zzz = new RegExp(eval(xxx)+eval(yyy));
will give "//abcd//efgh//" which is not the intended result.
Using source like
var zzz = new RegExp(xxx.source+yyy.source);
will give "/abcdefgh/" and that is correct.
Logicaly there is no need to EVALUATE, you know your EXPRESSION. You just need its SOURCE or how it is written not necessarely its value. As for the flags, you just need to use the optional argument of RegExp.
In my situation, I do run in the issue of ^ and $ being used in several expression I am trying to concatenate together! Those expressions are grammar filters used accross the program. Now I wan't to use some of them together to handle the case of PREPOSITIONS.
I may have to "slice" the sources to remove the starting and ending ^( and/or )$ :)
Cheers, Alex.

Problem If the regexp contains back-matching groups like \1.
var r = /(a|b)\1/ // Matches aa, bb but nothing else.
var p = /(c|d)\1/ // Matches cc, dd but nothing else.
Then just contatenating the sources will not work. Indeed, the combination of the two is:
var rp = /(a|b)\1(c|d)\1/
rp.test("aadd") // Returns false
The solution:
First we count the number of matching groups in the first regex, Then for each back-matching token in the second, we increment it by the number of matching groups.
function concatenate(r1, r2) {
var count = function(r, str) {
return str.match(r).length;
}
var numberGroups = /([^\\]|^)(?=\((?!\?:))/g; // Home-made regexp to count groups.
var offset = count(numberGroups, r1.source);
var escapedMatch = /[\\](?:(\d+)|.)/g; // Home-made regexp for escaped literals, greedy on numbers.
var r2newSource = r2.source.replace(escapedMatch, function(match, number) { return number?"\\"+(number-0+offset):match; });
return new RegExp(r1.source+r2newSource,
(r1.global ? 'g' : '')
+ (r1.ignoreCase ? 'i' : '')
+ (r1.multiline ? 'm' : ''));
}
Test:
var rp = concatenate(r, p) // returns /(a|b)\1(c|d)\2/
rp.test("aadd") // Returns true

Providing that:
you know what you do in your regexp;
you have many regex pieces to form a pattern and they will use same flag;
you find it more readable to separate your small pattern chunks into an array;
you also want to be able to comment each part for next dev or yourself later;
you prefer to visually simplify your regex like /this/g rather than new RegExp('this', 'g');
it's ok for you to assemble the regex in an extra step rather than having it in one piece from the start;
Then you may like to write this way:
var regexParts =
[
/\b(\d+|null)\b/,// Some comments.
/\b(true|false)\b/,
/\b(new|getElementsBy(?:Tag|Class|)Name|arguments|getElementById|if|else|do|null|return|case|default|function|typeof|undefined|instanceof|this|document|window|while|for|switch|in|break|continue|length|var|(?:clear|set)(?:Timeout|Interval))(?=\W)/,
/(\$|jQuery)/,
/many more patterns/
],
regexString = regexParts.map(function(x){return x.source}).join('|'),
regexPattern = new RegExp(regexString, 'g');
you can then do something like:
string.replace(regexPattern, function()
{
var m = arguments,
Class = '';
switch(true)
{
// Numbers and 'null'.
case (Boolean)(m[1]):
m = m[1];
Class = 'number';
break;
// True or False.
case (Boolean)(m[2]):
m = m[2];
Class = 'bool';
break;
// True or False.
case (Boolean)(m[3]):
m = m[3];
Class = 'keyword';
break;
// $ or 'jQuery'.
case (Boolean)(m[4]):
m = m[4];
Class = 'dollar';
break;
// More cases...
}
return '<span class="' + Class + '">' + m + '</span>';
})
In my particular case (a code-mirror-like editor), it is much easier to perform one big regex, rather than a lot of replaces like following as each time I replace with a html tag to wrap an expression, the next pattern will be harder to target without affecting the html tag itself (and without the good lookbehind that is unfortunately not supported in javascript):
.replace(/(\b\d+|null\b)/g, '<span class="number">$1</span>')
.replace(/(\btrue|false\b)/g, '<span class="bool">$1</span>')
.replace(/\b(new|getElementsBy(?:Tag|Class|)Name|arguments|getElementById|if|else|do|null|return|case|default|function|typeof|undefined|instanceof|this|document|window|while|for|switch|in|break|continue|var|(?:clear|set)(?:Timeout|Interval))(?=\W)/g, '<span class="keyword">$1</span>')
.replace(/\$/g, '<span class="dollar">$</span>')
.replace(/([\[\](){}.:;,+\-?=])/g, '<span class="ponctuation">$1</span>')

It would be preferable to use the literal syntax as often as possible. It's shorter, more legible, and you do not need escape quotes or double-escape backlashes. From "Javascript Patterns", Stoyan Stefanov 2010.
But using New may be the only way to concatenate.
I would avoid eval. Its not safe.

You could do something like:
function concatRegex(...segments) {
return new RegExp(segments.join(''));
}
The segments would be strings (rather than regex literals) passed in as separate arguments.

You can concat regex source from both the literal and RegExp class:
var xxx = new RegExp(/abcd/);
var zzz = new RegExp(xxx.source + /efgh/.source);

Use the constructor with 2 params and avoid the problem with trailing '/':
var re_final = new RegExp("\\" + ".", "g"); // constructor can have 2 params!
console.log("...finally".replace(re_final, "!") + "\n" + re_final +
" works as expected..."); // !!!finally works as expected
// meanwhile
re_final = new RegExp("\\" + "." + "g"); // appends final '/'
console.log("... finally".replace(re_final, "!")); // ...finally
console.log(re_final, "does not work!"); // does not work

No, the literal way is not supported. You'll have to use RegExp.

the easier way to me would be concatenate the sources, ex.:
a = /\d+/
b = /\w+/
c = new RegExp(a.source + b.source)
the c value will result in:
/\d+\w+/

I prefer to use eval('your expression') because it does not add the /on each end/ that ='new RegExp' does.