This question already has answers here:
JavaScript regex pattern concatenate with variable
(3 answers)
Closed 4 years ago.
I am trying to create a regular expression with a character class that has a specific quantifier which is a variable for example:
var str = "1234.00";
var quantifier = 3;
str = str.replace(/(\d)(\d{quantifier}\.)/,"$1,$2");
//str should be "1,234.00"
This works as follows (without a variable):
var str = "1234.00";
str = str.replace(/(\d)(\d{3}\.)/,"$1,$2");
//str == "1,234.00"
However it does not have the same functionality with a quoted pattern instead of a slash-delimited pattern as follows:
var str = "1234.00";
str = str.replace("(\d)(\d{3}\.)","$1,$2");
//str == "1234.00" - not "1,234.00"
//quote symbol choice does not change this
str = str.replace('(\d)(\d{3}\.)',"$1,$2");
//str == "1234.00" - not "1,234.00"
edit: to be more clear I have added a summary question which was answered below:
How do I create a regular expression with an interpolated variable from a quoted string?
Although my preference would be to use interpolation, it seems that is not available (at least in this context), and is not necessary.
I have also tried to come up with a way to concatenate/join some regex literals to achieve the same result, but have been unable to do so for this use case.
As a side note - I am familiar with this type of regular expression in perl:
my $str = "1234.00";
my $quantifier = 3;
$str =~ s/(\d)(\d{$quantifier}\.)/$1,$2/;
# $str eq "1,234.00"
Which can be made useful as follows:
my $str = "1234567890.00";
for my $quantifier (qw(9 6 3)) {
$str =~ s/(\d)(\d{$quantifier}\.)/$1,$2/;
}
# $str eq "1,234,567,890.00"
With the suggestions/answers provided I have created a sample currency string prototype as follows:
String.prototype.toCurrency = function() {
var copy = parseFloat(this).toFixed(2);
for (var times = parseInt(copy.length/3); times > 0; times--) {
var digits = times * 3;
var re = new RegExp("(\\d)(\\d{" + digits + "}\\.)");
copy = copy.replace(re,"$1,$2");
}
return '$'+copy;
};
str = "1234567890";
str.toCurrency();
// returns "$1,234,567,890.00"
There are two problems with this statement:
str.replace("(\d)(\d{3}\.)","$1,$2");
The first is that you are passing a string and not a regular expression object, and the second is that within a string literal the backslash has a special meaning to escape certain things (e.g., "\n" is a newline) so to have an actual backslash in your string literal you need to double it as "\\". Using the RegExp() constructor to create a regex object from a string you get this:
str.replace(new RegExp("(\\d)(\\d{3}\\.)"),"$1,$2");
So from there you can do this:
var quantifier = 3
str = str.replace(new RegExp("(\\d)(\\d{" + quantifier + "}\\.)"),"$1,$2");
In JavaScript, you can't concatenate or interpolate into regex literals, but you can create a regex from a string by using the RegExp constructor:
str = str.replace(new RegExp('(\\d)(\\d{' + quantifier + '}\\.'), "$1,$2");
Note, by the way, that this:
str.replace(..., ...);
has no effect, because replace doesn't modify a string, but rather, it returns a copy of the string with the replacements made. So you need to write this:
str = str.replace(..., ...);
instead.
You can create a RegExp object:
var str = "1234.00";
var digits = 2;
var re = new RegExp("(\\d)(\\d{" + digits + "})");
var str2 = str.replace(re,"$1,$2-");
str2 would contain 1,23-4.00.
Working example:
http://jsfiddle.net/JuZtc/
Note that you need to escape \ in strings, thus \\.
Hope this helps.
Related
This question already has answers here:
How do you use a variable in a regular expression?
(27 answers)
Closed 2 years ago.
So for example:
function(input){
var testVar = input;
string = ...
string.replace(/ReGeX + testVar + ReGeX/, "replacement")
}
But this is of course not working :)
Is there any way to do this?
const regex = new RegExp(`ReGeX${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
So as the title says I'd like to add to every character of a string a backslash, whether the string has special characters or not. The string should not be considered 'safe'
eg:
let str = 'dj%^3&something';
str = str.replace(x, y);
// str = '\d\j\%\^\3\&\s\o\m\e\t\h\i\n\g'
You could capture every character in the string with (.) and use \\$1 as replacement, I'm not an expert but basically \\ will render to \ and $1 will render to whatever (.) captures.
HIH
EDIT
please refer to Wiktor Stribiżew's comment for an alternative which will require less coding. Changes as follows:
str = str.replace(/(.)/g, '\\$1'); for str = str.replace(/./g, '\\$&');
Also, for future reference I strongly advice you to visit regexr.com when it comes to regular expressions, it's helped ME a lot
let str = 'dj%^3&something';
str = str.replace(/(.)/g, '\\$1');
console.log(str);
If you just want to display a string safely, you should just do:
let str = 'dj%^3&something';
let node = document.createTextNode(str);
let dest = document.querySelector('.whatever');
dest.appendChild(node);
And then you are guaranteed that it will be treated as text, and won't be able to execute a script or anything.
