Parse Ascii Escaped String in JavaScript - javascript

Say we're given a string that itself contains ASCII escape characters, such as "\x64" (note that this is the contents of the string, not what it would be in code). Is there an easy way in JavaScript to parse this to its actual character?
There's the old trick of using JSON.parse but that seems to only work with unicode escape sequences and can't recognize ASCII ones, as in:
JSON.parse(`"\\u0064"`); // Output: "d" (as expected)
JSON.parse(`"\\x64"`); // ERROR: Unexpected token x
Is there an equivalent that would work for the ASCII escape sequence? I know that using eval() works but I'd really rather not use eval.
EDIT: To clarify, some characters in the original string may not be escaped; as we may need to parse "Hello Worl\x64", for instance.

One solution is to replace all the \x** patterns in the string, using a callback to parse them as base 16 integers and pass the result to String.fromCharCode to convert to a character:
const str = "\\x48\\x65ll\\x6f\\x20W\\x6f\\x72\\x6c\\x64";
const res = str.replace(/\\x(..)|./g, (m, p1) => p1 ? String.fromCharCode(parseInt(p1, 16)) : m);
console.log(res);
If you don't mind overwriting the original string, this can be simplified slightly to:
let str = "\\x48\\x65ll\\x6f\\x20W\\x6f\\x72\\x6c\\x64";
str = str.replace(/\\x(..)/g, (m, p1) => String.fromCharCode(parseInt(p1, 16)));
console.log(str);

You can use the eval function to evaluate a string in the same way it would be evaluated from Javascript source code, you'll only need to make sure you quote its content as such:
eval("\"\\x64\"") // will return the javascript string: "d"

Related

Validate input value to allow only integer and decimal value using Js regex [duplicate]

I'm doing a small javascript method, which receive a list of point, and I've to read those points to create a Polygon in a google map.
I receive those point on the form:
(lat, long), (lat, long),(lat, long)
So I've done the following regex:
\(\s*([0-9.-]+)\s*,\s([0-9.-]+)\s*\)
I've tested it with RegexPal and the exact data I receive:
(25.774252, -80.190262),(18.466465, -66.118292),(32.321384, -64.75737),(25.774252, -80.190262)
and it works, so why when I've this code in my javascript, I receive null in the result?
var polygons="(25.774252, -80.190262),(18.466465, -66.118292),(32.321384, -64.75737),(25.774252, -80.190262)";
var reg = new RegExp("/\(\s*([0-9.-]+)\s*,\s([0-9.-]+)\s*\)/g");
var result = polygons.match(reg);
I've no javascript error when executing(with debug mode of google chrome). This code is hosted in a javascript function which is in a included JS file. This method is called in the OnLoad method.
I've searched a lot, but I can't find why this isn't working. Thank you very much!
Use a regex literal [MDN]:
var reg = /\(\s*([0-9.-]+)\s*,\s([0-9.-]+)\s*\)/g;
You are making two errors when you use RegExp [MDN]:
The "delimiters" / are should not be part of the expression
If you define an expression as string, you have to escape the backslash, because it is the escape character in strings
Furthermore, modifiers are passed as second argument to the function.
So if you wanted to use RegExp (which you don't have to in this case), the equivalent would be:
var reg = new RegExp("\\(\\s*([0-9.-]+)\\s*,\\s([0-9.-]+)\\s*\\)", "g");
(and I think now you see why regex literals are more convenient)
I always find it helpful to copy and past a RegExp expression in the console and see its output. Taking your original expression, we get:
/(s*([0-9.-]+)s*,s([0-9.-]+)s*)/g
which means that the expressions tries to match /, s and g literally and the parens () are still treated as special characters.
Update: .match() returns an array:
["(25.774252, -80.190262)", "(18.466465, -66.118292)", ... ]
which does not seem to be very useful.
You have to use .exec() [MDN] to extract the numbers:
["(25.774252, -80.190262)", "25.774252", "-80.190262"]
This has to be called repeatedly until the whole strings was processed.
Example:
var reg = /\(\s*([0-9.-]+)\s*,\s([0-9.-]+)\s*\)/g;
var result, points = [];
while((result = reg.exec(polygons)) !== null) {
points.push([+result[1], +result[2]]);
}
This creates an array of arrays and the unary plus (+) will convert the strings into numbers:
[
[25.774252, -80.190262],
[18.466465, -66.118292],
...
]
Of course if you want the values as strings and not as numbers, you can just omit the +.

