Extending @KooiInc answer, you can avoid manually escaping every special character by using the source
property of the RegExp
object.
Example:
var urlRegex= new RegExp(''
+ /(?:(?:(https?|ftp):)?\/\/)/.source // protocol
+ /(?:([^:\n\r]+):([^@\n\r]+)@)?/.source // user:pass
+ /(?:(?:www\.)?([^\/\n\r]+))/.source // domain
+ /(\/[^?\n\r]+)?/.source // request
+ /(\?[^#\n\r]*)?/.source // query
+ /(#?[^\n\r]*)?/.source // anchor
);
or if you want to avoid repeating the .source
property you can do it using the Array.map()
function:
var urlRegex= new RegExp([
/(?:(?:(https?|ftp):)?\/\/)/ // protocol
,/(?:([^:\n\r]+):([^@\n\r]+)@)?/ // user:pass
,/(?:(?:www\.)?([^\/\n\r]+))/ // domain
,/(\/[^?\n\r]+)?/ // request
,/(\?[^#\n\r]*)?/ // query
,/(#?[^\n\r]*)?/ // anchor
].map(function(r) {return r.source}).join(''));
In ES6 the map function can be reduced to:
.map(r => r.source)
You could convert it to a string and create the expression by calling new RegExp()
:
var myRE = new RegExp (['^(([^<>()[\]\\.,;:\\s@\"]+(\\.[^<>(),[\]\\.,;:\\s@\"]+)*)',
'|(\\".+\\"))@((\\[[0-9]{1,3}\\.[0-9]{1,3}\\.[0-9]{1,3}\\.',
'[0-9]{1,3}\])|(([a-zA-Z\-0-9]+\\.)+',
'[a-zA-Z]{2,}))$'].join(''));
Notes:
RegExp
accepts modifiers as a second parameter
/regex/g
=> new RegExp('regex', 'g')
[Addition ES20xx (tagged template)]
In ES20xx you can use tagged templates. See the snippet.
Note:
\s
, \s+
, \s{1,x}
, \t
, \n
etc).(() => {
const createRegExp = (str, opts) =>
new RegExp(str.raw[0].replace(/\s/gm, ""), opts || "");
const yourRE = createRegExp`
^(([^<>()[\]\\.,;:\s@\"]+(\.[^<>()[\]\\.,;:\s@\"]+)*)|
(\".+\"))@((\[[0-9]{1,3}\.[0-9]{1,3}\.[0-9]{1,3}\.[0-9]{1,3}\])|
(([a-zA-Z\-0-9]+\.)+[a-zA-Z]{2,}))$`;
console.log(yourRE);
const anotherLongRE = createRegExp`
(\byyyy\b)|(\bm\b)|(\bd\b)|(\bh\b)|(\bmi\b)|(\bs\b)|(\bms\b)|
(\bwd\b)|(\bmm\b)|(\bdd\b)|(\bhh\b)|(\bMI\b)|(\bS\b)|(\bMS\b)|
(\bM\b)|(\bMM\b)|(\bdow\b)|(\bDOW\b)
${"gi"}`;
console.log(anotherLongRE);
})();
Using strings in new RegExp
is awkward because you must escape all the backslashes. You may write smaller regexes and concatenate them.
Let's split this regex
/^foo(.*)\bar$/
We will use a function to make things more beautiful later
function multilineRegExp(regs, options) {
return new RegExp(regs.map(
function(reg){ return reg.source; }
).join(''), options);
}
And now let's rock
var r = multilineRegExp([
/^foo/, // we can add comments too
/(.*)/,
/\bar$/
]);
Since it has a cost, try to build the real regex just once and then use that.
Thanks to the wonderous world of template literals you can now write big, multi-line, well-commented, and even semantically nested regexes in ES6.
//build regexes without worrying about
// - double-backslashing
// - adding whitespace for readability
// - adding in comments
let clean = (piece) => (piece
.replace(/((^|\n)(?:[^\/\\]|\/[^*\/]|\\.)*?)\s*\/\*(?:[^*]|\*[^\/])*(\*\/|)/g, '$1')
.replace(/((^|\n)(?:[^\/\\]|\/[^\/]|\\.)*?)\s*\/\/[^\n]*/g, '$1')
.replace(/\n\s*/g, '')
);
window.regex = ({raw}, ...interpolations) => (
new RegExp(interpolations.reduce(
(regex, insert, index) => (regex + insert + clean(raw[index + 1])),
clean(raw[0])
))
);
Using this you can now write regexes like this:
let re = regex`I'm a special regex{3} //with a comment!`;
Outputs
/I'm a special regex{3}/
Or what about multiline?
'123hello'
.match(regex`
//so this is a regex
//here I am matching some numbers
(\d+)
//Oh! See how I didn't need to double backslash that \d?
([a-z]{1,3}) /*note to self, this is group #2*/
`)
[2]
Outputs hel
, neat!
"What if I need to actually search a newline?", well then use \n
silly!
Working on my Firefox and Chrome.
Okay, "how about something a little more complex?"
Sure, here's a piece of an object destructuring JS parser I was working on:
regex`^\s*
(
//closing the object
(\})|
//starting from open or comma you can...
(?:[,{]\s*)(?:
//have a rest operator
(\.\.\.)
|
//have a property key
(
//a non-negative integer
\b\d+\b
|
//any unencapsulated string of the following
\b[A-Za-z$_][\w$]*\b
|
//a quoted string
//this is #5!
("|')(?:
//that contains any non-escape, non-quote character
(?!\5|\\).
|
//or any escape sequence
(?:\\.)
//finished by the quote
)*\5
)
//after a property key, we can go inside
\s*(:|)
|
\s*(?={)
)
)
((?:
//after closing we expect either
// - the parent's comma/close,
// - or the end of the string
\s*(?:[,}\]=]|$)
|
//after the rest operator we expect the close
\s*\}
|
//after diving into a key we expect that object to open
\s*[{[:]
|
//otherwise we saw only a key, we now expect a comma or close
\s*[,}{]
).*)
$`
It outputs /^\s*((\})|(?:[,{]\s*)(?:(\.\.\.)|(\b\d+\b|\b[A-Za-z$_][\w$]*\b|("|')(?:(?!\5|\\).|(?:\\.))*\5)\s*(:|)|\s*(?={)))((?:\s*(?:[,}\]=]|$)|\s*\}|\s*[{[:]|\s*[,}{]).*)$/
And running it with a little demo?
let input = '{why, hello, there, "you huge \\"", 17, {big,smelly}}';
for (
let parsed;
parsed = input.match(r);
input = parsed[parsed.length - 1]
) console.log(parsed[1]);
Successfully outputs
{why
, hello
, there
, "you huge \""
, 17
,
{big
,smelly
}
}
Note the successful capturing of the quoted string.
I tested it on Chrome and Firefox, works a treat!
If curious you can checkout what I was doing, and its demonstration.
Though it only works on Chrome, because Firefox doesn't support backreferences or named groups. So note the example given in this answer is actually a neutered version and might get easily tricked into accepting invalid strings.
There are good answers here, but for completeness someone should mention Javascript's core feature of inheritance with the prototype chain. Something like this illustrates the idea:
RegExp.prototype.append = function(re) {
return new RegExp(this.source + re.source, this.flags);
};
let regex = /[a-z]/g
.append(/[A-Z]/)
.append(/[0-9]/);
console.log(regex); //=> /[a-z][A-Z][0-9]/g
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