For example: https://jsfiddle.net/s6udj03L/1/
You can split the string to an array, add a \ to each element to the array, then joint the array back to the string that you wanted.
var str = 'dj%^3&something';
var split = str.split(""); // split string into array
for (var i = 0; i < split.length; i++) {
split[i] = '\\' + split[i]; // add backslash to each element in array
}
var joint = split.join('') // joint array to string
console.log(joint);
If you don't care about creating a new string and don't really have to use a regex, why not just iterate over the existing one and place a \ before each char. Notice to you have to put \\ to escape the first \.
To consider it safe, you have to encode it somehow. You could replace typical 'non-safe' characters like in the encode() function below. Notice how & get's replaced by &
let str = 'dj%^3&something';
let out = "";
for(var i = 0; i < str.length; i++) {
out += ("\\" + str[i]);
}
console.log(out);
console.log(encode(out));
function encode(string) {
return String(string).replace(/&/g, '&').replace(/</g, '<').replace(/>/g, '>').replace(/"/g, '"');
}
I have a string which has some data with a few special characters, Need to remove the data between the desired special char in JavaScript.
The special char would be obtained in a variable.
var desiredChar = "~0~";
And Imagine this to be the Input string:
~0~1|0|20170807|45|111.00|~0~~1~1|0|20170807|50|666.00|~1~~2~1|0|20170807|55|111.00|~2~
So I'm supposed to remove the text in bold.
The desired output is supposed to be-
~1~1|0|20170807|50|666.00|~1~~2~1|0|20170807|55|111.00|~2~
I've tried using "Replace" and "Regex", but as the desired character is being passed in a variable and keeps changing I'm facing difficulties.
You can create your own regex based on whatever the bounding character(s) are that contain the text you want removed, and then replace any text that matches that regex with a blank string "".
The JS below should work for your use case (and it should work for multiple occurrences as well):
var originalText = "~0~1|0|20170807|45|111.00|~0~~1~1|0|20170807|50|666.00|~1~~2~1|0|20170807|55|111.00|~2~";
var desiredChar = "~0~";
var customRegex = new RegExp(desiredChar + ".*?" + desiredChar, "gi");
var processedText = originalText.replace(customRegex, "");
console.log(processedText);
You can build your regex from the constructor with a string input.
var desiredChar = "~0~";
// use the g flag in your regex if you want to remove all substrings between desiredChar
var myRegex = new Regex(desiredChar + ".*" + desiredChar, 'ig');
var testString = "~0~1|0|20170807|45|111.00|~0~~1~1|0|20170807|50|666.00|~1~~2~1|0|20170807|55|111.00|~2~";
testString = testString.replace(myRegex, "");
Given input string you can use .indexOf(), .lastIndexOf() and .slice().
Note, OR character | passed to RegExp constructor should be escaped to avoid RegExp created by passing string interpreting | character as OR | within resulting RegExp passed to .replace().
var desiredChar = "~0~";
var str = "~0~1|0|20170807|45|111.00|~0~~1~1|0|20170807|50|666.00|~1~~2~1|0|20170807|55|111.00|~2~";
var not = str.slice(str.indexOf(desiredChar), str.lastIndexOf(desiredChar) + desiredChar.length);
// escape OR `|`
var res = str.replace(new RegExp(not.replace(/[|]/g, "\\|")), "");
console.log(res)
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 at http://www.regular-expressions.info/javascript.html
Hi all ia m trying to replace all characters of "+" in a string by using the code below:
var findValue = "+";
var re = new RegExp(findValue, 'g');
searchValueParam = searchValueParam.replace(re, " ");
However i recieve this exception:
SyntaxError: Invalid regular expression: nothing to repeat
previously i applied just searchValueParam = searchValueParam.replace("+", " "); but that only replaces the first occurrence, not all.
Any suggestions?
For multiple replacements you need to use regex with the global (g) modifier, however + has a special meaning (the previous item 1 or more times), so it needs to be escaped.
searchValueParam = searchValueParam.replace(/\+/g,' ');
You need to escape the + sign:
searchValueParam.replace(/\+/g, " ");
If you want to keep the code you have, replace
var findValue = '+';
with
var findValue = '\\+';
Plus has a special meaning (quantifier) in a regular expression. This is why we need to escape it with a backslash: \+. However, when you place this in a string, the backslash itself has to be escaped as it has a special meaning in a string. This is how we end up with '\\+'.
In conclusion, this
var re = new RegExp('\\+', 'g')
is equivalent to this
var re = /\+/g;
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