Find character characters except when surrounded by specific characters

I have a string: "${styles.button} ${styles[color]} ${styles[size]} ${styles[_state]} ${iconOnly ? styles.iconOnly : ''}", and I'm trying to use regex to find all the spaces, except for spaces that are part of an interpolation string (${...}).
I'm willing to admit that regex might not be the right tool for this job, but I'm curious what I'm missing.
Essentially what I'm trying to do is replace the spaces with a newline character.
You can split the string in interpolation string and non-interpolation string sequences and then only modify the odd sequences (the resulting array always starts with a non-interpolation string, don't worry about that). This has to be done, because regular expressions are limited in the states they can remember (for more about that study CS). A solution would be:
var string = "${styles.button} ${styles[color]} ${styles[size]} ${styles[_state]} ${iconOnly ? styles.iconOnly : ''}";
var result = string
// split in non-interpolation string and interpolation string sequences
.split(/(\${[^}]*})/g)
// modify the sequences with odd indices ( non-interpolation)
.map((part, i) => (i % 2 ? part : part.replace(/ +/g, '')))
// concatenate the strings
.join('');
console.log(result);
But also mind the comment by ggorlen on your question:
Looks like you're trying to use regex to parse arbitrary JS template strings. That isn't an easy task in the general case and regex is probably the wrong tool for the job--it's likely an xy problem. Can you provide more context (why do you need to parse JS template strings in the first place?) and show an attempt? Thanks.
Assuming you have only have ${...} patterns separated by space as per your example you can apply this regex:
var str = "${styles.button} ${styles[color]} ${styles[size]} ${styles[_state]} ${iconOnly ? styles.iconOnly : ''}"
var re = /(\}) +(\$\{)/g;
var result = str.replace(re, "$1\n$2");
console.log('result: ' + result);
Result:
result: ${styles.button}
${styles[color]}
${styles[size]}
${styles[_state]}
${iconOnly ? styles.iconOnly : ''}
I tested with a simple find ' \$' (without quotes), replace with '\n$' (without quotes) - in sublime text regex search, works well

Wrong result when replacing with regex

I'm replacing a sub-string using replace function and regex expression.
However after character escape and replacement, I still have an extra '/' character. I'm not really familiar with regex can someone guide me.
I have implemented the escape character function found here: Is there a RegExp.escape function in Javascript?
RegExp.escape= function(s) {
return s.replace(/[-\/\\^$*+?.()|[\]{}]/g, '\\$&');
};
const latexConversions = [
["\\cdot", "*"],
["\\right\)", ")"],
["\\left\(", "("],
["\\pi", "pi"],
["\\ln\((.*?)\)", "log($1)"],
["stdev\((.*?)\)", "std($1)"],
["stdevp\((.*?)\)", "std(\[$1\], \"uncorrected\")"],
["mean\((.*?)\)", "mean($1)"],
["\\sqrt\((.*?)\)", "sqrt($1)"],
["\\log\((.*?)\)", "log10($1)"],
["\(e\)", "e"],
["\\exp\((.*?)\)", "exp($1)"],
["round\((.*?)\)", "round($1)"],
["npr\((.*?),(.*?)\)", "($1!/($1-$2)!)"],
["ncr\((.*?),(.*?)\)", "($1!/($2!($1-$2)!))"],
["\\left\|", "abs("],
["\\right\|", ")"],
];
RegExp.escape = function (s) {
var t = s.replace(/[-\/\\^$*+?.()|[\]{}]/g, '\\$&');
return t;
};
mathematicalExpression = "\\sqrt( )"
//Problem is here
mathematicalExpression = mathematicalExpression.replace(new RegExp(RegExp.escape(latexConversions[8][0]), 'g'), latexConversions[8][1]);
//Works
mathematicalExpression2 = mathematicalExpression.replace(/\\sqrt\((.*?)\)/g, "sqrt($1)");
alert("what I got: "+mathematicalExpression); // "\sqrt()"
alert("Supposed to be: "+ mathematicalExpression2); // "sqtr()"
I have a live example here: https://jsfiddle.net/nky342h5/2/
There are several misconceptions regarding the string literal "\\sqrt\((.*?)\)":
This string in raw characters is: \sqrt((.*?)). Note how there is no difference between the two opening parentheses: the backslash in the string literal was not very useful. In other words, "\(" === "("
Both opening parentheses will be escaped by RegExp.escape
Points 1 and 2 are equally true for the closing parentheses, for the dot, the asterisk and the question mark: they will be escaped by RegExp.escape.
In short, you have no way to distinguish that a character is intended as a literal or as a regex special symbol -- you are escaping all of them as if they were intended as literal characters.
The solution:
Since you already are encoding regex specific syntax in your strings (like (.*?)), you might as well use regex literals instead of string literals.
In the case you highlighted, instead of this:
["\\sqrt\((.*?)\)", "sqrt($1)"]
...use this:
[/\\sqrt\((.*?)\)/g, "sqrt($1)"]
And let your code do:
mathematicalExpression = mathematicalExpression.replace(...latexConversions[8]);
Alternative
If for some reason regex literals are a no-go, then define your own special syntax for (.*?). For instance, use the symbol µ to denote that particular regex syntax.
Then your array pair would look like this:
["\\sqrt(µ)", "sqrt($1)"],
...and code:
mathematicalExpression = mathematicalExpression.replace(
new RegExp(RegExp.escape(latexConversions[8][0]).replace(/µ/g, '(.*?)'), 'g'),
latexConversions[8][1]
);
Note how here the (.*?) is introduced in the string after RegExp.escape has done its job.
extra \ rather than escaping everything
replace ["\\sqrt\((.*?)\)", "sqrt($1)"], with ["\\\\sqrt\((.*?)\)", "sqrt($1)"],
and replace the final replace with
mathematicalExpression = mathematicalExpression.replace(new RegExp((latexConversions1[8][0]), 'g'), latexConversions1[8][1]);

Regex conversion from PHP to JavaScript [duplicate]

I'm doing a small javascript method, which receive a list of point, and I've to read those points to create a Polygon in a google map.
I receive those point on the form:
(lat, long), (lat, long),(lat, long)
So I've done the following regex:
\(\s*([0-9.-]+)\s*,\s([0-9.-]+)\s*\)
I've tested it with RegexPal and the exact data I receive:
(25.774252, -80.190262),(18.466465, -66.118292),(32.321384, -64.75737),(25.774252, -80.190262)
and it works, so why when I've this code in my javascript, I receive null in the result?
var polygons="(25.774252, -80.190262),(18.466465, -66.118292),(32.321384, -64.75737),(25.774252, -80.190262)";
var reg = new RegExp("/\(\s*([0-9.-]+)\s*,\s([0-9.-]+)\s*\)/g");
var result = polygons.match(reg);
I've no javascript error when executing(with debug mode of google chrome). This code is hosted in a javascript function which is in a included JS file. This method is called in the OnLoad method.
I've searched a lot, but I can't find why this isn't working. Thank you very much!
Use a regex literal [MDN]:
var reg = /\(\s*([0-9.-]+)\s*,\s([0-9.-]+)\s*\)/g;
You are making two errors when you use RegExp [MDN]:
The "delimiters" / are should not be part of the expression
If you define an expression as string, you have to escape the backslash, because it is the escape character in strings
Furthermore, modifiers are passed as second argument to the function.
So if you wanted to use RegExp (which you don't have to in this case), the equivalent would be:
var reg = new RegExp("\\(\\s*([0-9.-]+)\\s*,\\s([0-9.-]+)\\s*\\)", "g");
(and I think now you see why regex literals are more convenient)
I always find it helpful to copy and past a RegExp expression in the console and see its output. Taking your original expression, we get:
/(s*([0-9.-]+)s*,s([0-9.-]+)s*)/g
which means that the expressions tries to match /, s and g literally and the parens () are still treated as special characters.
Update: .match() returns an array:
["(25.774252, -80.190262)", "(18.466465, -66.118292)", ... ]
which does not seem to be very useful.
You have to use .exec() [MDN] to extract the numbers:
["(25.774252, -80.190262)", "25.774252", "-80.190262"]
This has to be called repeatedly until the whole strings was processed.
Example:
var reg = /\(\s*([0-9.-]+)\s*,\s([0-9.-]+)\s*\)/g;
var result, points = [];
while((result = reg.exec(polygons)) !== null) {
points.push([+result[1], +result[2]]);
}
This creates an array of arrays and the unary plus (+) will convert the strings into numbers:
[
[25.774252, -80.190262],
[18.466465, -66.118292],
...
]
Of course if you want the values as strings and not as numbers, you can just omit the +.

How to split a long regular expression into multiple lines in JavaScript?

I have a very long regular expression, which I wish to split into multiple lines in my JavaScript code to keep each line length 80 characters according to JSLint rules. It's just better for reading, I think.
Here's pattern sample:
var pattern = /^(([^<>()[\]\\.,;:\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,}))$/;
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)
[Edit 2022/08] Created a small github repository to create regular expressions with spaces, comments and templating.
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:
when converting the expression literal to a string you need to escape all backslashes as backslashes are consumed when evaluating a string literal. (See Kayo's comment for more detail.)
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:
Disadvantage here is that you can't use plain whitespace in the regular expression string (always use \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
The regex above is missing some black slashes which isn't working properly. So, I edited the regex. Please consider this regex which works 99.99% for email validation.
let EMAIL_REGEXP =
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(''));
To avoid the Array join, you can also use the following syntax:
var pattern = 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,}))$');
You can simply use string operation.
var pattenString = "^(([^<>()[\]\\.,;:\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,}))$";
var patten = new RegExp(pattenString);
I tried improving korun's answer by encapsulating everything and implementing support for splitting capturing groups and character sets - making this method much more versatile.
To use this snippet you need to call the variadic function combineRegex whose arguments are the regular expression objects you need to combine. Its implementation can be found at the bottom.
Capturing groups can't be split directly that way though as it would leave some parts with just one parenthesis. Your browser would fail with an exception.
Instead I'm simply passing the contents of the capture group inside an array. The parentheses are automatically added when combineRegex encounters an array.
Furthermore quantifiers need to follow something. If for some reason the regular expression needs to be split in front of a quantifier you need to add a pair of parentheses. These will be removed automatically. The point is that an empty capture group is pretty useless and this way quantifiers have something to refer to. The same method can be used for things like non-capturing groups (/(?:abc)/ becomes [/()?:abc/]).
This is best explained using a simple example:
var regex = /abcd(efghi)+jkl/;
would become:
var regex = combineRegex(
/ab/,
/cd/,
[
/ef/,
/ghi/
],
/()+jkl/ // Note the added '()' in front of '+'
);
If you must split character sets you can use objects ({"":[regex1, regex2, ...]}) instead of arrays ([regex1, regex2, ...]). The key's content can be anything as long as the object only contains one key. Note that instead of () you have to use ] as dummy beginning if the first character could be interpreted as quantifier. I.e. /[+?]/ becomes {"":[/]+?/]}
Here is the snippet and a more complete example:
function combineRegexStr(dummy, ...regex)
{
return regex.map(r => {
if(Array.isArray(r))
return "("+combineRegexStr(dummy, ...r).replace(dummy, "")+")";
else if(Object.getPrototypeOf(r) === Object.getPrototypeOf({}))
return "["+combineRegexStr(/^\]/, ...(Object.entries(r)[0][1]))+"]";
else
return r.source.replace(dummy, "");
}).join("");
}
function combineRegex(...regex)
{
return new RegExp(combineRegexStr(/^\(\)/, ...regex));
}
//Usage:
//Original:
console.log(/abcd(?:ef[+A-Z0-9]gh)+$/.source);
//Same as:
console.log(
combineRegex(
/ab/,
/cd/,
[
/()?:ef/,
{"": [/]+A-Z/, /0-9/]},
/gh/
],
/()+$/
).source
);
Personally, I'd go for a less complicated regex:
/\S+#\S+\.\S+/
Sure, it is less accurate than your current pattern, but what are you trying to accomplish? Are you trying to catch accidental errors your users might enter, or are you worried that your users might try to enter invalid addresses? If it's the first, I'd go for an easier pattern. If it's the latter, some verification by responding to an e-mail sent to that address might be a better option.
However, if you want to use your current pattern, it would be (IMO) easier to read (and maintain!) by building it from smaller sub-patterns, like this:
var box1 = "([^<>()[\]\\\\.,;:\s#\"]+(\\.[^<>()[\\]\\\\.,;:\s#\"]+)*)";
var box2 = "(\".+\")";
var host1 = "(\\[[0-9]{1,3}\\.[0-9]{1,3}\\.[0-9]{1,3}\\.[0-9]{1,3}\\])";
var host2 = "(([a-zA-Z\-0-9]+\\.)+[a-zA-Z]{2,})";
var regex = new RegExp("^(" + box1 + "|" + box2 + ")#(" + host1 + "|" + host2 + ")$");
#Hashbrown's great answer got me on the right track. Here's my version, also inspired by this blog.
function regexp(...args) {
function cleanup(string) {
// remove whitespace, single and multi-line comments
return string.replace(/\s+|\/\/.*|\/\*[\s\S]*?\*\//g, '');
}
function escape(string) {
// escape regular expression
return string.replace(/[-.*+?^${}()|[\]\\]/g, '\\$&');
}
function create(flags, strings, ...values) {
let pattern = '';
for (let i = 0; i < values.length; ++i) {
pattern += cleanup(strings.raw[i]); // strings are cleaned up
pattern += escape(values[i]); // values are escaped
}
pattern += cleanup(strings.raw[values.length]);
return RegExp(pattern, flags);
}
if (Array.isArray(args[0])) {
// used as a template tag (no flags)
return create('', ...args);
}
// used as a function (with flags)
return create.bind(void 0, args[0]);
}
Use it like this:
regexp('i')`
//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*/
`
To create this RegExp object:
/(\d+)([a-z]{1,3})/i

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